Class 12 Commerce Applied Mathematics - Determinants

If $D_x\, =\, -18$ and $D =\, 3$ are the values of the determinants for certain simultaneous equations in $x$ and $y$ , Find $-x$ .

Given:

$D_x = -18$ and

$D = 3$
Thus we know

$x = \dfrac{D_x}{D}$

$\Rightarrow x =$

$\dfrac{-18}{3}$

$= -6$

Therefore,

$-x$ will be equal to

$6$ .

If the value of determinants $\begin{vmatrix} x& -5\\ 3 & 4\end{vmatrix}$ is $31$ , then find the value of $x$

We have,

$\begin{vmatrix} x& -5\\ 3 & 4\end{vmatrix} = 31$
On solving,

$4x - (-5) (3) = 31$

$\Rightarrow 4x + 15 = 31$

$\Rightarrow 4x = 31 - 15$

$\Rightarrow x = \dfrac {16}{4} = 4$

$\therefore x = 4$

Find the value of the determinant: $\begin{vmatrix} 4& -2\\ 3 & 1\end{vmatrix}$

Given,

$\begin{vmatrix} 4& -2\\ 3 & 1\end{vmatrix}$

$= (4) (1) - (-2)(3)$

$= 4 + 6 = 10$

If $D=7,{D}_{x}=35$ and ${D}_{y}=42$ are the values of the determinants for certain simultaneous equations in $x$ and $y$ , then find the values of $x$ and $y$ .

Given,

$D=7,{D}_{x}=35$ and

${D}_{y}=42$
Now

$x=\cfrac { { D }_{ x } }{ D } =\cfrac { 35 }{ 7 } =5$
and

$y=\cfrac { { D }_{ y } }{ D } =\cfrac { 42 }{ 7 } =6$

$\therefore$ The value of

$x$ and

$y$ is

$5$ and

$6$ respectively.

Find the value of the following determinant
$\begin{vmatrix}7 & 2\\ 5 &4 \end{vmatrix}$

$\begin{vmatrix} 7& 2\\ 5 &4 \end{vmatrix} = (7 \times 4) - (2 \times 5)$

$= 28 - 10 = 18$

solve the following simultaneous equations by using Cramer's rule:
$x+y=7,x-y=5$

The given equations are

$x+y=7$

$x-y=5$

$D=\begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix}=(1\times -1)-(1\times 1)=-1-1=-2$

${D}_{x}=\begin{vmatrix} 7 & 1 \\ 5 & -1 \end{vmatrix}=(7\times -1)-(1\times 5)=-7-5=-12$

${D}_{y}=\begin{vmatrix} 1 & 7 \\ 1 & 5 \end{vmatrix}=(1\times 5)-(7\times 1)=5-7=-2$

$\therefore$ By cramer's rule we get

$\therefore$

$x=\cfrac { { D }_{ x } }{ D } =\cfrac { -12 }{ 2 } =6$ and

$y=\cfrac { { D }_{ y } }{ D } =\cfrac { -2 }{ 2 } =1$

If ${ D }_{ x }=18,{ D }_{ y }=15$ and $D=3$ are the values of the determinants for certain simultaneous equations in $x$ and $y$ , then find the values of $x$ and $y$ .

Given

${ D }_{ x }=18,{ D }_{ y }=15$ and

$D=3$
Now,

$x=\cfrac { { D }_{ x } }{ D } =\cfrac { 18 }{ 3 } =6$

$y=\cfrac { { D }_{ y } }{ D } =\cfrac { 15 }{ 3 } =5$

$\therefore$ The value of

$x$ and

$y$ is

$6$ and

$5$ respectively.

Find the value of the following determinant $\begin{vmatrix} 4 & 3\\2 & 7\end{vmatrix}$ .

$\begin{vmatrix}4&3\\2&7\end{vmatrix}=4\times7-2\times3=28-6=22$

Find the value of following determinant
$\begin{vmatrix} 5& 2\\ 7 & 4 \end{vmatrix}$

$\begin{vmatrix}5&2\\7&4\end{vmatrix}=5\times4-7\times2=20-14=6$

If ${D}_{x}=30,{D}_{y}=42$ and $D=6$ are the values of the determinants for certain simultaneous equations in $x$ and $y$ , then find the values of $x$ and $y$ .

Given

${ D }_{ x }=30,{ D }_{ y }=42$ and

$D=6$
now

$x=\cfrac { { D }_{ x } }{ D } =\cfrac { 30 }{ 6 } =5\quad y=\cfrac { { D }_{ y } }{ D } =\cfrac { 42 }{ 6 } =7$

$\therefore$ The value of

$x$ and

$y$ is

$5$ and

$7$ respectively.

Solve the following simultaneous equations by using Cramer's rule:
$x - 2y = -18$
$2x - y = 9$

Equations are

$x - 2y = -18 ...(i)$

$2x - y = 9 ...(ii)$

$D = \begin{vmatrix}1 &-2 \\ 2 & -1\end{vmatrix}$

$= (18 + 18) = 36$

$D_{y} = \begin{vmatrix} 1& -18\\ 2 & 9\end{vmatrix}$

$= 9 + 36 = 45$
By Cramer's rule, we get

$\therefore x = \dfrac {D_{x}}{D}$

$= \dfrac {36}{3} = 12$
and

$y = \dfrac {D_{y}}{D}$

$= \dfrac {45}{3} = 15$

$\therefore x = 12$ and

$y = 15$ is the solution of the given simultaneous equations.

Find $D_{x}$ for the following simultaneous equations: $3x + 4y = 8, x - 2y = 5$

In cramer's rule of

${ a }_{ 1 }x+{ b }_{ 1 }y={ c }_{ 1 },{ a }_{ 2 }x+{ b }_{ 2 }y={ c }_{ 2 }$

$x=\dfrac { { D }_{ x } }{ D } =\dfrac { \begin{vmatrix} { c }_{ 1 } & { b }_{ 1 } \\ { c }_{ 2 } & { b }_{ 2 } \end{vmatrix} }{ \begin{vmatrix} { a }_{ 1 } & { b }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } \end{vmatrix} }$

$y=\dfrac { { D }_{ y } }{ D } =\dfrac { \begin{vmatrix} { a }_{ 1 } & { c }_{ 1 } \\ { a }_{ 2 } & { c }_{ 2 } \end{vmatrix} }{ \begin{vmatrix} { a }_{ 1 } & { b }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } \end{vmatrix} }$

${ D }_{ x }=\begin{vmatrix} { c }_{ 1 } & { b }_{ 1 } \\ c_{ 2 } & { b }_{ 2 } \end{vmatrix}$

So here

${ D }_{ x }=\begin{vmatrix} 8 & 4 \\ 5 & -2 \end{vmatrix}=-36$

If ${ D }_{ x }=24,{ D }_{ y }=32$ are the values of the determinants for certain simultaneous equations in $x$ and $y$ , then find the values of $x$ and $y$

Given

${ D }_{ x }=24,{ D }_{ y }=32$ and

$D=8$
Now

$x=\cfrac { { D }_{ x } }{ D } =\cfrac { 24 }{ 8 } =3$

$y=\cfrac { { D }_{ y } }{ D } =\cfrac { 32 }{ 8 } =4$

$\therefore$ The value of

$x$ and

$y$ is

$3$ and

$4$ respectively.

Solve the following equations by Cramer's rule:
$3x+y+z=2,2x-4y+3z=-1,4x+y-3z=-11\quad$

To solve the system by Cramer's rule, we do the following steps:

1. Write the system as a matrix equation

$AX=b$ where:

$A = \begin{bmatrix}3 & 1 & 1\\ 2 & -4 & 3\\ 4 & 1 & -3 \end{bmatrix}$

$X = \begin{bmatrix}x\\y\\z \end{bmatrix}$

$b = \begin{bmatrix}2\\-1\\-11 \end{bmatrix}$

2. Next, we find the following determinants:

$\Delta = \begin{vmatrix}3 & 1 & 1\\ 2 & -4 & 3\\ 4 & 1 & -3 \end{vmatrix} = |A| = 63$

$\Delta_1 = \begin{vmatrix}2 & 1 & 1\\ -1 & -4 & 3\\ -11 & 1 & -3 \end{vmatrix} = -63$

$\Delta_2 = \begin{vmatrix}3 & 2 & 1\\ 2 & -1 & 3\\ 4 & -11 & -3 \end{vmatrix} = 126$

$\Delta_3 = \begin{vmatrix}3 & 1 & 2\\ 2 & -4 & -1\\ 4 & 1 & -11 \end{vmatrix} = 189$

3. The solution is now given by:

$x = \dfrac{\Delta_1}{\Delta} = -1$

$y = \dfrac{\Delta_2}{\Delta} = 2$

$z = \dfrac{\Delta_3}{\Delta} = 3$

Find the ratios of $x : y : z$ from the equations
$7x = 4y + 8z, 3z = 12x + 11y$ .

By transposition we have

$7x - 4y - 8z = 0, 12x + 11y - 3z = 0$ .

Write down the coefficients, thus

$x=\begin{vmatrix} -4 & -8 \\ 11 & -3\end{vmatrix}$

$y=\begin{vmatrix} -8 & 7 \\ -3 & 12\end{vmatrix}$

$z=\begin{vmatrix} 7 & -4 \\ 12 & 11\end{vmatrix}$

whence we obtain the products

$(-4) \times (-3) - 11 \times (-8) , (-8) \times 12 - (-3) \times 7, 7 \times 11 \times -12 \times (-4) ,$

$100, -75, 125$ ;

$\therefore \dfrac {x}{100} = \dfrac {y}{-75} = \dfrac {z}{125} ,$

that is ,

$\dfrac {x}{4} = \dfrac {y}{-3} = \dfrac {z}{5} .$

If $A=\left[ \begin{matrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{matrix} \right]$ verify that $A(adj\ A)=\left| A \right| I$ .

$\left| A \right| =+1(9+2)=11$

$adj(A)$ =Transfer co factor matrix

${ c }_{ 11 }=\begin{vmatrix} 0 & -2 \\ 0 & 3 \end{vmatrix}=0, { c }_{ 23 }=\begin{vmatrix} 1 & -1 \\ 1 & 0 \end{vmatrix}=-1\\ { c }_{ 12 }=\begin{vmatrix} 3 & -2 \\ 1 & 3 \end{vmatrix}=11,\ { c }_{ 31 }=\begin{vmatrix} -1 & 2 \\ 0 & -2 \end{vmatrix}=2\\ { c }_{ 13 }=\begin{vmatrix} 3 & 0 \\ 1 & 0 \end{vmatrix}=0, { c }_{ 32 }=-\begin{vmatrix} 1 & 2 \\ 3 & -2 \end{vmatrix}=8\\ { c }_{ 21 }=\begin{vmatrix} -1 & 2 \\ 0 & 3 \end{vmatrix}=3, { c }_{ 33 }=\begin{vmatrix} 1 & -1 \\ 3 & 0 \end{vmatrix}=3\\ { c }_{ 22 }=\begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix}=1$

Co factor matrix

$\begin{vmatrix} 0 & -11 & 0 \\ 3 & 1 & -1 \\ 2 & 8 & 3 \end{vmatrix}$

$adj(A)$ =

$\begin{vmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{vmatrix}$

$A(adj(A))$ =

$\begin{vmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{vmatrix}\begin{vmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{vmatrix}$

$\begin{vmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{vmatrix}=\left| 11 \right| I=AI$

If
$x + y + z = 1$
$3x + 5y + 6z = 4$
$9x + 2y - 36z = 17$
Then find $y$

1926340_1003861_ans_3da0afd5480c4cc598a132c9681aa544.jpg

Solve the equation $\dfrac{x}{a}\, + \,\dfrac{y}{b}\, = \,2\,,\,ax\, - \,by\, = \,{a^2} - \,{b^2}$

Consider the given equation.

$ax-by=a^2-b^2$

Multiplying by

$a$ in both sides,

$a^2x-yab=a^3-ab^2$ ......(1)

Since,

$\dfrac{x}{a}+\dfrac{y}{b}=2$

$xb+ya=2ab$

Multiplying by

$b$ in both sides,

$xb^2+yab=2ab^2$ ......(2)

On adding equation(1) and (2),

$x(a^2+b^2)=a^3+ab^2$

$x(a^2+b^2)=a(a^2+b^2)$

$x=a$

Substituting this value in equation (1),

$a\times a^2-yab=a^3-ab^2$

$-yab=ab^2$

$y=b.$

So, for given equation

$x=a, y=b.$

Solve the system of linear equation $3x-2y=13$ and $4x+y=21$

$a_1 =3 ,a_2 = 4,b_1 = -2,b_2 = 1,c_1 =13,c_2 =21$

$x = \cfrac{c_1 b_2 - c_2 b_1}{a_1 b_2 - a_2 b_1} = \cfrac{13*1 +21*2}{3*1 +8}=\cfrac{55}{11} = 5$

$y = \cfrac{c_2 a_1 - c_1 a_2}{a_1 b_2 - a_2 b_1} = \cfrac{21*3 -13*4}{3*1 +8}=\cfrac{11}{11} = 1$

Solve the following simu1taneous equations.
(1) $5x-3y=8;$ $3x+y=2$

Consider the equation

$3x+y=2$

Multiply the quation by 3 so that we can eliminate y

$3(3x+y=2)=3\times 2$

$9x+3y=6$

Adding it with

$5x-3y=8$

$9x+3y+5x-3y=6+8$

$14x=14$

$x=1$

Putting the value of x in equation

$3x+y=2$

$3\times 1+y=2$

$3+y=2$

$y=2-3$

$y=-1$

So, we have

$x=1\, and\, y=-1$

Solve: (i) $4x+5y-62=0$
(ii) $3x+2y-36=0$

$4x+5y-62=0\\ \Rightarrow 4x+5y=62\longrightarrow (1)\\ 3x+2y-36=0\\ \Rightarrow 3x+2y=36\longrightarrow (2)\\ \because { a }_{ 1 }=4;\quad b_{ 1 }=5;\quad { c }_{ 1 }=62\\ { a }_{ 2 }=3;\quad { b }_{ 2 }=2;\quad { c }_{ 2 }=36\\ \therefore \quad x=\dfrac { \begin{vmatrix} { c }_{ 1 } & b_{ 1 } \\ { c }_{ 2 } & { b }_{ 2 } \end{vmatrix} }{ \begin{vmatrix} { a }_{ 1 } & b_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } \end{vmatrix} } =\dfrac { \begin{vmatrix} 62 & 5 \\ 36 & 2 \end{vmatrix} }{ \begin{vmatrix} 4 & 5 \\ 3 & 2 \end{vmatrix} } =8\\ y=\dfrac { \begin{vmatrix} { a }_{ 1 } & \quad { c }_{ 1 } \\ { a }_{ 2 } & { c }_{ 2 } \end{vmatrix} }{ \begin{vmatrix} { a }_{ 1 } & b_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } \end{vmatrix} } =\dfrac { \begin{vmatrix} 4 & 62 \\ 3 & 36 \end{vmatrix} }{ \begin{vmatrix} 4 & 5 \\ 3 & 2 \end{vmatrix} } =6$

Solve.
$3x-2y=1$ and $2x+y=3$ .

Given,

$3x-2y=1$ ..........(1)

$2x+y=3$

$y=3-2x$

Put in (1), we get

$3x-2(3-2x)=1$

$3x-6+4x=1$

$7x=7$

$x=1$

Put in (1)

$3(1)-2y=1$

$1-2y=1$

$2y=2$

$y=1$

Solve the given pair of linear equations:
$(a-b)x+(a+b)y={a}^{2}-2ab-{b}^{2}$
$(a+b)(x+y)={a}^{2}+{b}^{2}$

$(a-b)x+(a+b)y=a^2-2ab-b^2$

$(a+b)x+(a+b)y=a^2+b^2$

$-$

____________________________

$(a-b-a-b)x=a^2-2ab-b^2a^2-b^2$

$+2bx=+2ab+2b^2$

$\Rightarrow x=a+b$

$(a-b)(a+b)+(a+b)y=a^2-2ab-b^2$

$a^2-b^2+(a+b)y=a^2-2ab-b^2$

$y=\dfrac{-2ab}{a+b}$

1084256_1173150_ans_4459fc5188e84bf280aa1f3eaf60605e.png

Find co-factors of elements of the matrices $\begin {bmatrix} -1 & 2 \\ -3 & 4\end {bmatrix}$

$\begin {bmatrix} -1 & 2 \\ -3 & 4\end {bmatrix}$

co-factor of

$-1$ and

$4$ is

$4$ and

$-1$

minor of 2 is -3 and co factor is

$3$

minor of -3 is 2 and co factor is

$-2$

$\therefore$ co factor matrix

$\Rightarrow$

$\begin {bmatrix} 4 & 3 \\ -2 & -1\end {bmatrix}$

Solve the following equation by cramer's method:
$7x+3y=15 , 12y-5x=39$

$7x + 3y = 15; 12y - 5x = 39$

$D = \begin{vmatrix} 7 & {3} \\ {-5} & 12 \end{vmatrix} = 7 \times 12 - (-5) \times 3 = 84 + 15 = 99$

$D_x = \begin{vmatrix} {15} & {3} \\ 39 & 12 \end{vmatrix} = 15 \times 12 - 39 \times 3 = 180 - 117 = 63$

$D_y = \begin{vmatrix} 7 & {15} \\ -5 & 39 \end{vmatrix} = 7 \times 39 - (-5) \times 15 = 273 + 75 = 348$

$x = \dfrac{D_x}{D} = \dfrac{63}{99} = \dfrac{7}{11}$

$y = \dfrac{D_y}{D} = \dfrac{348}{99} = \dfrac{116}{33}$

$(x, y) = (\dfrac{7}{11}, \dfrac{116}{33})$

Find the value of: $\begin{vmatrix} -3 & -5 \\ -2 & -1 \end{vmatrix}$

$\begin{vmatrix} -3 & -5 \\ -2 & -1 \end{vmatrix}$

$= \left( -3 \right) \times \left( -1 \right) - \left( -5 \right) \times \left( -2 \right)$

$= 3 - 10 = -7$

Find ${D}_{x}$ and $D$ for the equation $2x+3y=4;7x-5y=2$

1212525_1363769_ans_08d330d893204690ae31a7af99586b22.JPG

Solve the following simultaneous equations. 2x - y = 5 ; 3x + 2y = 11

2x - y = 5 .....(I)
3x + 2y = 11 ....(II)
Multiplying (I) with 2 we get
4x - 2y = 10 ....(III)
Adding (II) with (III)
7x = 21

⇒⇒ y = 1

Solve the following simultaneous equations. x + y = 11 ; 2x - 3 y =7

x + y = 11 .....(I)
2x - 3y = 7 ..... (II)
Multiply (I) with 3
3x + 3y = 33 ..... (III)
Adding (II) and (III)
5x = 40
$$\Rightarrow4$ x = 8
Putting the value of x in (I)
8 + y = 11
y = 3

Solve by Cramer's rule
$x+y+z=6$
$x-y+z=2$
$3x+2y-4z=-5$ . The value of x is

$\Delta = \begin{vmatrix} 1&1&1\\1&-1&1\\3&2&-4 \end{vmatrix} = 1(4-2)-1(-4-3)+1(3+2)=14$

$\Delta_x = \begin{vmatrix} 6&1&1\\2&-1&1\\-5&2&-4 \end{vmatrix} = 6(4-2)-1(-8+5)+1(4-5)=14$

Hence by cramer rule,

$x=\cfrac { { \Delta }_{ x } }{ \Delta } =1$

If $D_y\, =\, -15$ and $D\, =\, -5$ are the values of the determinants for certain simultaneous equations in $x$ and $y$ , find $y$ .

Given:

$D_y = -15$ and

$D = -5$
Thus, we know

$y = \frac{D_y}{D}$

$y =$

$\dfrac{-15}{-5}$

$= 3$

Therefore, the value of

$y$ is

$3$ .

If $\displaystyle a_{1}f_{1}\left ( x \right )+a_{2}f_{2}\left ( x \right )+a_{3}f_{3}\left ( x \right )=0$ where $a_1,a_2,a_3$ are constants (not all zero) and $\displaystyle f_{1},f_{2},f_{3}$ are twice differentiable functions.Then $\displaystyle D=\begin{vmatrix}f_{1}\left ( x \right ) &f_{2}\left ( x \right ) & f_{3}\left ( x \right )\\ Df_{1}\left ( x \right ) & Df_{2}\left ( x \right ) & Df_{3}\left ( x \right )\\ D^{2}f_{1} \left ( x \right )& D^{2}f_{2}\left ( x \right ) & D^{2}f_{3}\left ( x \right )\end{vmatrix}$ is equal to $\displaystyle Df_{1}\left ( x \right )=\frac{d}{dx}f_{1}$

${ f }_{ 1 }(x)$ are differentiable i.e.

${ { D }^{ 3 }f }_{ 1 }(x)$ =0

$D\triangle =\begin{vmatrix} { Df }_{ 1 }(x) & { Df }_{ 2 }(x) & { Df }_{ 3 }(x) \\ { Df }_{ 1 }(x) & { Df }_{ 2 }(x) & { Df }_{ 3 }(x) \\ { { D }^{ 2 }f }_{ 1 }(x) & { { D }^{ 2 }f }_{ 2 }(x) & { { D }^{ 3 }f }_{ 3 }(x) \end{vmatrix}+\begin{vmatrix} { f }_{ 1 }(x) & { f }_{ 2 }(x) & { f }_{ 3 }(x) \\ { { D }^{ 2 }{ f }_{ 1 }(x) } & { { D }^{ 2 }f }_{ 2 }(x) & { { D }^{ 2 }f }_{ 3 }(x) \\ { { D }^{ 2 }{ f }_{ 1 }(x) } & { { D }^{ 2 }f }_{ 2 }(x) & { { D }^{ 2 }f }_{ 3 }(x) \end{vmatrix}\\ \\ \begin{vmatrix} { f }_{ 1 }(x) & { f }_{ 2 }(x) & { f }_{ 3 }(x) \\ { Df }_{ 1 }(x) & { Df }_{ 2 }(x) & { Df }_{ 3 }(x) \\ { { D }^{ 3 }f }_{ 1 }(x) & { { D }^{ 3 }f }_{ 2 }(x) & { { D }^{ 3 }f }_{ 3 }(x) \end{vmatrix}\\ \\ D\triangle =0+0+\begin{vmatrix} { f }_{ 1 }(x) & { f }_{ 2 }(x) & { f }_{ 3 }(x) \\ { Df }_{ 1 }(x) & { Df }_{ 2 }(x) & { Df }_{ 3 }(x) \\ 0 & 0 & 0 \end{vmatrix}=0\\ \\$

If the polynomial $\displaystyle f\left ( x \right )=2x^{3}+mx^{2}+nx-14$ has $\displaystyle \left ( x-1 \right )$ and $\displaystyle \left ( x+2 \right )$ as its factors find the value of $\displaystyle m\times n$

Since (x-1) and (x+2) are the factors of f(x) so f(1)=0 and f(-2)=0 Hence m+n-12=0 and 2 m-n-15=0

$\displaystyle \Rightarrow m=9, n=3$ Thus

$\displaystyle m\times n=27$

For what value of $k$ will the following pair of linear equations have no solution?
$2x + 3y = 1$ and $(3k - 1)x +(1 - 2k) y = 2k + 3$

Given equations are,

$2x + 3y = 1$ ....(1)
and

$(3k - 1)x + (1 - 2k)y = 2k + 3$ .....(2)
The pairs of linear equations have no solution, if

$\displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\neq \dfrac {c_1}{c_2}$

$\Rightarrow \dfrac{2}{3k-1}=\dfrac{3}{1-2k}$

$\Rightarrow 2(1-2k)=3(3k-1)$

$\Rightarrow 2-4k=9k-3$

$\displaystyle \Rightarrow 13k=5$

$\Rightarrow k=\dfrac {5}{13}$

Therefore, for

$k=\dfrac {5}{13}$ , the equations have no solution.

If show that ${ A }^{ 2 }-5A+7I=O$ . Hence find ${ A }^{ -1 }$

$A=\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$

To Prove :

$A^2-5A+7I=O$

Now,

$A^{ 2 }=\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$

$\Rightarrow A^{ 2 }=\begin{bmatrix} 9-1 & 3+2 \\ -3-2 & -1+4 \end{bmatrix}$

$\Rightarrow A^2 =\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$

Consider ,

$A^2-5A+7I$

$=\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}-5\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}+7\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$=\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}+\begin{bmatrix} -15 & -5 \\ 5 & -10 \end{bmatrix}+\begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$

$=\begin{bmatrix} 8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7 \end{bmatrix}$

$=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

$\Rightarrow A^2-5A+7I=O$ ....(1)

$A^{-1}$ :

Multiplying eqn (1) by

$A^{-1}$

$A^2 A^{-1} -5AA^{-1}+7IA^{-1} =O.A^{-1}$

$\Rightarrow A-5I+7A^{-1}=O$ (

$\because AA^{-1}=I , O.A=O$ )

$\Rightarrow 7A^{-1}=5I-A$

$=5\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}-\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$

$=\begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}-\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$

$=\begin{bmatrix} 5-3 & 0-1 \\ 0-(-1) & 5-2 \end{bmatrix}$

$\Rightarrow 7A^{-1}=\begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$

$\Rightarrow A^{-1}=\cfrac{1}{7}\begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$

If verify that ${ A }^{ 3 }-{ 6A }^{ 2 }+9A-4I=0$ . Hence find ${ A }^{ -1 }$

$A=\begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix}$

To Prove : $A^3-6A^2+9A-4I=O$

Consider,

${ A }^{ 2 }=\begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix}\begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix}$

$=\begin{bmatrix} 4+1+1 & -2-2-1 & 2+1+2 \\ -2-2-1 & 1+4+1 & -1-2-2 \\ 2+1+2 & -1-2-2 & 1+1+4 \end{bmatrix}$

$\Rightarrow A^{ 2 }=\begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix}$

Now,

${ A }^{ 3 }=A^{ 2 }A$

$=\begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix}\begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix}$

$=\begin{bmatrix} 12+5+5 & -6-10-5 & 6+5+10 \\ -10-6-5 &5+12+5 & -5-6-10 \\ 10+5+6 & -5-10-6 & 5+5+12 \end{bmatrix}$

$A^3=\begin{bmatrix} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22\end{bmatrix}$

Consider,

$A^{ 3 }-6A^{ 2 }+9A-4I$

$=\begin{bmatrix} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{bmatrix}-6\begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix}+9\begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix}-4\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$=\begin{bmatrix} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{bmatrix}-\begin{bmatrix} 36 & -30 & 30 \\ -30 & 36 & -30 \\ 30 & -30 & 36 \end{bmatrix}+\begin{bmatrix} 18 & -9 & 9 \\ -9 & 18 & -9 \\ 9 & -9 & 18 \end{bmatrix}-\begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}$

$=\begin{bmatrix} 22-36+18-4 & -21+30-9 & 21-30+9 \\ -21+30-9 & 22-36+18-4 & -21+30-9 \\ 21-30+9 & -21+30-9 & 22-36+18-4 \end{bmatrix}$

$=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

$\Rightarrow A^{ 3 }-6A^{ 2 }+9A-4I=O$ .....(1)

$A^{-1}$ :

Multiplying eqn (1) by

$A^{-1}$ , we get

$A^{ 3 }{ A }^{ -1 }-6A^{ 2 }{ A }^{ -1 }+9A{ A }^{ -1 }-4I{ A }^{ -1 }=O{ A }^{ -1 }$

$\Rightarrow A^{ 2 }-6A+9I-4{ A }^{ -1 }=O$

$\Rightarrow 4{ A }^{ -1 }=A^{ 2 }-6A+9I$

Substituting the values in above eqn

$\Rightarrow 4{ A }^{ -1 }=\begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix}-6\begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix}+9\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$\Rightarrow 4{ A }^{ -1 }=\begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix}-\begin{bmatrix} 12 & -6 & 6 \\ -6 & 12 & -6 \\ 6 & -6 & 12 \end{bmatrix}+\begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix}$

$\Rightarrow 4{ A }^{ -1 }=\begin{bmatrix} 6-12+9 & -5+6 & 5-6 \\ -5+6 & 6-12+9 & -5+6 \\ 5-6 & -5+6 & 6-12+9 \end{bmatrix}$

$\Rightarrow { A }^{ -1 }=\cfrac { 1 }{ 4 } \begin{bmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{bmatrix}$

For the matrix show that ${ A }^{ 3 }-{ 6A }^{ 2 }+5A+11I=0$ . Hence find ${ A }^{ -1 }$

$A=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}$

To Prove : $A^3-6A^2+5A+11I=O$

Consider,

${ A }^{ 2 }=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}$

$=\begin{bmatrix} 1+1+2 & 1+2-1 & 1-3+3 \\ 1+2-6 & 1+4+3 & 1-6-9 \\ 2-1+6 & 2-2-3 & 2+3+9 \end{bmatrix}$

$\Rightarrow A^{ 2 }=\begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix}$

Now,

${ A }^{ 3 }=A^{ 2 }A$

$=\begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix}\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}$

$=\begin{bmatrix} 4+2+2 & 4+4-1 & 4-6+3 \\ -3+8-28 & -3+16+14 & -3-24-42 \\ 7-3+28 & 7-6-14 & 7+9+42 \end{bmatrix}$

$A^3=\begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{bmatrix}$

Consider,

$A^{ 3 }-6A^{ 2 }+5A+11I$

$=\begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{bmatrix}-6\begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix}+5\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}+11\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$=\begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{bmatrix}-\begin{bmatrix} 24 & 12 & 6 \\ -18 & 48 & -84 \\ 42 & -18 & 84 \end{bmatrix}+\begin{bmatrix} 5 & 5 & 5 \\ 5 & 10 & -15 \\ 10 & -5 & 15 \end{bmatrix}+\begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}$

$=\begin{bmatrix} 8-24+5+11 & 7-12+5 & 1-6+5 \\ -23+18+5 & 27-48+10+11 & -69+84-15 \\ 32-42+10 & -13+18-5 & 58-84+15+11 \end{bmatrix}$

$=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

$\Rightarrow A^{ 3 }-6A^{ 2 }+5A+11I=O$ .....(1)

$A^{-1}$ :

Multiplying eqn (1) by

$A^{-1}$ , we get

$A^{ 3 }{ A }^{ -1 }-6A^{ 2 }{ A }^{ -1 }+5A{ A }^{ -1 }+11I{ A }^{ -1 }=O{ A }^{ -1 }$

$\Rightarrow A^{ 2 }-6A+5I+11{ A }^{ -1 }=O$

$\Rightarrow 11{ A }^{ -1 }=-A^{ 2 }+6A-5I$

Substituting the values in above eqn

$11A^{-1}=-\begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix}+6\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}-5\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$=\begin{bmatrix} -4 & -2 & -1 \\ 3 & -8 & 14 \\ -7 & 3 & -14 \end{bmatrix}+\begin{bmatrix} 6 & 6 & 6 \\ 6 & 12 & -18 \\ 12 & -6 & 18 \end{bmatrix}-\begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}$

$11{ A }^{ -1 }=\begin{bmatrix} -4+6-5 & -2+6 & -1+6 \\ 3+6 & -8+12-5 & 14-18 \\ -7+12 & 3-6 & -14+18-5 \end{bmatrix}$

$\Rightarrow { A }^{ -1 }=\displaystyle \frac { 1 }{ 11 } \begin{bmatrix} -3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \end{bmatrix}$

Given, $-3x + 4y = 20$ and $6x +3y =15$
If $(x, y)$ is the solution to the system of equations above, what is the value of $x$ ?

Let

$-3x + 4y = 20$ ....(1)

$6x + 3y = 15$ ....(2)
Multiplying equation (1) with

$2$ , we get

$-6x + 8y = 40$ ....(3)
Adding equations

$1$ and

$3$ , we get

$11y = 55$

$\Rightarrow y = 5$
Substituting in equation (1), we get

$-3x + 4(5) = 20$

$\Rightarrow -3x + 20 = 20$

$\Rightarrow -3x = 0$

$\Rightarrow x = 0$

Find ' $m+n$ ' for the following simultaneous equations:
$3m+4n=7$
and $4m+3n=14$

$3m+4n=7..........(1)$

$4m+3n=14........(2)$
Adding equations

$(1)$ and

$(2)$

$3m+4n=7$

$4m+3n=14$
-------------------

$7m+7n=21$

$\therefore$

$7(m+n)=21$

$\therefore$

$m+n=\cfrac{21}{7}$

$\therefore$

$m+n=3$

Solve the following simultaneous equations by using Cramer's Rule
$3x + y = 1$ ,
$2x - 11 y = 3$

The given simultaneous equations are

$3x + y = 1$ ....... (1)

$2x- 11 y = 3$ ........ (2)

$D = \begin{vmatrix}3 & 1\\ 2 & -11\end{vmatrix}$

$= 3 \times (-11) - 1 \times 2$

$= -33 -2 = -35$

$D_x = \begin{vmatrix}1 & 1\\ 3 & -11\end{vmatrix}$

$= 1 \times (-11) - 1 \times 3$

$= -11 - 3 = - 14$

$D_y = \begin{vmatrix} 3& 1\\ 2 & 3\end{vmatrix}$

$= 3 \times 3 - 1 \times 2 = 9 - 2 = 7$

$\therefore$ By Cramer's rule, we get

$x = \dfrac{D_x}{D}=\dfrac{-14}{35} = -\dfrac25$ and

$y = \dfrac{D_y}{D}=\dfrac{7}{35} = \dfrac15$

If $A=\begin{bmatrix} \cos^2\theta & \cos\theta \sin\theta\\ \cos\theta\sin\theta & \sin^2\theta\end{bmatrix}$ and $B=\begin{bmatrix}\cos^2\Phi & \cos\Phi \sin\Phi\\ \cos\Phi \sin\Phi & \sin^2\Phi\end{bmatrix}$ , show that AB is a zero matrix if $\theta$ and $\Phi$ differ by an odd multiple of $\displaystyle\frac{\pi}{2}$ .

$AB=\begin{bmatrix} \cos^2\theta&\cos\theta\sin\theta\\\cos\theta\sin\theta&\sin^2\theta\end{bmatrix}\begin{bmatrix} \cos^2\phi&\cos\phi\sin\phi\\\cos\phi\sin\phi&\sin^2\phi\end{bmatrix}$

$\therefore AB=\begin{bmatrix} \cos^2\theta\cos^2\phi+\cos\theta\sin\theta\cos\phi\sin\phi & \cos^2\theta\cos\phi\sin\phi+\cos\theta\sin\theta\sin^2\phi \\ \cos\theta\sin\theta\cos^2\phi+\sin^2\theta\cos\phi\sin\phi & \cos\theta\sin\theta\cos\phi\sin\phi+\sin^2\theta\sin^2\phi \end{bmatrix}$

For

$AB$ to be a zero matrix, each element of matrix should be 0.

$\therefore$ from 1st term and 4th term, we have

$\cos^2\theta\cos^2\phi+\cos\theta\sin\theta\cos\phi\sin\phi+\cos\theta\sin\theta\cos\phi\sin\phi+\sin^2\theta\sin^2\phi=0$

$\therefore \cos^2\theta\cos^2\phi+2\cos\theta\sin\theta\cos\phi\sin\phi+\sin^2\theta\sin^2\phi=0$

$\therefore (\cos\theta\cos\phi+\sin\theta\sin\phi)^2=0$

$\therefore \cos\theta\cos\phi+\sin\theta\sin\phi=0$

$\therefore \cos(\theta-\phi)=0$

$\therefore \theta-\phi=n\pi+\dfrac{\pi}{2}$

Hence, proved.

Solve the following simultaneous equations using Cramer's rule.
$3x-2y=3$ ;
$2x+y=16$ .

Writing the system of equations in matrix form

$AX = B$

where

$A$ is the

$2 \times 2$ coefficient matrix,

$X = \begin{pmatrix} x\\y\end{pmatrix}$ and

$B$ is the

$2\times 1$ matrix of constants.

Here,

$A = \begin{pmatrix} 3 & -2\\ 2 & 1 \end{pmatrix}$ and

$B = \begin{pmatrix} 3\\16\end{pmatrix}$

We can write,

$X = A^{-1}B$ , where

$A^{-1} = \dfrac{1}{|A|} adj(A)$

$A^{-1} = \dfrac{1}{7} \times \begin{pmatrix} 1 & 2\\ -2 & 3\end{pmatrix}$

Substituting the values we get,

$\begin{pmatrix} x\\y \end{pmatrix} =$

$\dfrac{1}{7} \times \begin{pmatrix} 1 & 2\\ -2 & 3\end{pmatrix}$

$\times \begin{pmatrix} 3\\16\end{pmatrix}$

$\begin{pmatrix} x\\y \end{pmatrix} =$

$\dfrac{1}{7} \times \begin{pmatrix} 3\times1 + 2\times16\\ (-2)\times 3 + 3 \times 16\end{pmatrix}$

$\begin{pmatrix} x\\y \end{pmatrix} =$

$\begin{pmatrix} 5\\ 6\end{pmatrix}$ .

Hence,

$x = 5$ and

$y = 6$ .

Solve $x + y + z = 11\\ 2x - 6y - z = 0\\ 3x + 4y + 2z = 0$ .

1311867_1003858_ans_a6e42ddd8e1d4f72a0c889c0c155e290.jpg

Solve the equations
$x + 2y + 3z = 14$
$2x - y + 5z = 15$
$3x - 2y - 4z = -13$

$x = 1, y = 2, z = 3$

Find x and y using Cramer's Rule if : $\dfrac{1}{2^x}+\dfrac{2}{3^y}=10,\dfrac{3}{2^x}-\dfrac{5}{3^y}=-3$ ?

$\dfrac { 1 }{ { 2 }^{ x } } +\dfrac { 2 }{ { 3 }^{ y } } =10\longrightarrow (1)\\ \dfrac { 3 }{ { 2 }^{ x } } -\dfrac { 5 }{ { 3 }^{ y } } =-3\longrightarrow (2)\\$

Let

${ 2 }^{ x }=p\quad$ and

${ 3 }^{ y }=q$

Equation becomes;

$\dfrac { 1 }{ p } +\dfrac { 2 }{ q } =10\longrightarrow (3)\\ \dfrac { 3 }{ p } -\dfrac { 5 }{ q } =-3\longrightarrow (4)\\$

again let,

$\dfrac { 1 }{ p } =u$ and

$\dfrac { 1 }{ q } =v$

Equation (3) and (4) becomes

$u+2v=10\longrightarrow \left( 5 \right) \\ 3u-5v=-3\longrightarrow \left( 6 \right)$

By Cramer's rule;

$u=\dfrac { \begin{vmatrix} { c }_{ 1 } & { b }_{ 1 } \\ { c }_{ 2 } & { b }_{ 2 } \end{vmatrix} }{ \begin{vmatrix} { a }_{ 1 } & { b }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } \end{vmatrix} } =\dfrac { \begin{vmatrix} 10 & 2 \\ -3 & -5 \end{vmatrix} }{ \begin{vmatrix} 1 & 2 \\ 3 & -5 \end{vmatrix} } =4$

and,

$v=\dfrac { \begin{vmatrix} { a }_{ 1 } & { c }_{ 1 } \\ { a }_{ 2 } & { c }_{ 2 } \end{vmatrix} }{ \begin{vmatrix} { a }_{ 1 } & { b }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } \end{vmatrix} } =\dfrac { \begin{vmatrix} 1 & 10 \\ 3 & -3 \end{vmatrix} }{ \begin{vmatrix} 1 & 2 \\ 3 & -5 \end{vmatrix} } =3$

$\because \quad \dfrac { 1 }{ p } =u;\quad \dfrac { 1 }{ q } =v\\ \Rightarrow p=\dfrac { 1 }{ 4 } ;\quad q=\dfrac { 1 }{ 3 } \\$

again,

${ 2 }^{ x }=p;\quad { 3 }^{ y }=q\\ \Rightarrow { 2 }^{ x }=\dfrac { 1 }{ 4 } ;\quad { 3 }^{ y }=\dfrac { 1 }{ 3 } \\ \Rightarrow x\log 2=\log\dfrac { 1 }{ 4 } ;\quad y\log 3=\log\dfrac { 1 }{ 3 } \\ \Rightarrow x=\log\dfrac { 1 }{ 8 } ;\quad y=\log\dfrac { 1 }{ 9 } \quad$

Find x and y using Cramer's Rule,if $\dfrac{1}{x}-\dfrac{2}{y}=6,\dfrac{3}{x}+\dfrac{1}{y}=8$ ?

Given the two equations,

$\dfrac { 1 }{ x } -\dfrac { 2 }{ y } =6\quad \quad \longrightarrow (1)$

$\dfrac { 3 }{ x } +\dfrac { 1 }{ y } =8\quad \quad \longrightarrow (2)$

Let,

$p=\dfrac { 1 }{ x } ,q=\dfrac { 1 }{ y }$ .

Then, equation

$(1)$ becomes

$\Rightarrow p-2q=6\quad \quad \longrightarrow (3)$

and equation

$(2)$ becomes

$\Rightarrow 3p+q=8\quad \quad \longrightarrow (4)$ .

Now, solving equation

$(3)$ and

$(4)$ by Cramer's rule,

Let,

$A=\begin{pmatrix} 1 & -2 \\ 3 & 1 \end{pmatrix}$

Finding the determinant of the matrix

$A$ ,

$\left| A \right| =\begin{vmatrix} 1 & -2 \\ 3 & 1 \end{vmatrix}$

$=1+6=7$

Now,for matrix

${ A }_{ p }$ , we have to replace the first column of the matrix

$A$ with the right hand side values of equation

$(3)$ and

$(4)$ ,

${ A }_{ p }=\begin{pmatrix} 6 & -2 \\ 8 & 1 \end{pmatrix}$

$\therefore \left| { A }_{ p } \right| =\begin{vmatrix} 6 & -2 \\ 8 & 1 \end{vmatrix}$

$=6+16=22$

$\therefore \quad p=\dfrac { \left| { A }_{ p } \right| }{ \left| A \right| } =\dfrac { 22 }{ 7 }$

Again, for matrix

${ A }_{ q }$ , we have to replace the second column of the matrix

$A$ with the right hand side values of equation

$(3)$ and

$(4)$ ,

$\quad { A }_{ q }=\begin{pmatrix} 1 & 6 \\ 3 & 8 \end{pmatrix}$

$\therefore \left| { A }_{ q } \right| =\begin{vmatrix} 1 & 6 \\ 3 & 8 \end{vmatrix}=8-18=-10$

$\therefore q=\dfrac { \left| { A }_{ q } \right| }{ \left| A \right| } =\dfrac { -10 }{ 7 }$

Thus,

$\dfrac { 1 }{ x } =\dfrac { 22 }{ 7 } ,\dfrac { 1 }{ y } =\dfrac { -10 }{ 7 }$

$\Rightarrow x=\dfrac { 7 }{ 22 } ,y=-\dfrac { 7 }{ 10 }$

Hence, the solutions are

$x=\dfrac { 7 }{ 22 } \quad and\quad y=-\dfrac { 7 }{ 10 }$ .

If $A = \left[ \begin{array} { c c c } { 5 } & { 8 } & { 1 } \\ { 0 } & { 2 } & { 1 } \\ { 4 } & { 3 } & { - 1 } \end{array} \right] , B = \left[ \begin{array} { c } { 2 } \\ { - 1 } \\ { 3 } \end{array} \right]$ and $A X = B$ then find $X$

$A = \begin{bmatrix} 5 & 8 & 1 \\ 0 & 2 & 1 \\ 4 & 3 & -1 \end{bmatrix} \, B = \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}$

$AX = B$

$\begin{bmatrix} 5 & 8 & 1 \\ 0 & 2 & 1 \\ 4 & 3 & -1 \end{bmatrix} \, \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}$

$5x + 8y + z = 2$ .................(i)

$2y + z = -1 \Rightarrow z = -(1 + 2y)$

$4x + 3y - z = 3$ ...................(ii)

$5x + 8y - 1 - 2y = 2$ ...............replacing value of

$z$ in eq. (i)

$4x + 3y + 1 + 2y = 3$ ...............replacing value of

$z$ in eq. (ii)

$5x + 6y = 3$ .... (iii)

$4x + 5y = 2$ ....(iv)

Now, (iii)

$\times$ 5-(iv)

$\times$ 6

we get,

$x=3$

Replacing value of

$x$ in eq. i

we get,

$y = -2$

$z = -(1 + 2(-2))$

$z = 3$

$x = 3; \, y = -2$ and

$z = 3$

$\begin{bmatrix} 3 \\ -2 \\ 3 \end{bmatrix} = x$ .

Find $A^2$ where A is given $A= \begin {bmatrix}1 & 2 & 3 \\ 1 & 2 & 3\\ -1 & 2 & -3 \end{bmatrix}$

$A= \begin {bmatrix}1 & 2 & 3 \\ 1 & 2 & 3\\ -1 & 2 & -3 \end{bmatrix}$

$A^{2} = A\times A$

$= \begin {bmatrix}1 & 2 & 3 \\ 1 & 2 & 3\\ -1 & 2 & -3 \end{bmatrix} \times \begin {bmatrix}1 & 2 & 3 \\ 1 & 2 & 3\\ -1 & 2 & -3 \end{bmatrix} = \begin {bmatrix}0 & 12 & 0 \\ 0 & 12 & 0 \\ 4 & -4 & 0 \end{bmatrix}$

Solve $\left| {\begin{array}{*{20}{c}}1 & 1 & 1\\a & b & c\\{{a^3}} & {{b^3}} & {{c^3}}\end{array}} \right| = \left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)\left( {a + b + c} \right)$

$\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{vmatrix}$

$c_1 \rightarrow c_1 - c_2, \, c_2 \rightarrow c_2 - c_3$

$= \begin{vmatrix} 0 & 0 & 1 \\ a - b & b - c & c \\ a^3 - b^3 & b^3 - c^3 & c^3 \end{vmatrix}$

$= \begin{vmatrix} 0 & 0 & 1 \\ a - b & b - c & c \\ (a-b)(a^2 + b^2 + ab) & (b - c)(b^2 + c^2 + bc) & c^3 \end{vmatrix}$

$\{(A^3 - B^3) = (A - B) (A^2 + B^2 + AB) \}$

taking

$(a - b)$ from

$c_1$ and

$(b - c)$ from

$c_2$

$\Rightarrow (a - b)(b - c) \begin{vmatrix} 0 & 0 & 1 \\ 1 & 1 & c \\ a^2 + b^2 +ab & b^2+ c^2 + bc & c^3 \end{vmatrix}$

$\Rightarrow (a - b) (b - c) [0 - 0 + (b^2 + c^2 + bc - a^2 - b^2 - an)]$

$\Rightarrow (a - b) (b - c) [c^2 - a^2 + b(c - a)]$

$\Rightarrow (a - b)(b - c) [(c - a) (c + a) + b (c - a)]$

$\Rightarrow (a - b)(b - c)(c - a) ( a+ b + c)$

For what value of k , the following pair of linear equations has infinite number of solutions :
$5x + 2y = k ; 10x + 4y = 3$

The given equations are :

$5x+2y-k=0$

$10x+4y-3=0$

For infinitely many solutions,

$\dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} = \dfrac{c_{1}}{c_{2}}$

Therefore,

$\dfrac{5}{10}=\dfrac{2}{4}=\dfrac{-k}{-3}$

$\dfrac{k}{3}=\dfrac{1}{2}$

$k=\dfrac{3}{2}$

If 3m +5n =9 and 5m +3n =7 then find the value of 8m +8n.

Given 3m+5n=9 -----(1)

5m+3n=7 -----(3)

multiply eq (1) by 5 and eq (2)by (3) we get

15m+25n=45

15m+9n=21

___________

16n=34.

$[n= \frac{17}{8}]$

now form eq (1)

$3m= 9-\frac{5\times 17}{8}=\frac{72-85}{8}=\frac{-13}{8}.$

$m=\frac{-13}{24}$

Now

$8m+8n=8(m+n)=8(\frac{-13}{24}+\frac{17}{8})=\frac{51-13}{24}$

$[8m+8n=\frac{38}{3}]$

1192823_1376959_ans_7a4533746cd44aa9ae3d4fa86e174760.JPG

Solve the following
$\frac{x}{3}+\frac{y}{4}=4, \frac{x}{2}-\frac{y}{4}=1$

$\frac{x}{3}+\frac{y}{4}=4$ ----(1)

$\frac{x}{2}-\frac{y}{4}=1$ ----(2)

4x+3y=48 form eq (1)

4x-2y=8 form eq (2)

(-) (+) (-)

_______

5y=40

[y = 8] and 4x = 8 + 2y = 24

[x=6]

Hence [x = 6] and [y = 8]

1192824_1376966_ans_24a9453b84ba4d80bd6227e1c241a50f.JPG

Solve the following simultaneous equation using cramer's rule
$4x+3y-4=0: 6x=1$

1208402_1543116_ans_d8a80c102d8a4ce78dbad981b006a631.jpg

Solve the following simultaneous equation using cramer's rule
$6x-4y=-12: 8x-3y=-2$

1208414_1543135_ans_ac4a4ad01d6e42d0b596a9c9dfbb064b.jpg

Solve the following simultaneous equation using cramer's rule
$4m+6n=54 : 3m +2n =28$

1208422_1543195_ans_4a1a8b4bc8d44789a3e95f0a5cdc85bc.jpg

Solve the simultaneous equations.
$\dfrac { x } { 3 } + \dfrac { y } { 4 } = 4 ; \quad \dfrac { x } { 2 } - \dfrac { y } { 4 } = 1$

$\dfrac{x}{3}+\dfrac{y}{4}=4$ ......(I)

$\dfrac{x}{2}-\dfrac{y}{4}=1$ ......(II)
Multiplying (I) with LCM of 3 and 4 which is 12 we get
4x+3y=48 .....(Ill)
Multiplying (I) with LCM of 3 and 4 which is 12 we get
2x - y = 4 .....(IV)
Multiplying (IV) with 2
4x - 2y = 8 .....(V)
Subtracting (V) from (III)
5y = 40

$\Rightarrow$ y = 8
Putting this value of y in (IV) we get
2x - 8 = 4

$\Rightarrow$ 2x = 12

$\Rightarrow$ x = 6

Solve the following simultaneous equations using Cramer's rule :
$3x+y=7; \ 2x-11y=3.$

2042802_1534569_ans_49dbafce071e44608c3a5a60e47b286b.jpg

Solve the following simultaneous equation using cramer's rule.
$3x-4y=10:4x+3y=5$

$x=2$ but

$y=\dfrac{15-40}{25}=\dfrac{-25}{25}=-1$ .
1208397_1543108_ans_e0cf2280af9345e6a6e646cd299ff248.jpg

Solve the following simultaneous equation using cramer's rule
$x+2y=-1: 2x-3y=12$

1208405_1543126_ans_f3a03f92c5b84395a6a53e5dcde2321e.jpg

Solve the following simultaneous equation using cramer's rule
2x + 3y = 2: x- $\frac { y }{ 2 } =\frac{1}{2}$

2011591_1543203_ans_7e155efce1cf49d3b2d4db01e4f1b9f0.jpg

If $A = \begin{bmatrix} 3& 2\\ 2 & 1\end{bmatrix}$ , verify that $A^{2} - 4A - I = O$ , and hence find $A^{-1}$ .

1537914_1705458_ans_9313de82f8fc476c99a815153cdedb2e.jpg

Show that the matrix $A =\begin{bmatrix} -8& 5\\ 2 & 4\end{bmatrix}$ satisifies the equation $A^{2} + 4A - 42I = 0$ and hence find $A^{-1}$ .

1538687_1705460_ans_02587ff7638645e8b0153083323eaf7b.jpg

If $A = \begin{bmatrix} -1& -1\\ 2 & -2\end{bmatrix}$ , show that $A^{2} + 3A + 4I_{2} = O$ and hence find $A^{-1}$ .

1537966_1705463_ans_c6a93042f4cf4552a52e74581d981b56.jpg

If $A = \begin{bmatrix} 3& -2\\ 4 & -2\end{bmatrix}$ , find the value of $\lambda$ so that $A^{2} = \lambda A - 2I$ . Hence, find $A^{-1}$ .

1538104_1705469_ans_48e8bcdc12044091b1bf3d8a20792e82.jpg

Show that the $A = \begin{bmatrix}1 & 0 & -2\\ -2 & -1 & 2\\ 3 & 4 & 1\end{bmatrix}$ satisfied the equation $A^{3} - A^{2} - 3A - I = O$ , and hence find $A^{-1}$ .

1538285_1705471_ans_1df07f4387774df7b3b10444949ea752.jpg

Solve the following equation:
$\begin{vmatrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{vmatrix}=0$

1552570_1705735_ans_53cf178037b14441a7090ea699375286.jpg

Show that $\begin{vmatrix} a^2+\lambda & ab & ac & ad \\ ab & b^2+\lambda & bc & bd \\ ac & bc & c^2+\lambda & cd \\ ad & bd & cd & d^2+\lambda \end{vmatrix}$
is divisible by $\lambda^{3}$ and find the other factor.

1578860_1747407_ans_b129007cbbaa4a44980ac2565eb95409.JPG

Absolute value of the sum of roots of the equation
$\begin{vmatrix} x+2 & 2x+3 & 3x+4 \\ 2x+3 & 3x+4 & 4x+5 \\ 3x+5 & 5x+8 & 10x+17 \end{vmatrix}=0$ is

$\Delta =\begin{vmatrix} x+2 & 2x+3 & 3x+4 \\ 2x+3 & 3x+4 & 4x+5 \\ 3x+5 & 5x+8 & 10x+17 \end{vmatrix}=0$

applying

$({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 };{ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 })$

$\Delta =\begin{vmatrix} x+2 & 2x+3 & 3x+4 \\ x+1 & 3x+4 & 4x+5 \\ 3x+5 & 5x+8 & 10x+17 \end{vmatrix}=0\quad$

$\therefore \Delta =(x+1)(x+2)\begin{vmatrix} x+2 & 2x+3 & 3x+4 \\ 1 & 1 & 1 \\ 1 & 2 & 6 \end{vmatrix}=0$

$\therefore \Delta =\left( x+1 \right) \left( x+2 \right) \left( x+2 \right) .4-\left( 2x+3 \right) .5+\left( 3x+4 \right) .1=0\Rightarrow { \left( x+1 \right) }^{ 2 }\left( x+2 \right) =0\Rightarrow x=-1,-1,-2$

$\begin{vmatrix} { x }^{ n } & { x }^{ n+2 } & { x }^{ n+4 } \\ { y }^{ n } & { y }^{ n+2 } & { y }^{ n+4 } \\ { z }^{ n } & { z }^{ n+2 } & { z }^{ n+4 } \end{vmatrix}=\left( \cfrac { 1 }{ { y }^{ 2 } } -\cfrac { 1 }{ { x }^{ 2 } } \right) \left( \cfrac { 1 }{ { z }^{ 2 } } -\cfrac { 1 }{ { y }^{ 2 } } \right) \left( \cfrac { 1 }{ { x }^{ 2 } } -\cfrac { 1 }{ { z }^{ 2 } } \right)$ then $n$ is _____.

Let

$\Delta = \begin{vmatrix} { x }^{ n } & { x }^{ n+2 } & { x }^{ n+4 } \\ { y }^{ n } & { y }^{ n+2 } & { y }^{ n+4 } \\ { z }^{ n } & { z }^{ n+2 } & { z }^{ n+4 } \end{vmatrix}$

$\Delta ={(xyz)}^{n}\begin{vmatrix} 1 & { x }^{ 2 } & { x }^{ 4 } \\ 1 & { y }^{ 2 } & { y }^{ 4 } \\ 1 & { z }^{ 2 } & { z }^{ 4 } \end{vmatrix}$

$={(xyz)}^{n}({x}^{2}-{y}^{2})({y}^{2}-{z}^{2})({z}^{2}-{x}^{2})$

Clearly when

$n=-4$

$\Delta =\left( \cfrac { 1 }{ { y }^{ 2 } } -\cfrac { 1 }{ { x }^{ 2 } } \right) \left( \cfrac { 1 }{ { z }^{ 2 } } -\cfrac { 1 }{ { y }^{ 2 } } \right) \left( \cfrac { 1 }{ { x }^{ 2 } } -\cfrac { 1 }{ { z }^{ 2 } } \right)$

If $x=-4$ is a root of $\Delta = \left| \begin{matrix} x & 2 & 3 \\ 1 & x & 1 \\ 3 & 2 & x \end{matrix} \right| = 0$ then find the other two roots.

Applying

$R_1 \rightarrow (R_1+R_2+R_3),$ we get

$\left| \begin{matrix} x+4 & x+4 & x+4 \\ 1 & x & 1 \\ 3 & 2 & x \end{matrix} \right|$

Taking

$(x+4)$ common from

$R_1$ , we get

$\Delta=(x+4 ) \left| \begin{matrix} 1 & 1 & 1 \\ 1 & x & 1 \\ 3 & 2 & x \end{matrix} \right|$

Applying

$C_2 \rightarrow C_2-C_1, C_3 \rightarrow C_3-C_1$ we get

$\Delta= (x+4) \left| \begin{matrix} 1 & 0 & 0 \\ 1 & x-1 & 0 \\ 3 & -1 & x-3 \end{matrix} \right|$

Expanding along

$R_1$ ,

$\Delta= (x+4)[(x-3)-0].$ Thus,

$\Delta =0$ implies

$x=-4,1,3$

If $x = - 9$ is root of $\left| \begin{matrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{matrix} \right| =0$ then other two roots are _________

Since ,

$\left| \begin{matrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{matrix} \right| =0$

expanding along

$R_1$

$x(x^2 - 12) - 3 (2x -14) +7 ( 12 -7x) = 0$

$\Rightarrow x^3 - 12x -6x +42 +84 - 49x = 0$

$\Rightarrow x^3 -67x +126 = 0$ ,.....(i)

Here

$126 \times 1 = 9 \times 2 \times 7$

For

$x = 2, 2^3 - 67 \times 2 +126 = 134 - 134 = 0$

Hence

$x = 2$ is root.

For

$x = 7 , 7^3 - 67 \times 7 + 126 = 469 -469 = 0$

hence

$x = 7$ is also a root.

If $f(x) = \left| \begin{matrix} (1+x)^{17}& (1+x)^{19} &(1+x)^{23} \\ (1+x)^{23} & (1+x)^{29} & (1+x)^{34} \\ (1+x)^{41} & (1+x)^{43} & (1+x)^{47} \end{matrix} \right| = A +Bx +Cx^2 + ........then A$ = ___________

Since

$f(x) = ( 1 +x)^{17} + (1+x)^{23} (1+x) 41 \left| \begin{matrix} 1 & (1+x)^{ 2 } & (1+x)^{ 6 } \\ 1 & (1+x)^{ 6 } & (1+x)^{ 11 } \\ 1 & (1+x)^{ 2 } & (1+x)^{ 6 } \end{matrix} \right| =0$

[ since

$R_1$ and

$R_3$ are identical ]

$\therefore A = 0$

Solve the following simultaneous equation.
$3a + 5b = 26; a + 5b = 22$

$3a + 5b = 26$ -------(I)

$a + 5b = 22$ ------------(II)
Subtracting (II) from (I)

$2a = 4$

$\Rightarrow a = 2$
Putting the value of

$a = 2$ in (II)

$5b = 22 - 2 = 20$

$\Rightarrow b = \dfrac{20}{5} = 4$
Thus,

$a = 2$ and

$b = 4$ .

Solve the following simultaneous equation.
$x + 7y = 10; 3x - 2y = 7$

$x + 7y = 10;$ ------------(I)

$3x - 2y = 7$ -----------(II)
multiplying (I) with 3

$3x + 21y = 30;$ ---------(III)

$3x - 2y = 7$ ---------------(IV)
Subtracting (IV) from (III) we get

$3x+21y-3x+2y=30-7$

$23y = 23$

$\Rightarrow 1$
Putting the value of y in (IV) we get

$3x - 2(1) = 7$

$\Rightarrow 3x = 7 + 2 = 9$

$\Rightarrow 3x = 9$

$\Rightarrow x = 3$
Thus,

$(x, y) = (3, 1)$

Solve the following simultaneous equation:
$5m - 3n = 19; m - 6n = -7$

$5m - 3n = 19$ ...........(I)

$m - 6n = - 7$ .............(II)
Multiplying (I) with 2 we get

$10m - 6n = 38$ ..........(III)
Subtracting (II) from (III) wqe get

$10m - m - 6n - (-6n) = 38 - (-7)$

$\Rightarrow 9m = 45$

$\Rightarrow m = \dfrac{45}{9} = 5$
Putting the value of

$m = 5$ in (II) we get

$5 - 6n = -7$

$\Rightarrow - 6n = -7 - 5$

$\Rightarrow - 6n = - 12$

$\Rightarrow n = \dfrac{-12}{-6} = 2$
Thus,

$(m, n) = (5, 2)$ .

Solve the following simultaneous equation:
$5x + 2y = -3; x + 5y = 4$

$5x + 2y = -3$ ........(I)

$x + 5y = 4$ ..........(II)
Multiply (II) with 5 we get

$5x + 25y = 20$ ............(III)
Subtracting (III) from (I) we get

$5x - 5x + 2y - 25y = -3 - 20$

$\Rightarrow - 23y = -23$

$\Rightarrow y = \dfrac{-23}{-23} = 1$
Putting the value of

$y =1$ in (II) we get

$x + 5(1) = 4$

$\Rightarrow x + 5 = 4$

$\Rightarrow x = 4 - 5 = -1$
Thus

$(x,y) = (-1,1)$

Solve the following simultaneous equations.
$\frac{7x - 2y}{xy} = 5 ;\frac{8x + 7y}{xy} = 15$

$\dfrac{7x - 2y}{xy} = 5 ;\dfrac{8x + 7y}{xy} = 15$

$\Rightarrow \dfrac{7}{y} - \dfrac{2}{x} = 5$ and

$\dfrac{8}{y} - \dfrac{7}{x} = 15$
Let

$\dfrac{1}{x} = u, \dfrac{1}{y} = v$

$7v - 2u = 5......(I)$

$8v + 7u = 15......(II)$
Multiply (I) with 7 and (II) with 2

$49v - 14u = 35....(III)$

$16v + 14u = 30.....(IV)$
Adding (III) and (IV)

$65v = 65$

$\Rightarrow$

$v = 1$
And

$\dfrac{1}{y} = v = 1$

$\Rightarrow y = 1$
Putting the value of v in (I)

$7(1) - 2u = 5$

$\Rightarrow$

$u = 1$

$\dfrac{1}{x} = u = 1$

$\Rightarrow x = 1$

$(x,y) = (1,1)$

Solve the following simultaneous equation.
$49x - 57y = 172; 57x - 49y = 252$

$49x - 57y = 172$ ...........(I)

$57x - 49y = 252$ ...........(II)
Adding (I) and (II)

$49x + 57x - 57y - 49y = 172 + 252$

$\Rightarrow 106x - 106y = 424$

$\Rightarrow x - y = 4$ ...............(III)
Subtracting (II) from (I) we have

$49x + 57x - 57y - (- 49y) = 252 - 172$

$\Rightarrow - 8x - 8y = -80$

$\Rightarrow - x - y = -10$

$\Rightarrow x + y = 10$ ...............(IV)
Adding (III) and (IV)

$x - y = 4$

$x + y = 10$

$\Rightarrow 2x = 14$

$\Rightarrow x = 7$
Putting the value of x = 7 in (IV) we get

$7 + y = 10$

$\Rightarrow y = 10 -7$

$\Rightarrow y = 3$
Thus,

$(x,y) = (7,3)$

Solve the following simultaneous equation.
$\dfrac{1}{3}x + y = \dfrac{10}{3}; 2x + \dfrac{1}{4}y = \dfrac{11}{4}$

$\dfrac{1}{3}x + y = \dfrac{10}{3}.$ .........(I)

$2x + \dfrac{1}{4}y = \dfrac{11}{4}$ ......(II)
Multiply (I) with 3 and (II) with 4

$x + 3y = 10$ .............(III)

$8x + y = 11$ ...........(IV)
Multiply (IV) with 3

$24x + 3y = 33$ ..............(V)
Subtracting (V) from (III)

$x - 24x + 3y - 3y = 10 - 33$

$\Rightarrow- 23x = -23$

$\Rightarrow x = 1$
Putting the value of

$x = 1$

$1 + 3y = 10$

$\Rightarrow 3y = 10 - 1 = 9$

$\Rightarrow y = \dfrac{9}{3} = 3$
Thus,

$(x,y) = (11,3)$

Solve the following equation by Cramer's method.
$3x - 2y = \frac{5}{2}; \frac{1}{3}x + 3y = - \frac{4}{3}$

$3x - 2y = \dfrac{5}{2}; \dfrac{1}{3}x + 3y = - \dfrac{4}{3}$

$D = \begin{vmatrix} 3 & {-2} \\ {\dfrac{1}{3}} & 3 \end{vmatrix} = 9 + \dfrac{2}{3} = \dfrac{20}{3}$

$D_x = \begin{vmatrix} \dfrac{5}{2} & {-2} \\ \dfrac{-4}{3} & 3 \end{vmatrix} = \dfrac{5}{2} - \dfrac{8}{3} = \dfrac{29}{6}$

$D_y = \begin{vmatrix} 3 & \dfrac{5}{2} \\ \dfrac{1}{3} & \dfrac{-4}{3} \end{vmatrix}$

$= -4 - \dfrac{5}{6} = \dfrac{-29}{6}$

$x = \dfrac{D_x}{D} = \dfrac{\dfrac{29}{6}}{\dfrac{29}{3}} = \dfrac{1}{2}$

$y = \dfrac{D_y}{D} = \dfrac{\dfrac{-29}{6}}{\dfrac{29}{3}} = \dfrac{-1}{2}$

$(x,y) = (\dfrac{1}{2}, \dfrac{-1}{2})$

Solve the following simultaneous equations
$\frac{148}{x} + \frac{231}{y} = \frac{527}{xy}; \frac{231}{x} + \frac{148}{y} = \frac {610}{xy}$

$\dfrac{148}{x} + \dfrac{231}{y} = \dfrac{527}{xy}; \dfrac{231}{x} + \dfrac{148}{y} = \dfrac {610}{xy}$
Multiply by

$xy$

$148y + 231x = 527.....(I)$

$231y + 148x = 610....(II)$
Adding (I) and (II)

$379y + 179x = 1137$

$\Rightarrow x + y = 3 ......(III)$

$(II) - (I)$

$83y - 83x = 83$

$\Rightarrow y - x = 1 .....(IV)$
(III) + (IV)

$2y = 4$

$\Rightarrow y = 2$
Putting the value of y in (IV)

$2 - x = 1$

$\Rightarrow$

$x = 1$

$(x, y) = (1,2)$

Solve the following equations by Cramer's method.
4m - 2n = -4; 4m + 3n = 16

$4m - 2n = -4; 4m + 3n = 16$

$D = \begin{vmatrix} 4 & {-2} \\ 4 & 3 \end{vmatrix} = 4 \times 4 - (-2) \times 4 = 12 + 8 = 20$

$D_x = \begin{vmatrix} {-4} & {-2} \\ 16 & 3 \end{vmatrix} = -4 \times 3 - (-2) \times 16 = - 12 + 32 = 20$

$D_y = \begin{vmatrix} 4 & {-4} \\ 4 & 16 \end{vmatrix} = 4 \times 16 - (-4) \times 4 = 64 + 16 = 80$

$x = \dfrac{D_x}{D} = \dfrac{20}{20} = 1$

$y = \dfrac{D_y}{D} = \dfrac{80}{20} = 4$

$(x, y) = (1, 4)$

Solve the following simultaneous equations.
$\frac{1}{2(3x + 4y)} + \frac{1}{5(2x - 3y)} = \frac{1}{4}; \frac{5}{(3x + 4y)} - \frac{2}{(2x - 3y)} = -\frac{3}{2}$

$\dfrac{1}{2(3x + 4y)} + \dfrac{1}{5(2x - 3y)} = \dfrac{1}{4}; \dfrac{5}{(3x + 4y)} - \dfrac{2}{(2x - 3y)} = -\dfrac{3}{2}$

$\dfrac{1}{3x + 4y} = u, \dfrac{1}{2x - 3y} = v$

$\dfrac{1}{2}u + \dfrac{1}{5}v = \dfrac{1}{4}$

$\Rightarrow$

$10u + 4v = 5.$ ....(I)

$5u - 2v = \dfrac{-3}{2}$

$\Rightarrow$

$10u - 4v = -3$ . ......(II)
(I) + (II)

$20u = 2$

$\Rightarrow u = \dfrac{1}{10}$
Putting the value of u in (II)

$10 \times \dfrac{1}{10} -4v = -3$

$\Rightarrow 1 + 3 = 4v$

$\Rightarrow v = 1$

$\dfrac{1}{3x + 4y} = u = \dfrac{1}{10}$

$\Rightarrow 3x + 4y = 10$ ....(III)

$\dfrac{1}{2x - 3y} = v = 1$

$\Rightarrow 2x - 3y = 1$ ...........(VI)
Multiply (III) with 2 and (VI) with 3

$6x + 8y = 20 ....(V)$

$6x - 9y = 3 .......(VI)$

$(V) - (VI)$

$17y = 17$

$\Rightarrow y = 1$
Putting the value of y in (VI)

$6x - 9 = 3$

$\Rightarrow 6x = 12$

$\Rightarrow x = 2$

$(x,y) = (2,1)$

Solve the following equations by Cramer's methd.
$6x - 3y = -10; 3x + 5y - 8 = 0$

$6x - 3y = -10; 3x + 5y - 8 = 0$

$D = \begin{vmatrix} 6 & {-3} \\ 3 & 5 \end{vmatrix} = 6 \times 5 - (-3) \times 3 = 30 + 9 = 39$

$D_x = \begin{vmatrix} {10} & {-3} \\ 8 & 5 \end{vmatrix} = -10 \times 5 - (-3) \times 8 = - 50 + 24 = -26$

$D_y = \begin{vmatrix} 6 & {-10} \\ 3 & 8 \end{vmatrix} = 6 \times 8 - (-10) \times 3 = 48 + 30 = 78$

$x = \dfrac{D_x}{D} = \dfrac{-26}{39} = \dfrac{-2}{3}$

$y = \dfrac{D_y}{D} = \dfrac{78}{39} = 2$

$(x, y) = (\dfrac{-2}{3}, 2)$

Solve the following
simultaneous equation using Cramer's rule.
$6x - 4y = -12$
$8x - 3y = -2$

$6x - 4y = -12$
$8x - 3y = -2$

$D = \begin{vmatrix} 6&-4 \\ 8&-3 \end{vmatrix} = 6 \times (-3) - (-4) \times 8 = (-18) + 32 = 14$

${D}_{x} = \begin{vmatrix} -12&-4 \\ -2&-3 \end{vmatrix} = (-12) \times (-3) - (-4) \times (-2) = 36 - 8 = 28$

${D}_{y} = \begin{vmatrix} 6&-12 \\ 8&-2 \end{vmatrix} = 6 \times (-2) - (-12) \times 8 = -12 + 96 = 84$

$y = \dfrac {{D}_{y}}{D} = \dfrac {84}{14} = 6$

$(x,y) = (2,6)$

Solve the following simultaneous equations.
2x + y = - 2; 3x - y =7

2x + y = -2 .....(I)
3x - y = 7 .....(II)
Adding (I) and (II)
5x = 5

⇒4⇒4 x= 1
Putting the value of x in (I)
2 x 1 + y = -2

⇒⇒ 2 + y = -2

⇒⇒ y = - 4

Solve the following sets of simultaneous equations.
2y -x =0; 10x + 15y = 105

Let the length be x and the breadth be y.
Perimeter of the rectangle = 2(1 + b) = 26
2(1 + b) = 26

⇒⇒2[( x - 5) + (y - 5)] = 26

⇒⇒ 2 [ x + y - 10] = 26

⇒⇒ x + y = 23
Hence, the correct answer is option (C).

Solve the following simultaneous equations. x -2 y = -1 ; 2x - y =7

x + y = 11 .....(I)
2x - 3y = 7 ..... (II)
Multiply (I) with 3
3x + 3y = 33 ..... (III)
Adding (II) and (III)
5x = 40
$$\Rightarrow4$ x = 8
Putting the value of x in (I)
8 + y = 11
y = 3

Solve the following simultaneous equations. x - 2y = - 2 ; x +2y =10

x - 2y = - 2 .....(I)
x + 2y = 10 .......(II)
Adding I and II
2x = 8

$\Rightarrow$ x = 4
Putting the value of x in (I) we get
4 - 2y = -2

$\Rightarrow$ - 2y = -6

$\Rightarrow$ y = 3

If $a,b,c$ are real, find the factors of the determination.
$\Delta =\begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix}$

${R}_{1}\Rightarrow {R}_{1}+{R}_{2}+{R}_{3}$

$\Delta =\begin{vmatrix} 2\left( a+b+c \right) & 2\left( a+b+c \right) & 2\left( a+b+c \right) \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix}$

$=2\left( a+b+c \right) \begin{vmatrix} 1 & 1 & 1 \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix}$

${c}_{2}\Rightarrow {c}_{2}-{c}_{1}$ ;

${c}_{3}\Rightarrow {c}_{3}-{c}_{1}$

$=2\left( a+b+c \right) \begin{vmatrix} 1 & 0 & 0 \\ c+a & b-c & b-a \\ a+b & c-a & c-b \end{vmatrix}=2\left( a+b+c \right) \left[ (b-c)(c-b)-(b-a)(c-a) \right]$

$=2\left( a+b+c \right) [bc-{b}^{2}-{c}^{2}_bc-bc+ba+ac-{a}^{2}]$

$=2\left( a+b+c \right) [ab+bc+ca-{a}^{2}-{b}^{2}-{c}^{2}]$

$=-2\times \cfrac{1}{2}(a+b+c)[{(a-b)}^{2}+{(b-c)}^{2}+{(c-a)}^{2}$
The factors are

$a+b+c$ and

${a}^{2}+{b}^{2}+{c}^{2}-ab-bc-ca$

$=1(a+b+c)[{(a-b)}^{2}+{(b-c)}^{2}+{(c-a)}^{2}$
If

$\Delta =0$ then

$(a+b+c)[{(a-b)}^{2}+{(b-c)}^{2}+{(c-a)}^{2}]=0$
then

$a+b+c=0$ or

$a=b=c$

Solve the following system of equation by Cramer's rule:
$2x-7y-13=0, 5x+6y-9=0$

Given equations

$2x-7y-13=0, 5x+6y-9=0$

here,

$\Delta = \begin{vmatrix} 2 & -7 \\ 5 & 6 \end{vmatrix}=12+35=47$

$\Delta_1=\begin{vmatrix} 13 & -7 \\ 9 & 6 \end{vmatrix}=78+63=141$

$\Delta _2=\begin{vmatrix} 2 & 13 \\ 5 & 9 \end{vmatrix}=18-65=-47$

$\because \Delta \ne 0, \Delta_1 \ne 0$ and

$\Delta_2 \ne 0$

$\therefore$ Solution will be unique.

Using Cramer's rule

$x=\dfrac{\Delta_1}{\Delta}=\dfrac{141}{47}=3$

and

$y=\dfrac{\Delta_2}{\Delta}=\dfrac{-47}{47}=-1$

Solution of equation be

$x=3, y=-1$

Use Cramer's rule to solve the following system of equations:
$2x-y=17$
$3x+5y=6$

Given,

$2x-y=17$

$3x+5y=6$

Here,

$\Delta =\begin{vmatrix} 2 & -1 \\ 3 & 5 \end{vmatrix}=10+3=13\\$

$\Delta_a =\begin{vmatrix} 17 & -1 \\ 6 & 5 \end{vmatrix}=85+6=91\\$

$\Delta_2 =\begin{vmatrix} 2 & 17 \\ 3 & 6 \end{vmatrix}=12-51=-39\\$

$\because \Delta \ne 0, \Delta_1 \ne 0$ and

$\Delta_2 \ne 0$

$\therefore$ Solution will be unique

By using Cramer's rule

$x=\dfrac{\Delta_1}{\Delta}=\dfrac{91}{13}=7, y=\dfrac{\Delta_2}{\Delta}=\dfrac{-39}{13}=-3$

Hence, solution is

$x=7, y=-3$

Solve the following system of equation by Cramer's rule:
$2x+3y=9, 3x-2y=7$

Given equations

$2x+3y=9, 3x-2y=7$
here,

$x=\dfrac{\Delta_1}{\Delta}=\dfrac{-39}{-13}=3$
and

$y=\dfrac{\Delta_2}{\Delta}=\dfrac{-13}{-13}=1$
solution of equation

$x=3, y=1$

Use Cramer's rule to solve the following system of equations:
$3x+ay=4$
$2x+ay=2, a \ne 0$

Given,

$3x+ay=4$

$2x+ay=2, a \ne 0$

$a \ne 0$

where,

$\Delta =\begin{vmatrix} 3 & a \\ 2 & a \end{vmatrix}=3a-2a=a\\$

$\Delta_1 =\begin{vmatrix} 4 & a \\ 2 & a \end{vmatrix}=4a-2a=2a\\$

$\Delta_2 =\begin{vmatrix} 3 & 4\\ 2 & 2 \end{vmatrix}=6-8=-2\\$

$\because \Delta \ne 0, \Delta_1 \ne 0$ and

$\Delta_2 \ne 0$

$\therefore$ Solution will be unique

By using Cramer's rule

$x=\dfrac{\Delta_1}{\Delta}=\dfrac{2a}{a}=2, y=\dfrac{\Delta_2}{\Delta}=\dfrac{-2}{a}=\dfrac{-2}{a}$

Hence, solution is

$x=1, y=\dfrac{-2}{a}$

A determinant of second order is made with the elements 0 andFind the number of determinants with non-negative values.

$\textbf{Step -1: Finding the number of possible second order determinants with elements 0 an 1.}$

$\text{A determinant of second order has four elements and the choice for one element is 2, }$

$\text{i.e. either 0 or 1.}$

$\therefore \text{Number of possible second order determinants with elements 0 and 1}=2^4=16.$

$\textbf{Step -2: Finding the number of second order determinants with negative values.}$

$\begin{vmatrix}0&1\\1&0\end{vmatrix},\begin{vmatrix}0&1\\1&1\end{vmatrix}\text{ and }\begin{vmatrix}1&1\\1&0\end{vmatrix}\text{ are the only determinants of second order with negative values.}$

$\therefore\text{Number of second order determinants with negative values}=3.$

$\textbf{Step -3: Finding the number of second order determinants with non-negative values.}$

$\text{Number of second order determinants with non-negative values}=16-3$

$=13$

$\textbf{ Hence, the number of determinants with non-negative values are 13.}$

Let $A =\begin{bmatrix} 2&3 \\-1 &5 \end{bmatrix}$ . If $A^{-1}=xA+yI$ , find $x + 2y$ .

$A\quad =\quad \begin{bmatrix} 2 & 3 \\ -1 & 5 \end{bmatrix}\\$

$A^{ -1 }\quad =\quad xA\quad +\quad yI\\$

$\Rightarrow \quad A^{ -1 }A\quad =\quad xAA\quad +\quad yIA\\$

$\Rightarrow \quad I\quad =\quad xA^{ 2 }\quad +\quad yA\\$

$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\quad =\quad x\begin{bmatrix} 2 & 3 \\ -1 & 5 \end{bmatrix}\begin{bmatrix} 2 & 3 \\ -1 & 5 \end{bmatrix}\quad +\quad y\begin{bmatrix} 2 & 3 \\ -1 & 5 \end{bmatrix}\\$

$\quad \quad \quad \quad \quad =\quad x\begin{bmatrix} 1 & 21 \\ -7 & \quad 22\quad \end{bmatrix}\quad +\quad y\begin{bmatrix} 2 & 3 \\ -1 & 5 \end{bmatrix}\\$

$comparing\quad elements\quad on\quad both\quad sides\\$

$x+2y\quad =\quad 1\quad and\quad -7x-y\quad =\quad 0\\$

$solving\quad above\quad equations\quad gives\\$

$x\quad =\quad \displaystyle\frac { -1 }{ 13 } \quad and\quad y\quad =\quad \displaystyle\frac { 7 }{ 13 }$

$\Rightarrow x+2y=1$

If $A =\begin{bmatrix} 1&0 &0 \\0 &1 &1 \\0 &-2 &4 \end{bmatrix}$ , $I=\begin{bmatrix} 1&0 &0 \\0 &1 &0 \\0 &0 &1 \end{bmatrix}$ and
$\displaystyle A^{-1}=\frac{1}{6}(A^2+\alpha A+\beta I)$ , find $\beta/11$

$If\quad A=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4 \end{bmatrix},\quad I=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\quad and\quad A^{ -1 }=\dfrac { 1 }{ 6 } (A^{ 2 }+\alpha A+\beta I)\\ \Rightarrow \quad A^{ -1 }A\quad =\quad \frac { 1 }{ 6 } \left( A^{ 2 }A+\alpha AA+\beta IA \right) \\ \Rightarrow \quad 6I\quad =\quad A^{ 3 }+\alpha A^{ 2 }+\beta A\quad -------\boxed { 1 } \\ A^{ 2 }\quad =\quad \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4 \end{bmatrix}\quad =\quad \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 5 \\ 0 & -10 & 14 \end{bmatrix}\\ A^{ 3 }\quad =\quad \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 5 \\ 0 & -10 & 14 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4 \end{bmatrix}\quad =\quad \begin{bmatrix} 1 & 0 & 0 \\ 0 & -11 & 19 \\ 0 & -38 & 46 \end{bmatrix}\\ from\quad equation\quad ----\boxed { 1 } \\ \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}\quad =\quad \begin{bmatrix} 1 & 0 & 0 \\ 0 & -11 & 19 \\ 0 & -38 & 46 \end{bmatrix}+\alpha \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 5 \\ 0 & -10 & 14 \end{bmatrix}+\beta \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4 \end{bmatrix}\\ comparing\quad elements\quad on\quad both\quad sides\quad gives\\ \alpha +\beta \quad =\quad 5\quad and\quad -\alpha +\beta \quad =\quad 17\\ Hence,\quad \beta \quad =\quad 11\\$

The digit in the tens place of a two-digit number is three times that in the units place. If the digits are reversed, the new number will be $36$ less than the original number. Find the original number. Check your solution.

There are two possibility such as-

$\Rightarrow \ 3\ 1$

$\Rightarrow \ 6\ 2$

$\Rightarrow \ 9\ 3$

if we reversed the digit, only

$62$ will given the resent as per question

$62-26 \Rightarrow \ 36$

Hence original No is

$62$

Find $x$ and $y$ using Cramer's Rule if :
$\dfrac{1}{2x}+\dfrac{2}{3y}=10,\dfrac{3}{2x}-\dfrac{5}{3y}=-3$

$\dfrac { 1 }{ 2x } +\dfrac { 2 }{ 3y } =10\longrightarrow (1)\\ \dfrac { 3 }{ 2x } -\dfrac { 5 }{ 3y } =-3\longrightarrow (2)\\ Let,\quad \dfrac { 1 }{ 2x } =p\quad and\quad \dfrac { 1 }{ 3y } =q\\ \therefore \quad equation\quad becomes\quad \\ p+2q=10\longrightarrow (3)\\ 3p-5q=-3\longrightarrow (4)\\ By\quad Cramer's\quad rule\\ \begin{bmatrix} 1 & 2 \\ 3 & -5 \end{bmatrix}\left[ \begin{matrix} p \\ q \end{matrix} \right] =\left[ \begin{matrix} 10 \\ -3 \end{matrix} \right] \\ \therefore p=\dfrac { \begin{vmatrix} 10 & 2 \\ -3 & -5 \end{vmatrix} }{ \begin{vmatrix} 1 & 2 \\ 3 & -5 \end{vmatrix} } \quad and\quad q=\dfrac { \begin{vmatrix} 1 & 10 \\ 3 & -3 \end{vmatrix} }{ \begin{vmatrix} 1 & 2 \\ 3 & -5 \end{vmatrix} } \\ \Rightarrow p=4\quad and\quad q=3\\ \because p=\dfrac { 1 }{ 2x } =4\\ \Rightarrow x=\dfrac { 1 }{ 8 } \\ and\quad q=\dfrac { 1 }{ 3y } =3\\ \Rightarrow y=\dfrac { 1 }{ 9 }$

Prove that one root of the equation is x=2 and hence find the remaining roots
$\left| \begin{matrix} x \\ 2 \\ -3 \end{matrix}\begin{matrix} -6 \\ -3x \\ 2x \end{matrix}\begin{matrix} -1 \\ x-3 \\ x+2 \end{matrix} \right| =0$

$\begin{vmatrix} x & -6 & -1\\ 2 & -3x & x-3 \\ -3 & 2x & x+2 \end{vmatrix}=0$

$\Rightarrow x[-3x(x+2)-2x(x-3)]+6[2(x+2)+3(x-3)]-1[2(2x)-9x]=0$

$\Rightarrow x[-3x^2 - 6 - 2x^2+ 6]+6[2x+4+3x-9]-[4x-9x]=0$

$\Rightarrow x(-5x^2)+6(5x-5) + 5x = 0$

$\Rightarrow -5x^3 + 30(x-1)+5x=0$

$\Rightarrow -x^3 + 6x-6+x = 0$

$\Rightarrow -x^3 + 7x - 6 = 0$

$\Rightarrow x^3-7x+6=0$

Let

$f(x) = x^3 - 7x + 6$

$f(2) = 2^3 - 7.2 + 6$

$= 8-14 + 6$

$= 14 - 14$

$=0$

Hence

$x=2$ is a root of the equation

$x^3 - 7x + 6 = 0$

[see image]

$(x-2) [x(x+3) - 1(x+3) ] =0$

$(x-2)[x^2 + 3x - x - 3]=0$

$(x-2)[x(x+3) - 1(x+3)]=0$

$(x-2)(x-1)(x+3)=0$

$\therefore x = 2, 1, -3$

Answer: Roots of the equation are

$x=2, 1, -3$

2047422_1161645_ans_c596b71418ce4cff8d7d058a008753c4.png

If S the set of distinct values of 'b' for which the following system of linear equations
x + y + z = 1
x + ay + z = 1
ax + by + z = 1

2039531_1296476_ans_5edf15181a33443697582dad58ff68ee.jpg

Use cramer's rule and solve
$3x+4y+5z=18$
$2x-y+8z=13$
$5x-2y+7z=20$

Consider the system of equations

$3x+4y+5z=18$

$2x-y+8z=13$

$5x-2y+7z=20$

Cramer's rule is applicable when determinant

$\neq 0$

$D=\left|\begin{matrix}3 & 4 & 5 \\ 2 &-1 & 8 \\ 5 & -2 & 7 \end{matrix}\right|$

$=3\left(-7+16\right)-4\left(14-40\right)+5\left(-4+5\right)$

$=27+104+5=136\neq 0$

$\therefore$ Cramer's rule is aplicable.

${D}_{x}=\left|\begin{matrix}18 & 4 & 5 \\ 13 &-1 & 8 \\ 20 & -2 & 7 \end{matrix}\right|$ by replacing the coefficents of

${C}_{1}$ by the Constants.

$=18\left(-7+16\right)-4\left(91-160\right)+5\left(-26+20\right)$

$=81+276-30=321$

${D}_{y}=\left|\begin{matrix}3 & 18 & 5 \\ 2 &13 & 8 \\ 5 & 20 & 7 \end{matrix}\right|$ by replacing the coefficents of

${C}_{2}$ by the Constants.

$=3\left(91-160\right)-18\left(35-40\right)+5\left(40-65\right)$

$=-207+90-125=-242$

${D}_{z}=\left|\begin{matrix}3 & 4 & 18 \\ 2 &-1 & 13 \\ 5 & -2 & 20 \end{matrix}\right|$ by replacing the coefficents of

${C}_{3}$ by the Constants.

$=3\left(-20+26\right)-4\left(40-65\right)+8\left(-10+5\right)$

$=18+100-40=78$

The values of

$x,y,z$ are calculated as

$x=\dfrac{{D}_{x}}{D}=\dfrac{321}{136}$

$y=\dfrac{{D}_{y}}{D}=\dfrac{-242}{136}=\dfrac{-121}{68}$

$z=\dfrac{{D}_{z}}{D}=\dfrac{78}{136}=\dfrac{39}{68}$

Solve the following simultaneous equations using corner's rule.
(3) $x+2y=-1;2x-3y=12$

1961049_1491772_ans_1703b5c0cd7a4d1d8012ebea9c1c2247.jpg

The value of $\begin{vmatrix} 2{ x }_{ 1 }{ y }_{ 1 } & { x }_{ 1 }{ y }_{ 2 }+{ x }_{ 2 }{ y }_{ 1 } & { x }_{ 1 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 1 } \\ { x }_{ 1 }{ y }_{ 2 }+{ x }_{ 2 }{ y }_{ 1 } & 2{ x }_{ 2 }{ y }_{ 2 } & { x }_{ 2 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 2 } \\ { x }_{ 1 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 1 } & { x }_{ 2 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 2 } & 2{ x }_{ 3 }{ y }_{ 3 } \end{vmatrix}$ is

Given

$D=\left| \begin{matrix} 2{ x }_{ 1 }{ y }_{ 1 } & { x }_{ 1 }{ y }_{ 2 }+{ x }_{ 2 }{ y }_{ 1 } & { x }_{ 1 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 1 } \\ { x }_{ 1 }{ y }_{ 2 }+{ x }_{ 2 }{ y }_{ 1 } & 2{ x }_{ 2 }{ y }_{ 2 } & { x }_{ 2 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 2 } \\ { x }_{ 1 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 1 } & { x }_{ 2 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 2 } & 2{ x }_{ 3 }{ y }_{ 3 } \end{matrix} \right|$

$\therefore D=\left| \begin{matrix} { x }_{ 1 }{ y }_{ 1 }+{ x }_{ 1 }{ y }_{ 1 } & { x }_{ 1 }{ y }_{ 2 }+{ x }_{ 2 }{ y }_{ 1 } & { x }_{ 1 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 1 } \\ { x }_{ 1 }{ y }_{ 2 }+{ x }_{ 2 }{ y }_{ 1 } & { x }_{ 2 }{ y }_{ 2 }+{ x }_{ 2 }{ y }_{ 2 } & { x }_{ 2 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 2 } \\ { x }_{ 1 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 1 } & { x }_{ 2 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 2 } & { x }_{ 3 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 3 } \end{matrix} \right|$

$\therefore D=\left| \begin{matrix} { x }_{ 1 }{ y }_{ 1 } & { x }_{ 1 }{ y }_{ 2 }+{ x }_{ 2 }{ y }_{ 1 } & { x }_{ 1 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 1 } \\ { x }_{ 1 }{ y }_{ 2 } & { x }_{ 2 }{ y }_{ 2 }+{ x }_{ 2 }{ y }_{ 2 } & { x }_{ 2 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 2 } \\ { x }_{ 1 }{ y }_{ 3 } & { x }_{ 2 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 2 } & { x }_{ 3 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 3 } \end{matrix} \right| +\left| \begin{matrix} { x }_{ 1 }{ y }_{ 1 } & { x }_{ 1 }{ y }_{ 2 }+{ x }_{ 2 }{ y }_{ 1 } & { x }_{ 1 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 1 } \\ { x }_{ 2 }{ y }_{ 1 } & { x }_{ 2 }{ y }_{ 2 }+{ x }_{ 2 }{ y }_{ 2 } & { x }_{ 2 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 2 } \\ { x }_{ 3 }{ y }_{ 1 } & { x }_{ 2 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 2 } & { x }_{ 3 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 3 } \end{matrix} \right|$

$\therefore D={ x }_{ 1 }\left| \begin{matrix} { y }_{ 1 } & { x }_{ 1 }{ y }_{ 2 }+{ x }_{ 2 }{ y }_{ 1 } & { x }_{ 1 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 1 } \\ { y }_{ 2 } & { x }_{ 2 }{ y }_{ 2 }+{ x }_{ 2 }{ y }_{ 2 } & { x }_{ 2 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 2 } \\ { y }_{ 3 } & { x }_{ 3 }{ y }_{ 2 }+{ x }_{ 2 }{ y }_{ 3 } & { x }_{ 3 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 3 } \end{matrix} \right| +{ y }_{ 1 }\left| \begin{matrix} { x }_{ 1 } & { x }_{ 1 }{ y }_{ 2 }+{ x }_{ 2 }{ y }_{ 1 } & { x }_{ 1 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 1 } \\ { x }_{ 2 } & { x }_{ 2 }{ y }_{ 2 }+{ x }_{ 2 }{ y }_{ 2 } & { x }_{ 2 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 2 } \\ { x }_{ 3 } & { x }_{ 3 }{ y }_{ 2 }+{ x }_{ 2 }{ y }_{ 3 } & { x }_{ 3 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 3 } \end{matrix} \right|$

$\therefore D={ x }_{ 1 }\left| \begin{matrix} { y }_{ 1 } & { x }_{ 1 }{ y }_{ 2 } & { x }_{ 1 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 1 } \\ { y }_{ 2 } & { x }_{ 2 }{ y }_{ 2 } & { x }_{ 2 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 2 } \\ { y }_{ 3 } & { x }_{ 3 }{ y }_{ 2 } & { x }_{ 3 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 3 } \end{matrix} \right| +{ x }_{ 1 }\left| \begin{matrix} { y }_{ 1 } & { x }_{ 2 }{ y }_{ 1 } & { x }_{ 1 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 1 } \\ { y }_{ 2 } & { x }_{ 2 }{ y }_{ 2 } & { x }_{ 2 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 2 } \\ { y }_{ 3 } & { x }_{ 2 }{ y }_{ 3 } & { x }_{ 3 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 3 } \end{matrix} \right| +{ y }_{ 1 }\left| \begin{matrix} { x }_{ 1 } & { x }_{ 1 }{ y }_{ 2 } & { x }_{ 1 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 1 } \\ { x }_{ 2 } & { x }_{ 2 }{ y }_{ 2 } & { x }_{ 2 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 2 } \\ { x }_{ 3 } & { x }_{ 3 }{ y }_{ 2 } & { x }_{ 3 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 3 } \end{matrix} \right| +{ y }_{ 1 }\left| \begin{matrix} { x }_{ 1 } & { x }_{ 2 }{ y }_{ 1 } & { x }_{ 1 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 1 } \\ { x }_{ 2 } & { x }_{ 2 }{ y }_{ 2 } & { x }_{ 2 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 2 } \\ { x }_{ 3 } & { x }_{ 2 }{ y }_{ 3 } & { x }_{ 3 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 3 } \end{matrix} \right|$

$\therefore D={ x }_{ 1 }{ y }_{ 2 }\left| \begin{matrix} { y }_{ 1 } & { x }_{ 1 } & { x }_{ 1 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 1 } \\ { y }_{ 2 } & { x }_{ 2 } & { x }_{ 2 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 2 } \\ { y }_{ 3 } & { x }_{ 3 } & { x }_{ 3 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 3 } \end{matrix} \right| +{ x }_{ 1 }{ x }_{ 2 }\left| \begin{matrix} { y }_{ 1 } & { y }_{ 1 } & { x }_{ 1 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 1 } \\ { y }_{ 2 } & { y }_{ 2 } & { x }_{ 2 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 2 } \\ { y }_{ 3 } & { y }_{ 3 } & { x }_{ 3 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 3 } \end{matrix} \right| +{ y }_{ 1 }{ y }_{ 2 }\left| \begin{matrix} { x }_{ 1 } & { x }_{ 1 } & { x }_{ 1 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 1 } \\ { x }_{ 2 } & { x }_{ 2 } & { x }_{ 2 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 2 } \\ { x }_{ 3 } & { x }_{ 3 } & { x }_{ 3 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 3 } \end{matrix} \right| +{ y }_{ 1 }{ x }_{ 2 }\left| \begin{matrix} { x }_{ 1 } & { y }_{ 1 } & { x }_{ 1 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 1 } \\ { x }_{ 2 } & { y }_{ 2 } & { x }_{ 2 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 2 } \\ { x }_{ 3 } & { y }_{ 3 } & { x }_{ 3 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 3 } \end{matrix} \right|$

$\therefore D={ x }_{ 1 }{ y }_{ 2 }\left| \begin{matrix} { y }_{ 1 } & { x }_{ 1 } & { x }_{ 1 }{ y }_{ 3 } \\ { y }_{ 2 } & { x }_{ 2 } & { x }_{ 2 }{ y }_{ 3 } \\ { y }_{ 3 } & { x }_{ 3 } & { x }_{ 3 }{ y }_{ 3 } \end{matrix} \right| +{ x }_{ 1 }{ y }_{ 2 }\left| \begin{matrix} { y }_{ 1 } & { x }_{ 1 } & { x }_{ 3 }{ y }_{ 1 } \\ { y }_{ 2 } & { x }_{ 2 } & { x }_{ 3 }{ y }_{ 2 } \\ { y }_{ 3 } & { x }_{ 3 } & { x }_{ 3 }{ y }_{ 3 } \end{matrix} \right| +{ y }_{ 1 }{ x }_{ 2 }\left| \begin{matrix} { x }_{ 1 } & { y }_{ 1 } & { x }_{ 1 }{ y }_{ 3 } \\ { x }_{ 2 } & { y }_{ 2 } & { x }_{ 2 }{ y }_{ 3 } \\ { x }_{ 3 } & { y }_{ 3 } & { x }_{ 3 }{ y }_{ 3 } \end{matrix} \right| +{ y }_{ 1 }{ x }_{ 2 }\left| \begin{matrix} { x }_{ 1 } & { y }_{ 1 } & { x }_{ 3 }{ y }_{ 1 } \\ { x }_{ 2 } & { y }_{ 2 } & { x }_{ 3 }{ y }_{ 2 } \\ { x }_{ 3 } & { y }_{ 3 } & { x }_{ 3 }{ y }_{ 3 } \end{matrix} \right|$

$\therefore D={ x }_{ 1 }{ y }_{ 2 }{ y }_{ 3 }\left| \begin{matrix} { y }_{ 1 } & { x }_{ 1 } & { x }_{ 1 } \\ { y }_{ 2 } & { x }_{ 2 } & { x }_{ 2 } \\ { y }_{ 3 } & { x }_{ 3 } & { x }_{ 3 } \end{matrix} \right| +{ x }_{ 1 }{ y }_{ 2 }{ x }_{ 3 }\left| \begin{matrix} { y }_{ 1 } & { x }_{ 1 } & { y }_{ 1 } \\ { y }_{ 2 } & { x }_{ 2 } & { y }_{ 2 } \\ { y }_{ 3 } & { x }_{ 3 } & { y }_{ 3 } \end{matrix} \right| +{ y }_{ 1 }{ x }_{ 2 }{ y }_{ 3 }\left| \begin{matrix} { x }_{ 1 } & { y }_{ 1 } & { x }_{ 1 } \\ { x }_{ 2 } & { y }_{ 2 } & { x }_{ 2 } \\ { x }_{ 3 } & { y }_{ 3 } & { x }_{ 3 } \end{matrix} \right| +{ y }_{ 1 }{ x }_{ 2 }{ x }_{ 3 }\left| \begin{matrix} { x }_{ 1 } & { y }_{ 1 } & { y }_{ 1 } \\ { x }_{ 2 } & { y }_{ 2 } & { y }_{ 2 } \\ { x }_{ 3 } & { y }_{ 3 } & { y }_{ 3 } \end{matrix} \right|$

$\therefore D={ x }_{ 1 }{ y }_{ 2 }{ y }_{ 3 }\left( 0 \right) +{ x }_{ 1 }{ y }_{ 2 }{ x }_{ 3 }\left( 0 \right) +{ y }_{ 1 }{ x }_{ 2 }{ y }_{ 3 }\left( 0 \right) +{ y }_{ 1 }{ x }_{ 2 }{ x }_{ 3 }\left( 0 \right)$

$\therefore D=0$

Shew that $\begin{vmatrix} yz-x^2 & zx-y^2 & xsy-z^2 \\ zx-y^2 & xy-z^2 & yz-x^2 \\ xy-z^2 & yz-x^2 & zx-y^2 \end{vmatrix}=\begin{vmatrix} r^2 & u^2 & u^2 \\ u^2 & r^2 & u^2 \\ u^2 & u^2 & r^2 \end{vmatrix}$
where $r^{2}=x^{2}+y^{2}+z^{2}$ and $u^{2}=yz+zx+xy$

1986609_1747426_ans_096cfa2e7c4a47409e84fcf105f4df5d.jpg

Prove that $\left| \begin{matrix} bc-a^2 & ca-b^2 & ab-c^2 \\ ca-b^2 & ab-c^2 & bc-a^2 \\ ab-c^2 & bc-a^2 & ca-b^2 \end{matrix} \right|$ i divisible by $( a+b+ c)$ and find the quotient.

Let

$D=\left| \begin{matrix} bc-{ a }^{ 2 } & ca-{ b }^{ 2 } & ab-{ c }^{ 2 } \\ ca-{ b }^{ 2 } & ab-{ c }^{ 2 } & bc-{ a }^{ 2 } \\ ab-{ c }^{ 2 } & bc-{ a }^{ 2 } & ca-{ b }^{ 2 } \end{matrix} \right|$

This matrix is cofactor matrix of

$\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} \right|$

Thus, by property of cofactor matrix,

$\left| \begin{matrix} bc-{ a }^{ 2 } & ca-{ b }^{ 2 } & ab-{ c }^{ 2 } \\ ca-{ b }^{ 2 } & ab-{ c }^{ 2 } & bc-{ a }^{ 2 } \\ ab-{ c }^{ 2 } & bc-{ a }^{ 2 } & ca-{ b }^{ 2 } \end{matrix} \right| ={ \left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} \right| }^{ 2 }$

$\therefore \left| \begin{matrix} bc-{ a }^{ 2 } & ca-{ b }^{ 2 } & ab-{ c }^{ 2 } \\ ca-{ b }^{ 2 } & ab-{ c }^{ 2 } & bc-{ a }^{ 2 } \\ ab-{ c }^{ 2 } & bc-{ a }^{ 2 } & ca-{ b }^{ 2 } \end{matrix} \right| ={ \left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} \right| }{ \left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} \right| }$

$\therefore \left| \begin{matrix} bc-{ a }^{ 2 } & ca-{ b }^{ 2 } & ab-{ c }^{ 2 } \\ ca-{ b }^{ 2 } & ab-{ c }^{ 2 } & bc-{ a }^{ 2 } \\ ab-{ c }^{ 2 } & bc-{ a }^{ 2 } & ca-{ b }^{ 2 } \end{matrix} \right| =-{ \left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} \right| }{ \left| \begin{matrix} a & c & b \\ b & a & c \\ c & b & a \end{matrix} \right| }$

$\therefore \left| \begin{matrix} bc-{ a }^{ 2 } & ca-{ b }^{ 2 } & ab-{ c }^{ 2 } \\ ca-{ b }^{ 2 } & ab-{ c }^{ 2 } & bc-{ a }^{ 2 } \\ ab-{ c }^{ 2 } & bc-{ a }^{ 2 } & ca-{ b }^{ 2 } \end{matrix} \right| ={ \left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} \right| }{ \left| \begin{matrix} a & -c & b \\ b & -a & c \\ c & -b & a \end{matrix} \right| }$

$=\left| \begin{matrix} { a }^{ 2 }-bc+bc & ab-ab+{ c }^{ 2 } & ac-{ b }^{ 2 }+ac \\ ab-{ c }^{ 2 }+ab & { b }^{ 2 }-ac+ac & bc-bc+{ a }^{ 2 } \\ ac-ac+{ b }^{ 2 } & bc-{ a }^{ 2 }+bc & { c }^{ 2 }-ab+ab \end{matrix} \right|$

$=\left| \begin{matrix} { a }^{ 2 } & { c }^{ 2 } & 2ac-{ b }^{ 2 } \\ 2ab-{ c }^{ 2 } & { b }^{ 2 } & { a }^{ 2 } \\ { b }^{ 2 } & 2bc-{ a }^{ 2 } & { c }^{ 2 } \end{matrix} \right|$

Determinants - Class 12 Commerce Applied Mathematics - Extra Questions

If Dx=−18D_x\, =\, -18 and D=3 D =\, 3 are the values of the determinants for certain simultaneous equations in xx and yy, Find −x-x.

If the value of determinants |x−534|\begin{vmatrix} x& -5\\ 3 & 4\end{vmatrix} is 3131, then find the value of xx

Find the value of the determinant: |4−231|\begin{vmatrix} 4& -2\\ 3 & 1\end{vmatrix}

If D=7,Dx=35D=7,{D}_{x}=35 and Dy=42{D}_{y}=42 are the values of the determinants for certain simultaneous equations in xx and yy, then find the values of xx and yy.

Find the value of the following determinant|7254|\begin{vmatrix}7 & 2\\ 5 &4 \end{vmatrix}

solve the following simultaneous equations by using Cramer's rule:x+y=7,x−y=5x+y=7,x-y=5

If Dx=18,Dy=15{ D }_{ x }=18,{ D }_{ y }=15 and D=3D=3 are the values of the determinants for certain simultaneous equations in xx and yy, then find the values of xx and yy.

Find the value of the following determinant |4327|\begin{vmatrix} 4 & 3\\2 & 7\end{vmatrix}.

Find the value of following determinant|5274|\begin{vmatrix} 5& 2\\ 7 & 4 \end{vmatrix}

If Dx=30,Dy=42{D}_{x}=30,{D}_{y}=42 and D=6D=6 are the values of the determinants for certain simultaneous equations in xx and yy, then find the values of xx and yy.

Solve the following simultaneous equations by using Cramer's rule:x−2y=−18x - 2y = -182x−y=92x - y = 9

Find DxD_{x} for the following simultaneous equations: 3x+4y=8,x−2y=53x + 4y = 8, x - 2y = 5

If Dx=24,Dy=32{ D }_{ x }=24,{ D }_{ y }=32 are the values of the determinants for certain simultaneous equations in xx and yy, then find the values of xx and yy

Solve the following equations by Cramer's rule:3x+y+z=2,2x−4y+3z=−1,4x+y−3z=−113x+y+z=2,2x-4y+3z=-1,4x+y-3z=-11\quad

Find the ratios of x:y:zx : y : z from the equations7x=4y+8z,3z=12x+11y7x = 4y + 8z, 3z = 12x + 11y.

If A=[1−1230−2103]A=\left[ \begin{matrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{matrix} \right] verify that A(adj A)=|A|IA(adj\ A)=\left| A \right| I.

Ifx+y+z=1x + y + z = 13x+5y+6z=43x + 5y + 6z = 49x+2y−36z=179x + 2y - 36z = 17Then find yy

Solve the equation xa+yb=2,ax−by=a2−b2\dfrac{x}{a}\, + \,\dfrac{y}{b}\, = \,2\,,\,ax\, - \,by\, = \,{a^2} - \,{b^2}

Solve the system of linear equation 3x−2y=13 3x-2y=13 and 4x+y=21 4x+y=21

Solve the following simu1taneous equations. (1) 5x−3y=8;5x-3y=8; 3x+y=23x+y=2

Solve: (i)4x+5y−62=04x+5y-62=0 (ii)3x+2y−36=03x+2y-36=0

Solve.3x−2y=13x-2y=1 and 2x+y=32x+y=3.

Solve the given pair of linear equations:(a−b)x+(a+b)y=a2−2ab−b2(a-b)x+(a+b)y={a}^{2}-2ab-{b}^{2}(a+b)(x+y)=a2+b2(a+b)(x+y)={a}^{2}+{b}^{2}

Find co-factors of elements of the matrices [−12−34]\begin {bmatrix} -1 & 2 \\ -3 & 4\end {bmatrix}

Solve the following equation by cramer's method:7x+3y=15,12y−5x=397x+3y=15 , 12y-5x=39

Find the value of: |−3−5−2−1|\begin{vmatrix} -3 & -5 \\ -2 & -1 \end{vmatrix}

Find Dx{D}_{x} and DD for the equation 2x+3y=4;7x−5y=22x+3y=4;7x-5y=2

Solve the following simultaneous equations. 2x - y = 5 ; 3x + 2y = 11

Solve the following simultaneous equations. x + y = 11 ; 2x - 3 y =7

Solve by Cramer's rulex+y+z=6x+y+z=6x−y+z=2x-y+z=23x+2y−4z=−53x+2y-4z=-5. The value of x is

If Dy=−15D_y\, =\, -15 and D=−5D\, =\, -5 are the values of the determinants for certain simultaneous equations in xx and yy, find yy.

If the polynomial f(x)=2x3+mx2+nx−14\displaystyle f\left ( x \right )=2x^{3}+mx^{2}+nx-14 has (x−1)\displaystyle \left ( x-1 \right ) and (x+2)\displaystyle \left ( x+2 \right ) as its factors find the value of m×n\displaystyle m\times n

For what value of kk will the following pair of linear equations have no solution?2x+3y=12x + 3y = 1 and (3k−1)x+(1−2k)y=2k+3(3k - 1)x +(1 - 2k) y = 2k + 3

If A=[31−12]A=\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}, show that A2−5A+7I=O{ A }^{ 2 }-5A+7I=O. Hence find A−1{ A }^{ -1 }

If A=[2−11−12−11−12]A=\begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix}, verify that A3−6A2+9A−4I=0{ A }^{ 3 }-{ 6A }^{ 2 }+9A-4I=0. Hence find A−1 { A }^{ -1 }

For the matrix A=[11112−32−13]A=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}, show that A3−6A2+5A+11I=0{ A }^{ 3 }-{ 6A }^{ 2 }+5A+11I=0. Hence find A−1{ A }^{ -1 }

Given, −3x+4y=20-3x + 4y = 20 and 6x+3y=156x +3y =15If (x,y)(x, y) is the solution to the system of equations above, what is the value of xx ?

Find 'm+nm+n' for the following simultaneous equations:3m+4n=73m+4n=7and 4m+3n=144m+3n=14

Solve the following simultaneous equations by using Cramer's Rule3x+y=13x + y = 1,2x−11y=32x - 11 y = 3

Solve the following simultaneous equations using Cramer's rule.3x−2y=33x-2y=3;2x+y=162x+y=16.

Solve x+y+z=112x−6y−z=03x+4y+2z=0x + y + z = 11\\ 2x - 6y - z = 0\\ 3x + 4y + 2z = 0.

Solve the equationsx+2y+3z=14x + 2y + 3z = 142x−y+5z=152x - y + 5z = 153x−2y−4z=−133x - 2y - 4z = -13

Find x and y using Cramer's Rule if :12x+23y=10,32x−53y=−3\dfrac{1}{2^x}+\dfrac{2}{3^y}=10,\dfrac{3}{2^x}-\dfrac{5}{3^y}=-3 ?

Find x and y using Cramer's Rule,if 1x−2y=6,3x+1y=8\dfrac{1}{x}-\dfrac{2}{y}=6,\dfrac{3}{x}+\dfrac{1}{y}=8 ?

If A=[58102143−1],B=[2−13]A = \left[ \begin{array} { c c c } { 5 } & { 8 } & { 1 } \\ { 0 } & { 2 } & { 1 } \\ { 4 } & { 3 } & { - 1 } \end{array} \right] , B = \left[ \begin{array} { c } { 2 } \\ { - 1 } \\ { 3 } \end{array} \right] and AX=BA X = B then find XX

Find A2 A^2 where A is given A=[123123−12−3]A= \begin {bmatrix}1 & 2 & 3 \\ 1 & 2 & 3\\ -1 & 2 & -3 \end{bmatrix}

Solve |111abca3b3c3|=(a−b)(b−c)(c−a)(a+b+c)\left| {\begin{array}{*{20}{c}}1 & 1 & 1\\a & b & c\\{{a^3}} & {{b^3}} & {{c^3}}\end{array}} \right| = \left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)\left( {a + b + c} \right)

For what value of k , the following pair of linear equations has infinite number of solutions :5x+2y=k;10x+4y=35x + 2y = k ; 10x + 4y = 3

If 3m +5n =9 and 5m +3n =7 then find the value of 8m +8n.

Solve the following x3+y4=4,x2−y4=1\frac{x}{3}+\frac{y}{4}=4, \frac{x}{2}-\frac{y}{4}=1

Solve the following simultaneous equation using cramer's rule4x+3y−4=0:6x=14x+3y-4=0: 6x=1

Solve the following simultaneous equation using cramer's rule6x−4y=−12:8x−3y=−26x-4y=-12: 8x-3y=-2

Solve the following simultaneous equation using cramer's rule 4m+6n=54:3m+2n=284m+6n=54 : 3m +2n =28

Solve the simultaneous equations.x3+y4=4;x2−y4=1\dfrac { x } { 3 } + \dfrac { y } { 4 } = 4 ; \quad \dfrac { x } { 2 } - \dfrac { y } { 4 } = 1

Solve the following simultaneous equations using Cramer's rule : 3x+y=7; 2x−11y=3.3x+y=7; \ 2x-11y=3.

Solve the following simultaneous equation using cramer's rule.3x−4y=10:4x+3y=53x-4y=10:4x+3y=5

Solve the following simultaneous equation using cramer's rulex+2y=−1:2x−3y=12x+2y=-1: 2x-3y=12

Solve the following simultaneous equation using cramer's rule 2x + 3y = 2: x-y2=12\frac { y }{ 2 } =\frac{1}{2}

If A=[3221]A = \begin{bmatrix} 3& 2\\ 2 & 1\end{bmatrix}, verify that A2−4A−I=OA^{2} - 4A - I = O, and hence find A−1A^{-1}.

Show that the matrix A=[−8524]A =\begin{bmatrix} -8& 5\\ 2 & 4\end{bmatrix} satisifies the equation A2+4A−42I=0A^{2} + 4A - 42I = 0 and hence find A−1A^{-1}.

If A=[−1−12−2]A = \begin{bmatrix} -1& -1\\ 2 & -2\end{bmatrix}, show that A2+3A+4I2=OA^{2} + 3A + 4I_{2} = O and hence find A−1A^{-1}.

If A=[3−24−2]A = \begin{bmatrix} 3& -2\\ 4 & -2\end{bmatrix}, find the value of λ\lambda so that A2=λA−2IA^{2} = \lambda A - 2I. Hence, find A−1A^{-1}.

Show that the A=[10−2−2−12341]A = \begin{bmatrix}1 & 0 & -2\\ -2 & -1 & 2\\ 3 & 4 & 1\end{bmatrix} satisfied the equation A3−A2−3A−I=OA^{3} - A^{2} - 3A - I = O, and hence find A−1A^{-1}.

Solve the following equation:|x372x276x|=0\begin{vmatrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{vmatrix}=0

Show that |a2+λabacadabb2+λbcbdacbcc2+λcdadbdcdd2+λ|\begin{vmatrix} a^2+\lambda & ab & ac & ad \\ ab & b^2+\lambda & bc & bd \\ ac & bc & c^2+\lambda & cd \\ ad & bd & cd & d^2+\lambda \end{vmatrix}is divisible by λ3\lambda^{3} and find the other factor.

Absolute value of the sum of roots of the equation|x+22x+33x+42x+33x+44x+53x+55x+810x+17|=0\begin{vmatrix} x+2 & 2x+3 & 3x+4 \\ 2x+3 & 3x+4 & 4x+5 \\ 3x+5 & 5x+8 & 10x+17 \end{vmatrix}=0 is

If x=−4 x=-4 is a root of Δ=|x231x132x|=0 \Delta = \left| \begin{matrix} x & 2 & 3 \\ 1 & x & 1 \\ 3 & 2 & x \end{matrix} \right| = 0 then find the other two roots.

If x=−9 x = - 9 is root of |x372x276x|=0 \left| \begin{matrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{matrix} \right| =0 then other two roots are _________

Solve the following simultaneous equation.3a+5b=26;a+5b=223a + 5b = 26; a + 5b = 22

Solve the following simultaneous equation.x+7y=10;3x−2y=7x + 7y = 10; 3x - 2y = 7

Solve the following simultaneous equation:5m−3n=19;m−6n=−75m - 3n = 19; m - 6n = -7

Solve the following simultaneous equation:5x+2y=−3;x+5y=45x + 2y = -3; x + 5y = 4

Solve the following simultaneous equations.7x−2yxy=5;8x+7yxy=15\frac{7x - 2y}{xy} = 5 ;\frac{8x + 7y}{xy} = 15

Solve the following simultaneous equation.49x−57y=172;57x−49y=25249x - 57y = 172; 57x - 49y = 252

Solve the following simultaneous equation.13x+y=103;2x+14y=114\dfrac{1}{3}x + y = \dfrac{10}{3}; 2x + \dfrac{1}{4}y = \dfrac{11}{4}

Solve the following equation by Cramer's method.3x−2y=52;13x+3y=−433x - 2y = \frac{5}{2}; \frac{1}{3}x + 3y = - \frac{4}{3}

Solve the following simultaneous equations148x+231y=527xy;231x+148y=610xy\frac{148}{x} + \frac{231}{y} = \frac{527}{xy}; \frac{231}{x} + \frac{148}{y} = \frac {610}{xy}

Solve the following equations by Cramer's method.4m - 2n = -4; 4m + 3n = 16

Solve the following simultaneous equations.12(3x+4y)+15(2x−3y)=14;5(3x+4y)−2(2x−3y)=−32\frac{1}{2(3x + 4y)} + \frac{1}{5(2x - 3y)} = \frac{1}{4}; \frac{5}{(3x + 4y)} - \frac{2}{(2x - 3y)} = -\frac{3}{2}

If $D_x\, =\, -18$ and $D =\, 3$ are the values of the determinants for certain simultaneous equations in $x$ and $y$ , Find $-x$ .

If the value of determinants $\begin{vmatrix} x& -5\\ 3 & 4\end{vmatrix}$ is $31$ , then find the value of $x$

Find the value of the determinant: $\begin{vmatrix} 4& -2\\ 3 & 1\end{vmatrix}$

If $D=7,{D}_{x}=35$ and ${D}_{y}=42$ are the values of the determinants for certain simultaneous equations in $x$ and $y$ , then find the values of $x$ and $y$ .

Find the value of the following determinant
$\begin{vmatrix}7 & 2\\ 5 &4 \end{vmatrix}$

solve the following simultaneous equations by using Cramer's rule:
$x+y=7,x-y=5$

If ${ D }_{ x }=18,{ D }_{ y }=15$ and $D=3$ are the values of the determinants for certain simultaneous equations in $x$ and $y$ , then find the values of $x$ and $y$ .

Find the value of the following determinant $\begin{vmatrix} 4 & 3\\2 & 7\end{vmatrix}$ .

Find the value of following determinant
$\begin{vmatrix} 5& 2\\ 7 & 4 \end{vmatrix}$

If ${D}_{x}=30,{D}_{y}=42$ and $D=6$ are the values of the determinants for certain simultaneous equations in $x$ and $y$ , then find the values of $x$ and $y$ .

Solve the following simultaneous equations by using Cramer's rule:
$x - 2y = -18$
$2x - y = 9$

Find $D_{x}$ for the following simultaneous equations: $3x + 4y = 8, x - 2y = 5$

If ${ D }_{ x }=24,{ D }_{ y }=32$ are the values of the determinants for certain simultaneous equations in $x$ and $y$ , then find the values of $x$ and $y$

Solve the following equations by Cramer's rule:
$3x+y+z=2,2x-4y+3z=-1,4x+y-3z=-11\quad$

Find the ratios of $x : y : z$ from the equations
$7x = 4y + 8z, 3z = 12x + 11y$ .

If $A=\left[ \begin{matrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{matrix} \right]$ verify that $A(adj\ A)=\left| A \right| I$ .

If
$x + y + z = 1$
$3x + 5y + 6z = 4$
$9x + 2y - 36z = 17$
Then find $y$

Solve the equation $\dfrac{x}{a}\, + \,\dfrac{y}{b}\, = \,2\,,\,ax\, - \,by\, = \,{a^2} - \,{b^2}$

Solve the system of linear equation $3x-2y=13$ and $4x+y=21$

Solve the following simu1taneous equations.
(1) $5x-3y=8;$ $3x+y=2$

Solve: (i) $4x+5y-62=0$
(ii) $3x+2y-36=0$

Solve.
$3x-2y=1$ and $2x+y=3$ .

Solve the given pair of linear equations:
$(a-b)x+(a+b)y={a}^{2}-2ab-{b}^{2}$
$(a+b)(x+y)={a}^{2}+{b}^{2}$

Find co-factors of elements of the matrices $\begin {bmatrix} -1 & 2 \\ -3 & 4\end {bmatrix}$

Solve the following equation by cramer's method:
$7x+3y=15 , 12y-5x=39$

Find the value of: $\begin{vmatrix} -3 & -5 \\ -2 & -1 \end{vmatrix}$

Find ${D}_{x}$ and $D$ for the equation $2x+3y=4;7x-5y=2$

Solve by Cramer's rule
$x+y+z=6$
$x-y+z=2$
$3x+2y-4z=-5$ . The value of x is

If $D_y\, =\, -15$ and $D\, =\, -5$ are the values of the determinants for certain simultaneous equations in $x$ and $y$ , find $y$ .

If the polynomial $\displaystyle f\left ( x \right )=2x^{3}+mx^{2}+nx-14$ has $\displaystyle \left ( x-1 \right )$ and $\displaystyle \left ( x+2 \right )$ as its factors find the value of $\displaystyle m\times n$

For what value of $k$ will the following pair of linear equations have no solution?
$2x + 3y = 1$ and $(3k - 1)x +(1 - 2k) y = 2k + 3$

If show that ${ A }^{ 2 }-5A+7I=O$ . Hence find ${ A }^{ -1 }$

If verify that ${ A }^{ 3 }-{ 6A }^{ 2 }+9A-4I=0$ . Hence find ${ A }^{ -1 }$

For the matrix show that ${ A }^{ 3 }-{ 6A }^{ 2 }+5A+11I=0$ . Hence find ${ A }^{ -1 }$

Given, $-3x + 4y = 20$ and $6x +3y =15$
If $(x, y)$ is the solution to the system of equations above, what is the value of $x$ ?

Find ' $m+n$ ' for the following simultaneous equations:
$3m+4n=7$
and $4m+3n=14$

Solve the following simultaneous equations by using Cramer's Rule
$3x + y = 1$ ,
$2x - 11 y = 3$

Solve the following simultaneous equations using Cramer's rule.
$3x-2y=3$ ;
$2x+y=16$ .

Solve $x + y + z = 11\\ 2x - 6y - z = 0\\ 3x + 4y + 2z = 0$ .

Solve the equations
$x + 2y + 3z = 14$
$2x - y + 5z = 15$
$3x - 2y - 4z = -13$

Find x and y using Cramer's Rule if : $\dfrac{1}{2^x}+\dfrac{2}{3^y}=10,\dfrac{3}{2^x}-\dfrac{5}{3^y}=-3$ ?

Find x and y using Cramer's Rule,if $\dfrac{1}{x}-\dfrac{2}{y}=6,\dfrac{3}{x}+\dfrac{1}{y}=8$ ?

If $A = \left[ \begin{array} { c c c } { 5 } & { 8 } & { 1 } \\ { 0 } & { 2 } & { 1 } \\ { 4 } & { 3 } & { - 1 } \end{array} \right] , B = \left[ \begin{array} { c } { 2 } \\ { - 1 } \\ { 3 } \end{array} \right]$ and $A X = B$ then find $X$

Find $A^2$ where A is given $A= \begin {bmatrix}1 & 2 & 3 \\ 1 & 2 & 3\\ -1 & 2 & -3 \end{bmatrix}$

Solve $\left| {\begin{array}{*{20}{c}}1 & 1 & 1\\a & b & c\\{{a^3}} & {{b^3}} & {{c^3}}\end{array}} \right| = \left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)\left( {a + b + c} \right)$

For what value of k , the following pair of linear equations has infinite number of solutions :
$5x + 2y = k ; 10x + 4y = 3$

Solve the following
$\frac{x}{3}+\frac{y}{4}=4, \frac{x}{2}-\frac{y}{4}=1$

Solve the following simultaneous equation using cramer's rule
$4x+3y-4=0: 6x=1$

Solve the following simultaneous equation using cramer's rule
$6x-4y=-12: 8x-3y=-2$

Solve the following simultaneous equation using cramer's rule
$4m+6n=54 : 3m +2n =28$

Solve the simultaneous equations.
$\dfrac { x } { 3 } + \dfrac { y } { 4 } = 4 ; \quad \dfrac { x } { 2 } - \dfrac { y } { 4 } = 1$

Solve the following simultaneous equations using Cramer's rule :
$3x+y=7; \ 2x-11y=3.$

Solve the following simultaneous equation using cramer's rule.
$3x-4y=10:4x+3y=5$

Solve the following simultaneous equation using cramer's rule
$x+2y=-1: 2x-3y=12$

Solve the following simultaneous equation using cramer's rule
2x + 3y = 2: x- $\frac { y }{ 2 } =\frac{1}{2}$

If $A = \begin{bmatrix} 3& 2\\ 2 & 1\end{bmatrix}$ , verify that $A^{2} - 4A - I = O$ , and hence find $A^{-1}$ .

Show that the matrix $A =\begin{bmatrix} -8& 5\\ 2 & 4\end{bmatrix}$ satisifies the equation $A^{2} + 4A - 42I = 0$ and hence find $A^{-1}$ .

If $A = \begin{bmatrix} -1& -1\\ 2 & -2\end{bmatrix}$ , show that $A^{2} + 3A + 4I_{2} = O$ and hence find $A^{-1}$ .

If $A = \begin{bmatrix} 3& -2\\ 4 & -2\end{bmatrix}$ , find the value of $\lambda$ so that $A^{2} = \lambda A - 2I$ . Hence, find $A^{-1}$ .

Show that the $A = \begin{bmatrix}1 & 0 & -2\\ -2 & -1 & 2\\ 3 & 4 & 1\end{bmatrix}$ satisfied the equation $A^{3} - A^{2} - 3A - I = O$ , and hence find $A^{-1}$ .

Solve the following equation:
$\begin{vmatrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{vmatrix}=0$

Show that $\begin{vmatrix} a^2+\lambda & ab & ac & ad \\ ab & b^2+\lambda & bc & bd \\ ac & bc & c^2+\lambda & cd \\ ad & bd & cd & d^2+\lambda \end{vmatrix}$
is divisible by $\lambda^{3}$ and find the other factor.

Absolute value of the sum of roots of the equation
$\begin{vmatrix} x+2 & 2x+3 & 3x+4 \\ 2x+3 & 3x+4 & 4x+5 \\ 3x+5 & 5x+8 & 10x+17 \end{vmatrix}=0$ is

If $x=-4$ is a root of $\Delta = \left| \begin{matrix} x & 2 & 3 \\ 1 & x & 1 \\ 3 & 2 & x \end{matrix} \right| = 0$ then find the other two roots.

If $x = - 9$ is root of $\left| \begin{matrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{matrix} \right| =0$ then other two roots are _________

Solve the following simultaneous equation.
$3a + 5b = 26; a + 5b = 22$

Solve the following simultaneous equation.
$x + 7y = 10; 3x - 2y = 7$

Solve the following simultaneous equation:
$5m - 3n = 19; m - 6n = -7$

Solve the following simultaneous equation:
$5x + 2y = -3; x + 5y = 4$

Solve the following simultaneous equations.
$\frac{7x - 2y}{xy} = 5 ;\frac{8x + 7y}{xy} = 15$

Solve the following simultaneous equation.
$49x - 57y = 172; 57x - 49y = 252$

Solve the following simultaneous equation.
$\dfrac{1}{3}x + y = \dfrac{10}{3}; 2x + \dfrac{1}{4}y = \dfrac{11}{4}$

Solve the following equation by Cramer's method.
$3x - 2y = \frac{5}{2}; \frac{1}{3}x + 3y = - \frac{4}{3}$

Solve the following simultaneous equations
$\frac{148}{x} + \frac{231}{y} = \frac{527}{xy}; \frac{231}{x} + \frac{148}{y} = \frac {610}{xy}$

Solve the following equations by Cramer's method.
4m - 2n = -4; 4m + 3n = 16

Solve the following simultaneous equations.
$\frac{1}{2(3x + 4y)} + \frac{1}{5(2x - 3y)} = \frac{1}{4}; \frac{5}{(3x + 4y)} - \frac{2}{(2x - 3y)} = -\frac{3}{2}$

Solve the following equations by Cramer's methd.
$6x - 3y = -10; 3x + 5y - 8 = 0$

Solve the following
simultaneous equation using Cramer's rule.
$6x - 4y = -12$
$8x - 3y = -2$

Solve the following simultaneous equations.
2x + y = - 2; 3x - y =7

Solve the following sets of simultaneous equations.
2y -x =0; 10x + 15y = 105