Determinants - Class 12 Commerce Applied Mathematics - Extra Questions

Given: $$D_x = -18$$ and $$D = 3$$
Thus we know $$x = \dfrac{D_x}{D}$$
$$\Rightarrow x =$$ $$\dfrac{-18}{3}$$ $$= -6$$

We have, $$\begin{vmatrix} x& -5\\ 3 & 4\end{vmatrix} = 31$$
On solving, $$4x - (-5) (3) = 31$$
$$\Rightarrow 4x + 15 = 31$$
$$\Rightarrow 4x = 31 - 15$$
$$\Rightarrow x = \dfrac {16}{4} = 4$$
$$\therefore x = 4$$

Given, $$\begin{vmatrix} 4& -2\\ 3 & 1\end{vmatrix}$$
$$= (4) (1) - (-2)(3)$$
$$= 4 + 6 = 10$$

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.

$$\begin{vmatrix} 7& 2\\ 5 &4 \end{vmatrix} = (7 \times 4) - (2 \times 5)$$
$$= 28 - 10 = 18$$

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

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.

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

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

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.

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.

In cramer's rule of  $${ a }_{ 1 }x+{ b }_{ 1 }y={ c }_{ 1 },{ a }_{ 2 }x+{ b }_{ 2 }y={ c }_{ 2 }$$

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.

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

$$adj(A)$$ =Transfer co factor matrix

$$ax-by=a^2-b^2$$  

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

Consider the equation $$3x+y=2$$

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

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

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

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

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 

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 

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

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

Hence by cramer rule,
$$x=\cfrac { { \Delta  }_{ x } }{ \Delta  } =1$$

Given: $$D_y = -15$$ and $$D = -5$$
Thus, we know $$y = \frac{D_y}{D}$$
$$y =$$ $$\dfrac{-15}{-5}$$ $$= 3$$

$${ f }_{ 1 }(x)$$ are differentiable i.e. $${ { D }^{ 3 }f }_{ 1 }(x)$$ =0

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  $$\displaystyle m\times n=27$$

Given equations are,

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

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

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

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
$$-3x + 4(5) = 20$$
$$\Rightarrow  -3x + 20 = 20$$
$$\Rightarrow  -3x = 0$$
$$\Rightarrow  x = 0$$

$$3m+4n=7..........(1)$$
$$4m+3n=14........(2)$$
Adding equations $$(1)$$ and $$(2)$$
$$3m+4n=7$$
$$4m+3n=14$$
-------------------
$$7m+7n=21$$
$$\therefore$$ $$7(m+n)=21$$
$$\therefore$$ $$m+n=\cfrac{21}{7}$$
$$\therefore$$ $$m+n=3$$

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

Writing the system of equations in matrix form $$AX = B$$  

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

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

Given the two equations, $$\dfrac { 1 }{ x } -\dfrac { 2 }{ y } =6\quad \quad \longrightarrow (1)$$

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

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

The given equations are :

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

$$x=2$$ but $$y=\dfrac{15-40}{25}=\dfrac{-25}{25}=-1$$ .

$$3a + 5b = 26$$               -------(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 $$b = 4$$ .

$$x + 7y = 10;$$        ------------(I)
$$3x - 2y = 7$$              -----------(II)
multiplying (I) with 3
$$3x + 21y = 30;$$             ---------(III)
$$3x - 2y = 7$$        ---------------(IV)
Subtracting (IV) from (III) we get

$$5m - 3n = 19$$    ...........(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, $$(m, n) = (5, 2)$$ .

$$5x + 2y = -3$$    ........(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 $$(x,y) = (-1,1)$$

$$\dfrac{7x - 2y}{xy} = 5 ;\dfrac{8x + 7y}{xy} = 15$$
$$\Rightarrow \dfrac{7}{y} - \dfrac{2}{x} = 5$$ 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)
$$7(1) - 2u = 5$$
$$\Rightarrow$$ $$u = 1$$
$$\dfrac{1}{x} = u = 1$$
$$\Rightarrow x = 1$$
$$(x,y) = (1,1)$$

$$49x - 57y = 172$$ ...........(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, $$(x,y) = (7,3)$$

$$\dfrac{1}{3}x + y = \dfrac{10}{3}.$$ .........(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, $$(x,y) = (11,3)$$

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

$$\dfrac{148}{x} + \dfrac{231}{y} = \dfrac{527}{xy}; \dfrac{231}{x} + \dfrac{148}{y} = \dfrac {610}{xy}$$
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)
$$2 - x = 1$$
$$\Rightarrow$$ $$x = 1$$
$$(x, y) = (1,2)$$

$$4m - 2n = -4; 4m + 3n = 16$$  

$$\dfrac{1}{2(3x + 4y)} + \dfrac{1}{5(2x - 3y)} = \dfrac{1}{4}; \dfrac{5}{(3x + 4y)} - \dfrac{2}{(2x - 3y)} = -\dfrac{3}{2}$$
$$\dfrac{1}{3x + 4y} = u, \dfrac{1}{2x - 3y} = v$$
$$\dfrac{1}{2}u + \dfrac{1}{5}v = \dfrac{1}{4}$$
$$\Rightarrow$$ $$10u + 4v = 5.$$ ....(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)
$$6x - 9 = 3$$
$$\Rightarrow 6x = 12$$
$$\Rightarrow x = 2$$
$$(x,y) = (2,1)$$

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

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

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

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

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

$$x = \dfrac {{D}_{x}}{D} = \dfrac {28}{14} = 2$$

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

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

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 

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). 

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 

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 

$${R}_{1}\Rightarrow {R}_{1}+{R}_{2}+{R}_{3}$$
$$\Delta =\begin{vmatrix} 2\left( a+b+c \right)  & 2\left( a+b+c \right)  & 2\left( a+b+c \right)  \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix}$$
$$=2\left( a+b+c \right) \begin{vmatrix} 1 & 1 & 1 \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix}$$
$${c}_{2}\Rightarrow {c}_{2}-{c}_{1}$$ ; $${c}_{3}\Rightarrow {c}_{3}-{c}_{1}$$
$$=2\left( a+b+c \right) \begin{vmatrix} 1 & 0 & 0 \\ c+a & b-c & b-a \\ a+b & c-a & c-b \end{vmatrix}=2\left( a+b+c \right) \left[ (b-c)(c-b)-(b-a)(c-a) \right]$$
$$=2\left( a+b+c \right) [bc-{b}^{2}-{c}^{2}_bc-bc+ba+ac-{a}^{2}]$$
$$=2\left( a+b+c \right) [ab+bc+ca-{a}^{2}-{b}^{2}-{c}^{2}]$$
$$=-2\times \cfrac{1}{2}(a+b+c)[{(a-b)}^{2}+{(b-c)}^{2}+{(c-a)}^{2}$$
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 $$a=b=c$$

Given equations

Given,

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 $$x=3, y=1$$

Given,

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

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

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

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

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

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

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

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

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

                $$=13$$

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

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

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

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

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

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