If $$A = \begin{bmatrix} 1& 2\\ 3 & 4\end{bmatrix}$$, then $$A^{-1} =$$

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

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

MCQ Questions for Class 12 Commerce Maths Determinants Quiz 6

Value of p for which the points (-5, 1), (1, p) and (4, -2) are collinear is

0%
0
0%
2
0%
-1
0%
None of these

Explanation

Since the given points are collinear, they do not form a

triangle, which means area of the triangle is Zero.

Area

of a triangle with vertices

$({ x }_{ 1 },{ y }_{ 1 })$ ;

$({ x }_{ 2 },{ y }_{ 2 })$ and

$({ x }_{ 3 },{ y }_{ 3 })$ is

$\left| \dfrac { { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right|$

Hence, substituting the points

$({ x }_{ 1 },{ y }_{ 1 }) = (-5,1)$ ;

$({ x }_{ 2 },{ y }_{ 2 }) = (1,p)$ and

$({ x }_{ 3 },{ y }_{ 3 }) = (4,-2)$ in

the area formula, we get

$\left| \dfrac { -5(p+2) + 1(-2-1) + 4(1-p) } {2} \right| = 0$

$=> -9p-9 = 0$

$=>p = -1$

If the points (-2, -5), (2, -2) and (8, a) are collinear then value of a will be:

0%
$\displaystyle \frac { 1 }{ 2 }$
0%
$\displaystyle \frac { 3 }{ 2 }$
0%
$\displaystyle \frac { -5 }{ 2 }$
0%
$\displaystyle \frac { 5 }{ 2 }$

Explanation

Given points

$A(-2,-5),B(2,-2),C(8,a)$ are collinear

Hence area of triangle formed by A,B,C is zero

$Area =\dfrac{1}{2}\begin{vmatrix}x_{1}&y_{1}&1\\x_{2}&y_{2}&1\\x_{3}&y_{3}&1\end{vmatrix}$

$0=\dfrac{1}{2}\begin{vmatrix}-2&-5&1\\2&-2&1\\8&a&1\end{vmatrix}$

$0=-2(-2-a)+5(2-8)+1(2a+16)$

$4+2a-30+2a+16=0$

$4a=10\Rightarrow a=\dfrac{5}{2}$

The value of

$\displaystyle \begin{vmatrix}a - b - c & 2a & 2a\\ 2b & b - c- a & 2b\\ 2c & 2c & c- a -b\end{vmatrix}$ will be

0%
$(a+b+c)^2$
0%
$(a+b+c)^3$
0%
$(a-b-c)^2$
0%
$(a + b -c)^2$

Explanation

$\displaystyle \begin{vmatrix}a - b - c & 2a & 2a\\ 2b & b - c- a & 2b\\ 2c & 2c & c- a -b\end{vmatrix}$

$R_1 \rightarrow R_1 + (R_2 + R_3)$

$\begin{vmatrix}a+b+c & a+b+c & a+b+c\\2b &b - c-a & 2b\\2c & 2c & c-a-b\end{vmatrix}$

Taking common term (a + b + c) from the first row

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

Subtracting first column from second and third column

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

On expansion in terms of

$R_1 = (a + b + c) \begin{vmatrix}-b-c-a & 0 \\ 0 & -c-a-b\end{vmatrix}$

$= (a + b+c)\{(-b-c-a)(-c-a -b)-0\}$

$= (a + b+c)^3$

The points (-a, -b), (0, 0), (a, b) and (

$\displaystyle a^{2}$ , ab) are

0%
Collinear
0%
Vertices of a parallelogram
0%
Vertices of a rectangle
0%
None of these

Explanation

Let

$A (-a, -b), B (0,0), C(a,b)$ and

$D({a}^{2},ab)$

Distance between two points

$\left( { x }_{ 1 },{ y }_{ 1} \right)$ and

$\left( { x }_{ 2 },{ y }_{ 2 } \right)$ can be calculated using the formula

$\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2}+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } }$

Distance between the points A

$(-a,-b)$ and B

$(0,0) = \sqrt { \left( 0 + a \right) ^{ 2 }+\left( 0 + b \right) ^{ 2 } } = \sqrt { {a}^{2} + {b}^{2} }$

Distance between the points B

$(0,0)$ and C

$(a,b) = \sqrt { \left( a-0 \right) ^{ 2 }+\left( b-0\right) ^{ 2 } } = \sqrt { {a}^{2} + {b}^{2} }$

Distance between the points C $(a,b)$ and D $({a}^{2}, ab) = \sqrt { \left( {a}^{2} - a \right) ^{ 2 }+\left(ab - b \right) ^{ 2 } } = (a-1)\sqrt { {a}^{2} + {b}^{2} }$

Distance between the points A $(-a,-b)$ and D $({a}^{2}, ab) = \sqrt { \left( {a}^{2} + a \right) ^{ 2 }+\left(ab + b\right) ^{ 2 } } = (a +1)\sqrt { {a}^{2} + {b}^{2} }$

Now.

$AB+BC+CD = \sqrt { {a}^{2} + {b}^{2} } + \sqrt { {a}^{2} + {b}^{2} }+ (a-1)\sqrt { {a}^{2} + {b}^{2} } = (a +1)\sqrt { {a}^{2} + {b}^{2} } = AD$

Hence, the points are collinear.

The points (0, 8/3), (1, 3) and (82, 30) are the vertices of :

0%
obtuse angled triangle
0%
right angled triangle
0%
isosceles triangle
0%
None of these

Explanation

Let,

$P =(0, \dfrac {8}{3}),Q=(1,3),R=(82,30)$

Now, m1

$=$ slope of PQ

$= \dfrac{3 \dfrac {8}{3}}{10} = \dfrac {1}{3}$

m2

$=$ slope of QR

$= \dfrac {303}{821} = \dfrac {1}{3}$

Since,

$m1 = m2$ , the three points P, Q, R are collinear

Find the value of K if A(8, 1), B(K, -4), C(2, -5) are collinear :

0%
$2$
0%
$3$
0%
$4$
0%
$5$

Explanation

Since the given points are collinear, they do not form a

triangle, which means area of the triangle is Zero.

Area of a triangle with vertices $({ x }_{ 1 },{ y }_{ 1 })$ ; $({ x }_{ 2 },{ y }_{ 2 })$ and $({ x }_{ 3 },{ y }_{ 3 })$ is $\left| \frac { { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right|$

Hence, substituting the points

$({ x }_{ 1 },{ y }_{ 1 }) = (8,1)$ ;

$({ x }_{ 2 },{ y }_{ 2 }) = (k,-4)$ and

$({ x }_{ 3 },{ y }_{ 3 }) = (2,-5)$
the area formula, we get

$\left| \frac { 8(-4+5) + k(-5-1) + 2(1+4) }{ 2 } \right| = 0$

$=> 8 - 6k + 10 = 0$

$=> 6k = 18$

$=> k = 3$

If the points

$(k, 2k),\ ( 3k, 3k)$ and

$(3, 1)$ are collinear then the value of

$k$ is

0%
$\displaystyle \frac{7}{9}$
0%
$\displaystyle \frac{2}{3}$
0%
$\displaystyle \frac{-2}{3}$
0%
$\displaystyle \frac{-1}{3}$

Explanation

Since the given points are collinear, they do not form a triangle, which means area of the triangle is Zero
The area of the triangle with vertices

$({ x }_{ 1 },{ y }_{ 1 })$ ;

$({ x }_{ 2 },{ y}_{ 2 })$ and

$({ x }_{ 3 },{ y }_{ 3 })$ is

$\left| \frac { {x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right|$
Hence, substituting the points

$({ x }_{ 1 },{ y }_{ 1 }) = (k,2k)$ ;

$({ x}_{ 2 },{ y }_{ 2 }) = (3k,3k)$ and

$({ x }_{ 3 },{ y }_{ 3 }) = (3,1)$ in the area formula, we get

$\left| \frac { k(3k-1)+3k(1-2k)+3(2k-3k) }{ 2 } \right| =0$

$=> 3{k}^{2} -k + 3k-6{k}^{2} + 6k -9k = 0$

$=>-3{k}^{2} -k = 0$

$-k(3k+1) =0$

$k = 0$ or

$k = -\frac{1}{3}$

The points (-a, -b), (0, 0), (a, b) and (a

$^2$ , ab) are

0%
collinear
0%
vertices of a rectangle
0%
vertices of a parallelogram
0%
None of these

Explanation

Let the points be

$A (-a, -b), O (0, 0), B (a, b)$ and

$C ({a}^{2}, ab)$

Slope

of the line passing through points

$\left( { x }_{ 1 },{ y }_{ 1 } \right)$

and

$\left( { x }_{ 2 },{ y }_{ 2 } \right)$

$=$

$\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-x_{ 1 } }$

So, slope of the line joining OA $= \dfrac { -b-0 }{ -a-0 } = \dfrac {b}{a}$

Slope of the line joining OB

$= \dfrac { b-0 }{ a-0} = \dfrac {b}{a}$

Slope of the line joining OC $= \dfrac { ab-0 }{{a}^{2}-0 } = \dfrac {b}{a}$

Since slopes of the lines, OA, OB, OC are equal, they are all collinear.

Find the value of P for which the points A(-1, 3), B(2, P) and C(5, -1) are collinear :

0%
3
0%
1
0%
2
0%
4

Explanation

Since the given points are collinear, they do not form a

triangle, which means area of the triangle is Zero.

Area of a triangle with vertices $({ x }_{ 1 },{ y }_{ 1 })$ ; $({ x }_{ 2 },{ y }_{ 2 })$ and $({ x }_{ 3 },{ y }_{ 3 })$ is $\left| \dfrac { { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right|$

Hence, substituting the points

$({ x }_{ 1 },{ y }_{ 1 }) = (-1,3)$ ;

$({ x }_{ 2 },{ y }_{ 2 }) = (2,P)$ and

$({ x }_{ 3 },{ y }_{ 3 }) = (5,-1)$

in the area formula, we get

$\left| \dfrac { (-1)(P + 1) + 2(-1-3) + 5(3-P) }{ 2 } \right| = 0$

$=> -P - 1 - 8 + 15 - 5P = 0$

$=> 6P = 6$

$=> P = 1$

Find the value of K if (2, 3), (4, K) and (6, -3) are collinear :

0%
0
0%
1
0%
2
0%
3

Explanation

Since the given points are collinear, they do not form a

triangle, which means area of the triangle is Zero.

Hence, substituting the points

$({ x }_{ 1 },{ y }_{ 1 }) = (2,3)$ ;

$({ x }_{ 2 },{ y }_{ 2 }) = (4.k)$ and

$({ x }_{ 3 },{ y }_{ 3 }) = (6,-3)$

in the area formula, we get

$\left| \dfrac { 2 (k+3) + 4(-3-3) + 6(3-k) }{ 2 } \right| = 0$

$=> 2k + 6 -24 + 18 - 6k = 0$

$=> 4k = 0$

$=> k = 0$

$\displaystyle \begin{vmatrix} 2 & -4 \\ 9 & d-3 \end{vmatrix}=4$ then

$d=$

0%
$10$
0%
$-11$
0%
$12$
0%
$-13$

Explanation

Given, determinant of the matrix

$\begin{vmatrix} 2 & -4 \\ 9 & d-3 \end{vmatrix} = 4$

$=> 2(d-3) - (9)(-4) = 4$

$=> 2d-6+36 = 4$

$=> 2d = -26$

$=> d = -13$

If the coordinates of the vertices of a triangle are (0,0) , (0,2) and(3,1) , then area of the triangle is

0%
3 sq.units
0%
-3 sq. units
0%
2 sq. units
0%
1 sq.units

Explanation

Area of triangle

$=\frac{1}{2} \begin{vmatrix} 0 &0 &1 \\ 0&2 &1 \\3 &1 &1 \end{vmatrix}= \frac{1}{2}\times|-6|=3$

What is the value of y if

$(y, 3), (-5, 6)$ and

$(-8, 8)$ are collinear?

0%
$-1$
0%
$2$
0%
$\displaystyle \frac{1}{2}$
0%
$\displaystyle -\frac{1}{2}$

Explanation

Let A(y,3),B(-5,6) and C(-8,8) be the given points

Condition for collinear equation-

$x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0$

where

$x_1=y,x_2=3$

$x_2=-5,y_2=6$

$x_3=-8,y_3=8$

$\Rightarrow y(6-8)+-5(8-3)+-8(3-6)=0$

$\Rightarrow 6y-8y-40+15-24+48=0$

$\Rightarrow -2y-1=0$

$\Rightarrow -2y=1$

$\Rightarrow y=-\frac{1}{2}$

Which of the following points are collinear?

0%
(2a,0), (3a,0), (a,2a)
0%
(3a,0), (0,3b), (a,2b)
0%
(3a,b), (a,2b), (-a,b)
0%
(a,-6), (-a,3b), (-2a,-2b)

Explanation

Ans option B

Let A(3a,0),B(0,3b) and C(a,2b) be the given points

Condition for collinear equation-

$x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0$

where

$x_1=3a,y_1=0$

$x_2=0,y_2=3b$

$x_3=a,y_3=2b$

$\Rightarrow 3a(3b-2b)+0(2b-0)+a(0-3b)=0$

$\Rightarrow 3ab+0-3ab=0$

$\displaystyle \left[ \begin{matrix} \cos { \theta } \\ -\sin { \theta } \\ 0 \end{matrix}\begin{matrix} \sin { \theta } \\ \cos { \theta } \\ 0 \end{matrix}\begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right]$ then the value of

$\displaystyle { a }_{ 11 }{ A }_{ 11 }+{ a }_{ 12 }{ A }_{ 12 }+{ a }_{ 13 }{ A }_{ 13 }=$ where

$\displaystyle { A }_{ 11 },{ A }_{ 12 }{ A }_{ 13 }$ are cofactors of

$\displaystyle { a }_{ 11 },{ a }_{ 12 },{ a }_{ 13 }$ respectively

0%
$-1$
0%
$1$
0%
$0$
0%
$\displaystyle \frac { 1 }{ 2 }$

Explanation

$A_{11} = \begin{vmatrix} \cos{\theta} & 0 \\ 0 & 1 \end{vmatrix} = \cos{\theta}$

$A_{12} = - \begin{vmatrix} -\sin{\theta} & 0 \\ 0 & 1 \end{vmatrix} = \sin{\theta}$

$A_{13} = \begin{vmatrix} -\sin{\theta} & \cos{\theta} \\ 0 & 0 \end{vmatrix} = 0$

Now,

$a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13} = \cos^2{\theta} + \sin^2{\theta} = 1$

$z=\begin{vmatrix} 1 & 1+2i & -5i \\ 1-2i & -3 & 5+3i \\ 5i & 5-3i & 7 \end{vmatrix}$ , then

$i=\left( \sqrt { -1 } \right)$

0%
$z$ is purely real
0%
$z$ is purely imaginary
0%
$z+\overline { z } =0$
0%
$\left( z-\overline { z } \right) i$ is purely imaginery

Explanation

Given

$z=\begin{vmatrix} 1 & 1+2i & -5i \\ 1-2i & -3 & 5+3i \\ 5i & 5-3i & 7 \end{vmatrix}$

$=1(-21-(25+9))-(1+2i)(7-14i-(25i-15))-5i[(-1-13i)+15i]$

$-55-(1+2i)(22-39i)-5i(-1+2i)$

$=-55-100-5i+5i+10$

$=-145$
Hence,

$z$ is purely real.

$\Delta = \begin{vmatrix} -a & 2b & 0 \\ 0 & -a & 2 b \\ 2b & 0 &-a \end{vmatrix}=0$ , then

0%
$\dfrac{1}{b}$ is a cube root of unity
0%
$a$ is one of the cube roots of unity
0%
$b$ is one of the cube roots of $8$
0%
$\dfrac{a}{b}$ is a cube root of $8$

Explanation

$\begin{vmatrix} -a & 2b & 0 \\ 0 & -a & 2 b \\ 2b & 0 &-a \end{vmatrix}=0$

$\Rightarrow -a(a^2-0)-2b(0-4b^2)=0$ , Expand along 1st row

$\Rightarrow -a^2+8b^3=0\Rightarrow \left(\dfrac{a}{b}\right)^3=8$

$\Rightarrow \dfrac{a}{b}$ is a cube root of

$8$

Find the correct option regarding given points

$(1, 2), (2, 4)$ and

$(3, 6)$

0%
Non-colllinear
0%
Exists on parallel lines
0%
Collinear
0%
Exists on perpendicular lines

Explanation

Area of the triangle with vertices

$(x_1,y_1)$ ,

$(x_2,y_2)$ and

$(x_3,y_3)$ is:

$A=\dfrac { 1 }{ 2 } \left| x_{ 1 }(y_{ 2 }-y_{ 3 })+x_{ 2 }(y_{ 3 }-y_{ 2 })+x_{ 3 }(y_{ 1 }-y_{ 2 }) \right|$

Here the points are

$(1,2)$ ,

$(2,4)$ and

$(3,6)$ , therefore, the area of the triangle is:

$A=\dfrac { 1 }{ 2 } \left| x_{ 1 }(y_{ 2 }-y_{ 3 })+x_{ 2 }(y_{ 3 }-y_{ 1 })+x_{ 3 }(y_{ 1 }-y_{ 2 }) \right| \\ A=\dfrac { 1 }{ 2 } \left| 1(4-6)+2(6-2)+3(2-4) \right| \\ \Rightarrow A=\dfrac { 1 }{ 2 } \left| 1(-2)+2(4)+3(-2) \right| \\ \Rightarrow A=\dfrac { 1 }{ 2 } \left| -2+8-6 \right| \\ \Rightarrow A=\dfrac { 1 }{ 2 } \left| -8+8 \right| \\ \Rightarrow A=0$

Since the area of the triangle is zero,

Hence, the points are collinear.

Find the value of

$k$ for which the points

$(2, 3), (3, k)$ and

$(3, 7)$ are collinear.

0%
5
0%
6
0%
8
0%
7

Explanation

If three points are collinear, then the area of the triangle is zero.

Area of the triangle with vertices

$(x_1,y_1)$ ,

$(x_2,y_2)$ and

$(x_3,y_3)$ is:

$A=\dfrac { 1 }{ 2 } \left| x_{ 1 }(y_{ 2 }-y_{ 3 })+x_{ 2 }(y_{ 3 }-y_{ 1 })+x_{ 3 }(y_{ 1 }-y_{ 2 }) \right|$

Here the points are

$(2,3)$ ,

$(3,k)$ and

$(3,7)$ and it is given that these points are collinear therefore the area is zero that is:

$0=\dfrac { 1 }{ 2 } \left| 2(k-7)+3(7-3)+3(3-k) \right| \\ \Rightarrow 0=\left| 2k-14+3(4)+9-3k \right| \\ \Rightarrow 0=\left| -k-5+12 \right| \\ \Rightarrow 0=\left| -k+7 \right| \\ \Rightarrow k-7=0\\ \Rightarrow k=7$

Hence,

$k=7$

If the points

$(a,\,b),\,(3,\,-5)$ and

$(-5,\,-2)$ are collinear. Then find the value of

$3a+8b$

0%
$20$
0%
$-31$
0%
$10$
0%
None

Explanation

If three points are collinear then the area of the triangle is zero.

Area of the triangle passing through the points

$(x_1,y_1)$ ,

$(x_2,y_2)$ and

$(x_3,y_3)$ is given by:

$A=\dfrac { 1 }{ 2 } \left| x_{ 1 }(y_{ 2 }-y_{ 3 })+x_{ 2 }(y_{ 3 }-y_{ 1 })+x_{ 3 }(y_{ 1 }-y_{ 2 }) \right|$

Here the points are

$(a,b)$ ,

$(3,-5)$ and

$(-5,-2)$ and it is given that these points are collinear therefore the area is zero that is:

$A=\dfrac { 1 }{ 2 } \left| a(-5-(-2))+3(-2-b)+(-5)(b-(-5)) \right| \\ \Rightarrow 0=\dfrac { 1 }{ 2 } \left| a(-5+2)+3(-2-b)-5(b+5) \right| \\ \Rightarrow 0=\dfrac { 1 }{ 2 } \left| -3a-6-3b-5b-25 \right| \\ \Rightarrow 0=\dfrac { 1 }{ 2 } \left| -3a-8b-31 \right| \\ \Rightarrow 0=-3a-8b-31\\ \Rightarrow 3a+8b=-31$

Hence,

$3a+8b=-31$ .

$P = \begin{bmatrix} 1& 2 & 1\\ 1 & 3 & 1\end{bmatrix}, Q = PP^{T}$ , then the value of the determinant of

$Q$ is

0%
$2$
0%
$-2$
0%
$1$
0%
$0$

Explanation

Given,

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

$\therefore Q = PP^{T} = \begin{bmatrix} 1& 2 & 1\\ 1 & 3 & 1\end{bmatrix} \begin{bmatrix} 1& 1\\ 2 & 3\\ 1 & 1\end{bmatrix}$

$= \begin{bmatrix} 1\times 1 + 2\times 2 + 1\times 1& 1\times 1 + 2\times 3 + 1\times 1\\ 1\times 1 + 3\times 2 + 1\times 1 & 1\times 1 + 3\times 3 + 1\times 1\end{bmatrix}$

$= \begin{bmatrix} 1 + 4 + 1& 1 + 6 + 1\\ 1 + 6 + 1 & 1 + 9 + 1\end{bmatrix} = \begin{bmatrix}6 & 8\\ 8 & 11\end{bmatrix}$

$\therefore$ The determinant of

$Q = \begin{bmatrix} 6& 8\\ 8 & 11\end{bmatrix}$

$= 66 - 64 = 2$

Find the correct option regarding the given points

$(2, -1), (0, 2)$ and

$(3, 2)$ .

0%
Collinear
0%
Non-collinear
0%
Exists on parallel lines
0%
Exists on perpendicular lines

Explanation

Area of the triangle with vertices

$(x_1,y_1)$ ,

$(x_2,y_2)$ and

$(x_3,y_3)$ is:

$A=\dfrac { 1 }{ 2 } \left| x_{ 1 }(y_{ 2 }-y_{ 3 })+x_{ 2 }(y_{ 3 }-y_{ 1 })+x_{ 3 }(y_{ 1 }-y_{ 2 }) \right|$

Here the points are

$(2,-1)$ ,

$(0,2)$ and

$(3,2)$ , therefore, the area of the triangle is:

$A=\dfrac { 1 }{ 2 } \left| 2(2-2)+0(2+1)+3(-1-2) \right| \\ \Rightarrow A=\dfrac { 1 }{ 2 } \left| 2(0)+0(3)+3(-3) \right| \\ \Rightarrow A=\dfrac { 1 }{ 2 } \left| -9 \right| \\ \Rightarrow A=\dfrac { 9 }{ 2 } \neq 0$

Since the area of the triangle is not zero,

Hence, the points are non-collinear.

Find the correct option regarding given points

$(2, 2), (1, 2)$ and

$(3, 1)$ .

0%
Collinear
0%
Non-collinear
0%
Exists on parallel lines
0%
Exists on perpendicular lines

Explanation

Area of the triangle with vertices

$(x_1,y_1)$ ,

$(x_2,y_2)$ and

$(x_3,y_3)$ is:

$A=\dfrac { 1 }{ 2 } \left| x_{ 1 }(y_{ 2 }-y_{ 3 })+x_{ 2 }(y_{ 3 }-y_{ 1 })+x_{ 3 }(y_{ 1 }-y_{ 2 }) \right|$

Here the points are

$(2,2)$ ,

$(1,2)$ and

$(3,1)$ , therefore the area of the triangle is:

$A=\dfrac { 1 }{ 2 } \left| 2(2-1)+1(1-2)+3(2-2) \right| \\ \Rightarrow A=\dfrac { 1 }{ 2 } \left| 2(1)+1(-1)+3(0) \right| \\ \Rightarrow A=\dfrac { 1 }{ 2 } \left| 2-1 \right| \\ \Rightarrow A=\dfrac { 1 }{ 2 } \neq 0$

Since the area of the triangle is not zero,

Hence, the points are non-collinear.

Consider the three collinear points

$(3, p), (4, 4)$ and

$(5, 6)$ . Find the value of

$p$ .

0%
1
0%
2
0%
3
0%
4

Explanation

If three points are collinear then the area of the triangle is zero.

Area of the triangle with vertices

$(x_1,y_1)$ ,

$(x_2,y_2)$ and

$(x_3,y_3)$ is:

$A=\dfrac { 1 }{ 2 } \left| x_{ 1 }(y_{ 2 }-y_{ 3 })+x_{ 2 }(y_{ 3 }-y_{ 1 })+x_{ 3 }(y_{ 1 }-y_{ 2 }) \right|$

Here the points are

$(3,p)$ ,

$(4,4)$ and

$(5,6)$ and it is given that these points are collinear, therefore the area is zero that is:

$0=\dfrac { 1 }{ 2 } \left| 3(4-6)+4(6-p)+5(p-4) \right| \\ \Rightarrow 0=\left| 3(-2)+24-4p+5p-20 \right| \\ \Rightarrow 0=\left| -6+p+4 \right| \\ \Rightarrow 0=\left| p-2 \right| \\ \Rightarrow p-2=0\\ \Rightarrow p=2$

Hence,

$p=2$

The graph of

$f(x)$ is shown above in the

$xy$ -plane. The points

$(0,3), (5b, b)$ and

$(10b, -b)$ are on the line described by

$f(x)$ . If

$b$ is a positive constant, find the coordinates of point

$C$ .

0%
$(5, 1)$
0%
$(10, -1)$
0%
$(15, -0.5)$
0%
$(20, -2)$

Explanation

In given figure the line goes to point

$A(0,3) ,B(5b,b)$ and

$C(10b,-b)$

Then slope of line goes at point

$A(0,3)$ and

$B(5b,b)=$

$\dfrac{b-3}{5b-0}=\dfrac{b-3}{5b}$

And slope of line goes at point

$B(5b,b)$ and

$C(10b,-b)=$

$\dfrac{-b-b}{10b-5b}=\dfrac{-2b}{5b}$

Then given line is a st reline then both slopes are equal

$\therefore \dfrac{b-3}{5b}=\frac{-2b}{5b }$

$\Rightarrow -2b=b-3$

$\Rightarrow -3b=-3$

$\Rightarrow b=1$

Put the value of

$b=1$ in point C

We get

$C=(10,-1)$

If C =

$2 \cos \theta$ , then the value of the determinant to

$\Delta = \left |\begin{matrix} c & 1 & 0 \\ 1 & c & 1 \\ 6 & 1 & c \end{matrix}\right |$ is

0%
$\dfrac{2\,\sin^22\theta}{\sin\,\theta}$
0%
$8\, \cos^3\theta-4\,cos\theta+6$
0%
$\dfrac{2\,\sin\,2\theta}{\sin\, \theta}$
0%
$8\,\cos^3\theta+4\,\cos\theta+6$

Explanation

$C=2\cos { \theta }$

$\triangle =\begin{vmatrix} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{vmatrix}$

$\triangle =C\left( { C }^{ 2 }-1 \right) -1\left( C-6 \right)$

$\triangle ={ C }^{ 3 }-2C+6$

$C=2\cos { \theta }$

$\Rightarrow \triangle ={ \left( 2\cos { \theta } \right) }^{ 3 }-\left( 2\cos { \theta } \right) 2+6$

$\Rightarrow \triangle =8\cos ^{ 2 }{ \theta } -4\cos { \theta } +6$

If A =

$\begin{bmatrix}-8 & 5\\ 2 & 4\end{bmatrix}$ satisfies the equation

$x^2\, +\, 4x\, -\, p\, =\, 0$ , then p =

0%
64
0%
42
0%
36
0%
24

Explanation

$A= \begin{bmatrix}-8&5\\2&4\end{bmatrix}$

$A^2=\begin{bmatrix}-8&5\\2&4\end{bmatrix}\begin{bmatrix}-8&5\\2&4\end{bmatrix}=\begin{bmatrix}64+10&-40+20\\-16+8&10+16\end{bmatrix}=\begin{bmatrix}74&-20\\-8&26\end{bmatrix}$

$4A=\begin{bmatrix}-32&20\\8&16\end{bmatrix}$

$P=x^2+4x$ and

$A$ satisfies this equation

$So, P=\begin{bmatrix}74&-20\\-8&26\end{bmatrix}+\begin{bmatrix}-32&20\\8&16\end{bmatrix}$

$=\begin{bmatrix}42&0\\0&42\end{bmatrix}$

$=42\begin{bmatrix}1&0\\0&1\end{bmatrix}$

$=42I$

$=42$

Evaluate

$\begin{vmatrix} \cos 15^o& \sin 15^o\\ \sin 75^o & \cos 75^o\end{vmatrix}$

0%
$1$
0%
$0$
0%
$2$
0%
$3$

Explanation

Using fundamental theorem of determinant,

$\begin{vmatrix} \cos 15^o& \sin 15^o\\ \sin 75^o & \cos 75^o\end{vmatrix}=\cos15^o\cos75^o-\sin15^0\sin75^o$

$=\cos(75^o+15^o)=\cos90^o=0$ , [using sum to product formula]

We have used the Fundamental theorem of determinant is

$\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc$

Let

$A$ be a

$3 \times 3$ matrix and

$B$ be its adjoint matrix. If

$|B| = 64$ , then

$|A| =$

0%
$\pm 2$
0%
$\pm 4$
0%
$\pm 8$
0%
$\pm 12$

Explanation

$|\text{adj}\,A|=|A|^{n-1}$ where

$n$ is the order of the square matirx.
Hence

$|B|=|\text{adj}\,A|=|A|^{3-1}$

$=|A|^{2}$
Therefore

$|A|=\sqrt{|B|}=\sqrt{64}=\pm8$ .

The points

$(-a, -b), (a, b), (0, 0)$ and

$(a^2, ab), a\ne 0, b\ne 0$ are

0%
Collinear
0%
Vertices of a parallelogram
0%
Vertices of rectangle
0%
Lie on a circle

Explanation

Let the points are

$A(-a,-b), B(a,b), C(0,0)$ and

$D(a^2, ab)$ , then

slope of line

$AB =\dfrac{b}{a}=$ slope of line

$BC =$ slope of line

$CD =$ slope of line

$DA$

Hence all the given

$4$ points are collinear.

Note: Slope of line joining the points

$(x_1,y_1)$ and

$(x_2, y_2)$ is given by

$\dfrac{y_2-y_1}{x_2-x_1}$

$A(x)=\begin{vmatrix} x+1 & 2x+1 & 3x+1 \\ 2x+1 & 3x+1 & x+1 \\ 3x+1 & x+1 & 2x+1 \end{vmatrix}$
then

$\displaystyle \int _{ 0 }^{ 1 }{ A(x) } dx=$

0%
$-15$
0%
$\cfrac { -15 }{ 2 }$
0%
$-30$
0%
$-5$

Explanation

$\\ A(x)=\begin{vmatrix} x+1 & 2x+1 & 3x+1 \\ 2x+1 & 3x+1 & x+1 \\ 3x+1 & x+1 & 2x+1 \end{vmatrix}\\ =(x+1)((3x+1)(2x+1)-(x+1)(x+1))-(2x+1)((2x+1)(2x+1)-(x+1)(3x+1)) +\\\,\,\,\,(3x+1)((x+1)(2x+1)-(3x+1)(3x+1))\\ A(x)=-18{ x }^{ 3 }-9{ x }^{ 2 }\\\displaystyle \int _{ 0 }^{ 1 }{ A(x) } dx=\cfrac { -18 }{ 4 } { x }^{ 4 }-3{ x }^{ 3 }=\cfrac { -18 }{ 4 } -3=\cfrac {- 30 }{ 4 } =\cfrac { -15 }{ 2 } \\$

If a, b and c are in A, P., then the value of

$\begin{vmatrix} x+2 & x+3 &x+a \\ x+4& x+5 &x+b \\ x+6 & x+7 & x+c \end{vmatrix}$ is.

0%
$x-(a+b+c)$
0%
$9x^2+a+b+c$
0%
0
0%
$a+b+c$

Explanation

$\begin{vmatrix}x+2&x+3&x+a\\x+4&x+5&x+b\\x+6&x+7&x+c \end{vmatrix}$

$=\begin{vmatrix}2x+8&2x+10&2x+(a+c)\\x+4&x+5&x+b\\x+6&x+7&x+c \end{vmatrix}$ , applying

$R_1\to R_1+R_3$

$=\begin{vmatrix}2(x+4)&2(x+5)&2(x+b)\\x+4&x+5&x+b\\x+6&x+7&x+c \end{vmatrix}$ , since a,b,c are in A.P.

$a+c=2b$

$= 2\begin{vmatrix}x+4&x+5&x+b\\x+4&x+5&x+b\\x+6&x+7&x+c \end{vmatrix}$ , take 2 common from first row

$=2(0)=0$ , since first and second row of the determinant are same

If the points

$(2,5),(4,6)$ and

$(a,a)$ are collinear, then the value of

$a$ is equal to

0%
$-8$
0%
$4$
0%
$-4$
0%
$8$

Explanation

Consider the given points.

$(2, 5), (4, 6)$ and

$(a, a)$

Since, these points are collinear means that the area of triangle must me zero.

So,

$\dfrac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|=0$

where

$(x_1,y_1),(x_2,y_2),(x_3,y_3)$ are the points

Therefore,

$2(6-a)+4(a-5)+a(5-6)=0$

$12-2a+4a-20+5a-6a=0$

$a-8=0$

$a=8$

Hence, this is the answer.

$\Delta_r = \begin{vmatrix}2r - 1 & ^mC_r & 1\\ m^2 - 1 & 2^m & m+1\\ sin^2(m^2) & sin^2(m) & sin^2(m+1)\end{vmatrix}$ , then the value of

$\displaystyle \sum_{r = 0}^m \Delta_r$ , is

0%
1
0%
0
0%
2
0%
None of these

Explanation

$\Delta_r = \begin{vmatrix}2r -1 & ^mC_r & 1\\m^2 - 1 & 2^m & m+1\\ sin^2(m^2) & sin^2(m) & sin^2(m + 1)\end{vmatrix}$

$\therefore \sum_{r = 0}^m \Delta_r = \begin{vmatrix}\sum_{r = 0}^m (2r - 1) & \sum_{r = 0}^m { }^{m} C_r& \sum_{r = 0}^m 1\\ m^2 - 1 & 2^m & m+1\\ sin^2 (m^2) & sin^2(m) & sin^2(m + 1)\end{vmatrix}$

$= \begin{vmatrix} m^2 -1& 2^m & m+1\\ m^2 - 1 & 2^m & m+1\\ sin^2 (m^2) & sin^2(m) & sin^2(m + 1)\end{vmatrix}$

$=0$ (

$\because$ two rows are identical)

Hence, option B is correct.

The centre of a circle is

$(-6,4)$ . If one end of the diameter of the circle is at

$(-12,8)$ , then the other end is at

0%
$(-18,12)$
0%
$(-9,6)$
0%
$(-3,2)$
0%
$(0,0)$

Explanation

The centre is the midpont of the line joining the two points of the diameter.

Using Midpoint-formula

$\left(\alpha,\beta\right)\equiv\left(\dfrac{{x}_{1}+{x}_{2}}{2},\dfrac{{y}_{1}+{y}_{2}}{2}\right)$

From above we have

$\alpha=-6,\beta=4$

Let

${x}_{1}=x,{y}_{1}=y$

Given

${x}_{2}=-12,{y}_{2}=8$

We have

$\alpha=\dfrac{{x}_{1}+{x}_{2}}{2}$ and

$\beta=\dfrac{{y}_{1}+{y}_{2}}{2}$

$x+\dfrac{\left(-12\right)}{2} = -6$

$\Rightarrow x-6=-6$

$\therefore x = 0$

$\Rightarrow y+\dfrac{8}{2} = 4$

$\Rightarrow y+4=4$

$\therefore y = 0$

Hence, the other point is at

$\left(0,0\right)$

$a, b$ and

$c$ are in

$AP$ , then determinant

$\begin{vmatrix}x + 2 & x + 3 & x + 2a\\ x + 3 & x + 4 & x + 2b\\ x + 4 & x + 5 & x + 2c\end{vmatrix}$ is

0%
$0$
0%
$1$
0%
$x$
0%
$2x$

Explanation

Let

$\triangle = \begin{vmatrix}x + 2 & x + 3 & x + 2a\\ x + 3 & x + 4 & x + 2b\\ x + 4 & x + 5 & x + 2c\end{vmatrix}$

$= \dfrac {1}{2} \begin{vmatrix}x + 2 & x + 3 & x + 2a\\ 0 & 0 & 2(2b - a - c)\\ x + 4 & x + 5 & x + 2c\end{vmatrix}$ ...... (using

$R_{2}\rightarrow 2R_{2} - R_{1} - R_{2})$

But

$a, b$ and

$c$ are in

$AP$ using

$2b = a + c$ , we get

$\triangle = \dfrac {1}{2} \begin{vmatrix}x + 2 & x + 3 & x + 2a\\ 0 & 0 & 0\\ x + 4 & x + 5 & x + 2c\end{vmatrix}$

$=0$ ........ (Since, all elements of

$R_{2}$ are zero.)

Hence,

$\begin{vmatrix}x + 2 & x + 3 & x + 2a\\ x + 3 & x + 4 & x + 2b\\ x + 4 & x + 5 & x + 2c\end{vmatrix}=0$

The value of the determinant

$\begin{vmatrix}1 & \cos (\alpha - \beta) & \cos \alpha\\ \cos(\alpha - \beta) & 1 & \cos \beta\\ \cos \alpha & \cos \beta & 1\end{vmatrix}$ is

0%
$\alpha^{2} + \beta^{2}$
0%
$\alpha^{2} - \beta^{2}$
0%
$1$
0%
$0$

Explanation

On solving the determinant, we have

$1(1 - \cos^{2}\beta) - \cos (\alpha - \beta) [\cos (\alpha - \beta) - \cos \alpha \cdot \cos \beta] + \cos \alpha [\cos \beta \cdot \cos (\alpha - \beta) - \cos \alpha]$

$= 1 - \cos^{2}\beta - \cos^{2} \alpha - \cos^{2}(\alpha - \beta) + 2\cos \alpha \cdot \cos \beta . \cos (\alpha - \beta)$

$= 1 - \cos^{2}\beta - \cos^{2} \alpha + \cos (\alpha - \beta) [2\cos \alpha. \cos \beta - \cos (\alpha - \beta)]$

$= 1 - \cos^{2}\beta - \cos^{2} \alpha + \cos(\alpha - \beta) [\cos (\alpha + \beta) + \cos (\alpha - \beta) -\cos (\alpha - \beta)]$

$= 1 - \cos^{2}\beta - \cos^{2} \alpha - \cos^{2} \alpha . \cos^{2}\beta - \sin^{2} \alpha . \sin^{2}\beta$

$= 1 - \cos^{2} \beta - \cos^{2} \alpha (1 - \cos^{2}\beta) - \sin^{2} \alpha . \sin^{2}\beta$

$= 1 - \cos^{2} \beta - \cos^{2} \alpha . \sin^{2}\beta - \sin^{2}\alpha. \sin^{2}\beta$

$= (1 - \cos^{2}\beta) - \sin^{2} \beta(\sin^{2}\alpha + \cos^{2} \alpha)$

$= \sin^{2}\beta - \sin^{2}\beta = 0$

Hence, determinant is

$0$ .

The value of the determinant

$\begin{vmatrix}cos\, \alpha & -sin \,\alpha &1 \\ sin \, \alpha & cos \, \alpha & 1\\ cos (\alpha +\beta) & -sin (\alpha +\beta ) & 1\end{vmatrix}$ is

0%
Independent of $\alpha$
0%
Independent of $\beta$
0%
Independent of $\alpha$ and $\beta$
0%
None of the above

Explanation

Given,

$\begin{vmatrix}cos\, \alpha & sin \,\alpha &1 \\ sin \, \alpha & cos \, \alpha & 1\\ cos (\alpha +\beta) & -sin (\alpha +\beta ) & 1\end{vmatrix}$

[Applying

$R_3\rightarrow R_3-R_1(cos \,\beta )+R_2(sin \, \beta )$ ]

$\begin{vmatrix}cos\, \alpha & sin \,\alpha &1 \\ sin \, \alpha & cos \, \alpha & 1\\ 0 & 0 & 1+sin \,\beta -cos \,\beta \end{vmatrix}$

$=(1 + sin \,\beta -cos\,\beta )(cos^2\alpha +sin^2\alpha )$

$=1+sin\,\beta -cos \,\beta$ , which is independent of

$\alpha$ .

Hence, option A is correct.

$\triangle (r) = \begin{vmatrix}r & r^{3}\\ 1 & n(n + 1)\end{vmatrix}$ , then

$\displaystyle \sum_{r = 1}^{n} \triangle (r)$ is equal to

0%
$\displaystyle \sum_{r = 1}^{n} r^{2}$
0%
$\displaystyle \sum_{r = 1}^{n} r^{3}$
0%
$\displaystyle \sum_{r = 1}^{n} r$
0%
$\displaystyle \sum_{r = 1}^{n} r^{4}$

Explanation

$\triangle (r) = \begin{vmatrix}r & r^{3}\\ 1 & n(n + 1)\end{vmatrix}$

$\implies \displaystyle \sum_{r = 1}{n} \triangle (r) = \begin{vmatrix}\displaystyle \sum_{r = 1}^{n}r &\displaystyle \sum_{r = 1}^{n}r^{3} \\ 1 & n(n + 1)\end{vmatrix}$

$= \begin{vmatrix} \dfrac {n(n + 1)}{2}& \dfrac {[n(n + 1)]^{2}}{4}\\ 1 & n(n + 1)\end{vmatrix}$

$= \dfrac {[n(n + 1)]^{2}}{2} - \dfrac {[n(n + 1)]^{2}}{4}$

$= \dfrac {[n(n + 1)]^{2}}{4} = \displaystyle \sum_{r = 1}^{n} r^{3}$

$\begin{vmatrix} a & b & 0 \\ 0 & a & b \\ b & 0 & a\end{vmatrix}=0$ , then which one of the following is correct?

0%
$\displaystyle\frac{a}{b}$ is one of the cube roots of unity
0%
$\displaystyle\frac{a}{b}$ is one of the cube roots of $-1$ .
0%
a is one of the cube roots of unity
0%
b is one of the cube roots of unity.

Explanation

$\begin{vmatrix} a & b & 0 \\ 0 & a & b \\ b & 0 & a \end{vmatrix}=0$

$\Rightarrow a\left( { a }^{ 2 } \right) -b\left( -{ b }^{ 2 } \right) =0$

$\Rightarrow { a }^{ 3 }=-{ b }^{ 3 }$

$=\cfrac { a }{ b } =\sqrt [ 3 ]{ (-1) }$

Option B

If A is an invertible matrix, then what is

$det$

$(A^{-1})$ equal to?

0%
$det A$
0%
$\displaystyle\frac{1}{det A}$
0%
$1$
0%
None of the above

Explanation

We know

$A{ A }^{ -1 }=I$

$\Rightarrow \left| A{ A }^{ -1 } \right| =\left| I \right|$

$\Rightarrow \left| A \right| \left| { A }^{ -1 } \right| =1$

$\Rightarrow \left| { A }^{ -1 } \right| =\cfrac { 1 }{ \left| A \right| }$

$\Rightarrow \left| { A }^{ -1 } \right| =\left| A \right| ^{ -1 }$

Option B

$\begin{vmatrix}1 & 1 & 1\\ a & b & c\\ a^2 - bc & b^2 - ca & c^2 - ab\end{vmatrix} =$

0%
$0$
0%
$1$
0%
$abc$
0%
$(a -b), \,(b-c),\, (c-a)$

Explanation

$=\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ { a }^{ 2 }-bc & { b }^{ 2 }-ca & { c }^{ 2 }-ab \end{vmatrix}$

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

${ C }_{ 1 }\rightarrow { C }_{ 1 }-{ C }_{ 2\\ }$ and

${ C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 2 }$

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

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

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

$= 0$

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

$B = \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}$ then what is determinant of AB ?

0%
$0$
0%
$1$
0%
$10$
0%
$20$

Explanation

Given

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

$B=\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}$

$A\times B=\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}\times \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}=\begin{bmatrix} 1 \times 1+2 \times 0 & 1 \times 1+2 \times 0 \\ 2 \times 1+3 \times 0 & 2 \times 1+3 \times 0 \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 2 & 2 \end{bmatrix}\\$

$\left| A\times B \right| =\begin{vmatrix} 1 & 1 \\ 2 & 2 \end{vmatrix}=2 \times 1-2 \times 1=0$

$\begin{vmatrix} 8 & -5 & 1 \\ 5 & x & 3\\ 6 & 3 & 1 \end{vmatrix} = 2$ then what is the value of

$x$ ?

0%
$44$
0%
$55$
0%
$61$
0%
$84$

Explanation

$\begin{vmatrix} 8 & -5 & 1 \\ 5 & x & 3 \\ 6 & 3 & 1 \end{vmatrix}=2$

$8\left( x-9 \right) +5\left( 5-18 \right) +1\left( 15-6x \right) =2$

$8x-72-65+15-6x=2$

$2x=122$

$x=61$

The cofactor of the element

$4$ in the determinant

$\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 5 & 8 & 9 \end{vmatrix}$
is

0%
$2$
0%
$4$
0%
$6$
0%
$-6$

Explanation

Cofactor of element 4 in

$\left| \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix} \right|$ is

${(-1)}^3\left| \begin{matrix} 2 & 3 \\ 8 & 9 \end{matrix} \right| =6$

We got the matrice

$\left| \begin{matrix} 2 & 3 \\ 8 & 9 \end{matrix} \right|$ after deleting column 1 and row 2.

Consider the following statements in respect of the determinant

$\begin{vmatrix}{\cos}^2\dfrac{\alpha}{2}&{\sin}^2\dfrac{\alpha}{2}\\{\sin}^2\dfrac{\beta}{2}&{\cos}^2\dfrac{\beta}{2}\end{vmatrix}$ where

$\alpha , \beta$ are complementary angles
The value of the determinant is

$\dfrac{1}{\sqrt{2}}\cos \begin{pmatrix}\dfrac{\alpha - \beta}{2}\end{pmatrix}$ .
The maximum value of the determinant is

$\dfrac{1}{\sqrt{2}}$ .
Which of the above statements is/are correct?

0%
1 only
0%
2 only
0%
Both 1 and 2
0%
Neither 1 nor 2

Explanation

Solution:

We have,

$\triangle=\begin{bmatrix}\cos^2\cfrac{\alpha}{2}&\sin^2\cfrac{\alpha}2\\\sin^2\cfrac{\beta}2&\cos^2\cfrac{\beta}{2}\end{bmatrix}$

$=\cos^2\cfrac{\alpha}2\cos^2\cfrac{\beta}2-\sin^2\cfrac{\alpha}2\sin^2\cfrac{\beta}{2}$

$=\left(\cos\cfrac{\alpha}{2}\cos\cfrac{\beta}{2} +\sin\cfrac{\alpha}{2}\sin \cfrac{\beta}{2}\right)$

$\left(\cos\cfrac{\alpha}{2}\cos\cfrac{\beta}{2}-\sin\cfrac{\alpha}{2}\sin\cfrac{\beta}{2}\right)$

$=\cos\left(\cfrac{\alpha-\beta}{2}\right)\cos45^\circ$ ........

$[\alpha , \beta ]$ are complementry angles.

$=\cfrac1{\sqrt2}\cos\left(\cfrac{\alpha-\beta}{2}\right)$

Maximum value of

$\cos\left(\cfrac{\alpha-\beta}{2}\right)=1$

So, maximum value of

$\cfrac1{\sqrt2}\cos\left(\cfrac{\alpha-\beta}{2}\right)=\cfrac1{\sqrt2}$

Hence, C is the correct option.

Let

$A = \begin{pmatrix}x + 2 & 3x\\ 3 & x + 2\end{pmatrix}, B = \begin{pmatrix}x & 0\\ 5 & x + 2\end{pmatrix}$ . Then all solutions of the equation

$det (AB) = 0$ is

0%
$1, -1, 0, 2$
0%
$1, 4, 0, -2$
0%
$1, -1, 4, 3$
0%
$-1, 4, 0, 3$

Explanation

Given :

$A = \begin{pmatrix}x + 2 & 3x\\ 3 & x + 2\end{pmatrix}$ and

$B = \begin{pmatrix}x & 0\\ 5 & x + 2\end{pmatrix}$

$\implies AB = \begin{bmatrix}x^{2} + 17x & 3x^{2} + 6x\\ 8x + 10 & (x + 2)^{2}\end{bmatrix} = 0$

Now,

$det(AB) = \begin{vmatrix}x^{2} + 17x & 3x^{2} + 6x\\ 8x + 10 & (x + 2)^{2}\end{vmatrix} = 0$

$\implies (x^{2}+17x)(x+2)^{2}-(3x^{2}+6x)(8x+10)=0$

$\implies x(x+17)(x+2)^{2}-3x(x+2)(8x+10)=0$

$\implies x(x+2)[(x+17)(x+2)-3(8x+10)]=0$

$\implies x(x+2)[x^{2}+19x+34-24x-30]=0$

$\implies x(x+2)[x^{2}-5x+4]=0$

$\implies x(x + 2)(x - 4) (x - 1) = 0$

$\implies x = 0, -2, 1, 4$

$C = 2 \cos \theta$ , then the value of the determinant

$\triangle = \begin{vmatrix} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{vmatrix}$ is :

0%
$\dfrac { \sin 4 \theta } { \sin \theta}$
0%
$\dfrac { 2 \sin^2 \theta } { \sin \theta}$
0%
$4 \cos^2 \theta (2 \cos \theta - 1)$
0%
None of these above

Explanation

Let

$\triangle = \begin{vmatrix} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{vmatrix} = C (C^2-1) -1(C-6) = C^3 - 2C + 6$
Put

$C = 2 \cos \theta$ we get

$\triangle = (2 \cos \theta)^3 - 2(2 \cos \theta) + 6$

$= 8 \cos^3 \theta - 4 \cos \theta + 6$

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

$A^{-1} =$

0%
$\dfrac {-1}{2}\begin{bmatrix} 4& -2\\ -3 & 1\end{bmatrix}$
0%
$\dfrac {1}{2}\begin{bmatrix} 4& -2\\ -3 & 1\end{bmatrix}$
0%
$\begin{bmatrix} -2& 4\\ 1 & 3\end{bmatrix}$
0%
$\begin{bmatrix} 2& 4\\ 1 & 3\end{bmatrix}$

Explanation

For

$2 \times 2$ matrix

$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , we swap the positions of a and d, put negatives in front of b and c, and divide everything by the determinant

$(ad-bc).$

So for

$\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ inverse of matrice is

$ad-bc=1 \times 4-2 \times 3=-2$

$A^{-1}=$

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

$\alpha, \beta, \gamma$ are the roots of the equation

$x^{3} + px + q = 0$ then the value of the determinant

$\begin{vmatrix} \alpha& \beta & \gamma\\ \beta & \gamma & \alpha\\ \gamma & \alpha & \beta\end{vmatrix}$ is

0%
$q$
0%
$0$
0%
$p$
0%
$p^{2} - 2q$

Explanation

$\alpha,\beta,\gamma$ are the roots of given equation

$\implies (\alpha +\beta +\gamma )=0$

Now,

$\begin{vmatrix} \alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{vmatrix}$

$R_1\rightarrow R_1+R_2+R_3$

$= \begin{vmatrix} \alpha +\beta +\gamma & \alpha +\beta +\gamma & \alpha +\beta +\gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{vmatrix}$

$= (\alpha +\beta +\gamma )\begin{vmatrix} 1 & 1 & 1 \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{vmatrix}$

$= (\alpha +\beta +\gamma )(\alpha \beta +\alpha \gamma +\beta \gamma -\alpha ^{ 2 }-\beta ^{ 2 }-\gamma ^{ 2 })$

$\implies (\alpha +\beta +\gamma )(3(\alpha \beta +\alpha \gamma +\beta \gamma )-(\alpha +\beta +\gamma )^2)=0$ as

$(\alpha +\beta +\gamma )=0$

Hence, option B is correct.

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CBSE Questions for Class 12 Commerce Maths Determinants Quiz 6 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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