MCQ Questions for Class 12 Commerce Maths Matrices Quiz 7

Say true or false:

$A$ ,

$B$ be two matrices such that they commute, then

$A^2 - B^2 = (A - B)(A + B)$ .

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True
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False

Explanation

$A$ and

$B$ commute when

$AB=BA$
Hence

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

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

$(A-2I)(A-3I)=$

0%
$0$
0%
$A$
0%
$I$
0%
$5I$

Explanation

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

$\Rightarrow$

$(A-2I)(A-3I)={A}^{2}-5AI+6{I}^{2}$

${A}^{2}-5A+6I$
by Hemilton rule

$\begin{vmatrix} 4-\lambda & 2 \\ -1 & 1-\lambda \end{vmatrix}=0$

after expanding determinant we got;

$\Rightarrow 4-5\lambda +{ \lambda }^{ 2 }+2=0$

$\Rightarrow { \lambda }^{ 2 }-5\lambda +6=0$

$\Rightarrow { A }^{ 2 }-5A+6I=0$
Ans: 0

It can be done using scalar multiplication and followed by matrix multiplication.

$\displaystyle A\times \begin{bmatrix} 1 & 2 &3 \\ 4 & 5 & 6 \end{bmatrix}=\begin{bmatrix} 1 & 2 &3 \\ 3 & 2 & 1 \\ 3 & 1 & 2 \end{bmatrix}$ then the order of A is _______

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$\displaystyle 2\times 3$
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$\displaystyle 3\times 3$
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$\displaystyle 3\times 2$
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$\displaystyle 2\times 2$

Explanation

When a matrix of order

$m \times a$ is multiplied with a matrix of order

$a \times n$ , the resulting matrix is of order

$m \times n$

As the second matrix is of order

$2 \times 3$ and the product matrix is

$3 \times 3$ , we get

$a = 2 ; m = 3 ; n = 3$

Thus, the matrix A should have order

$3 \times 2$

Inverse of a diagonal matrix is

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Symmetric
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A skew-symmetric
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A diagonal matrix
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None of these

Explanation

Consider a diagonal

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

Minor of the matrix

$\begin{bmatrix} 4 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 2 \end{bmatrix}$

Co factor of matrix

$\begin{bmatrix} 4 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 2 \end{bmatrix}\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$

$\begin{bmatrix} 4 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 2 \end{bmatrix}$

Adjoint of matrix

$=$ Transpose of co factor

$=\begin{bmatrix} 4 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 2 \end{bmatrix}$

Set of matrix considered

$=6$

${ A }^{ -1 }=\cfrac { Adjoint\quad (A) }{ det(A) }$

$=\cfrac { 1 }{ 6 } \begin{bmatrix} 4 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 2 \end{bmatrix}$

Clearly

${ A }^{ -1 }$ is also diagonal matrix

$\therefore$ Inverse of a diagonal matrix is a diagonal matrix

Option C is correct

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

$A^2 - 5A$ is equal to

0%
$2I$
0%
$-2I$
0%
$3I$
0%
Null matrix

Explanation

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

Also,

$A^{2} = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 1\times1+2\times 3\,\,\,\,\,1\times 2+2\times 4 \\ 3\times 1+4\times 3\,\,\,\,\,3\times2+4\times 4 \end{bmatrix}$

$\Rightarrow A^{2} = \begin{bmatrix} 7 & 10 \\ 15 & 22 \end{bmatrix}$

$5A = 5\times \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 5 & 10 \\ 15 & 20 \end{bmatrix}$

$A^{2} -5A = \begin{bmatrix} 7 & 10 \\ 15 & 22 \end{bmatrix}-\begin{bmatrix} 5 & 10 \\ 15 & 20 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & +2 \end{bmatrix} = 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\therefore A^{2}-5A = 2I$

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

$\left( A+I \right) \cdot \left( A-I \right)$ is equal to

0%
$\begin{bmatrix} -5 & -4 \\ 8 & -9 \end{bmatrix}$
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$\begin{bmatrix} -5 & 4 \\ -8 & 9 \end{bmatrix}$
0%
$\begin{bmatrix} 5 & 4 \\ 8 & 9 \end{bmatrix}$
0%
$\begin{bmatrix} -5 & -4 \\ -8 & -9 \end{bmatrix}$

Explanation

Given,

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

$\therefore A-I=A+I-2I$

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

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

$\therefore \left( A+I \right) \cdot \left( A-I \right) =\begin{bmatrix} 3 & -2 \\ 4 & 1 \end{bmatrix}\begin{bmatrix} 1 & -2 \\ 4 & -1 \end{bmatrix}$

$=\begin{bmatrix} 3-8 & -6+2 \\ 4+4 & -8-1 \end{bmatrix}$

$=\begin{bmatrix} -5 & -4 \\ 8 & -9 \end{bmatrix}$

Two matrices

$A$ and

$B$ are multiplied to get

$AB$ if

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both are rectangular
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both have same order
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number of columns of $A$ is equal to rows of $B$
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number of rows of $A$ is equal to no of columns of $B$

Explanation

According to condition of multiplication of two matrices:

The columns in first matrix should be same in number as the rows in the second matrix [For

$AB$ to be defined, we must have columns in

$A=$ rows in

$B$ ]

Hence option C is true.

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

${ A }^{ 2 }$ is equal to

0%
Null matrix
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Itself $A$
0%
Unit matrix
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Scalar matrix

Explanation

Given,

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

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

$=\begin{bmatrix} 4+2-4 & -2-3+4 & 2+2-3 \\ -4-6+8 & 2+9-8 & -2-6+6 \\ -8-8+12 & 4+12-12 & -4-8+9 \end{bmatrix}$

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

$\displaystyle A=\left[ \begin{matrix} 2 \\ 0 \\ 0 \end{matrix}\,\,\,\begin{matrix} 0 \\ 2 \\ 0 \end{matrix}\,\,\,\begin{matrix} 0 \\ 0 \\ 2 \end{matrix} \right]$ then

$\displaystyle { A }^{ 5 }=$

0%
$\displaystyle 5A$
0%
$\displaystyle 10A$
0%
$\displaystyle 16A$
0%
$\displaystyle 32A$

Explanation

Given,

$A= \displaystyle \left[\begin{matrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end{matrix} \right]$

$\Rightarrow A= \displaystyle 2\left[\begin{matrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{matrix} \right]$

$\Rightarrow A= \displaystyle 2I$

$\Rightarrow A^5= \displaystyle 2^5I^5$

$\Rightarrow A^5= \displaystyle 2^5I$

$\Rightarrow A^5= \displaystyle 2^4.2I$

$\Rightarrow A^5= \displaystyle 2^4. \displaystyle 2\left[\begin{matrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{matrix} \right]$

$\Rightarrow A^5= \displaystyle 16\displaystyle \left[\begin{matrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end{matrix} \right]$ .

$\Rightarrow A^5= \displaystyle 16A$
Option

$C$ is correct.

$A$ and

$B$ are symmetric matrices, then

$ABA$ is

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Symmetric
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Skew-symmetric
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Diagonal
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Triangular

Explanation

Given

$A$ and

$B$ are symmetric matrices.

$\Rightarrow A^T = A$ and

$B^T = B$

Now, take

$(ABA)^T$

$\Rightarrow (ABA)^T = A^T B^T A^T$

$\Rightarrow (ABA)^T = ABA$

Hence,

$ABA$ is also a symmetric matrix.

$A$ is an

$m \times n$ matrix such that

$AB$ and

$BA$ are both defined, then order of

$B$ is

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$m \times n$
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$n \times m$
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$n \times n$
0%
$m \times m$

Explanation

Since

$AB$ exists, so the number of column in

$A=$ Number of column in

$B$ .

$B$ has

$n$ row.

Since

$BA$ exists, so the number of column in

$B=$ Number of column in

$A$ .

$B$ has

$m$ column.

$\therefore$ B is an

$n\times m$ matrix.

$\displaystyle A=\begin{bmatrix} 0 & 0 & 1\\ 0 & 1&0 \\ 1& 0 & 0\end{bmatrix}$ , then

$A^{-1}$ is.

0%
$-A$
0%
$A$
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$1$
0%
None of these

Explanation

We have,

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

$\Rightarrow |A|=0(0-0)-0(0-0)+1(0-1)$

$\Rightarrow |A|=-1$
and cofactors of A are

$A_{11}=0, A_{12}=0, A_{13}=-1,$

$A_{21}=0, A_{22}=-1, A_{23}=0,$

$A_{31}=-1, A_{32}=0, A_{33}=0$

$\therefore A^{-1}=\displaystyle\frac{adj(A)}{|A|}$

$=-\displaystyle\frac{1}{1}\begin{bmatrix} 0 & 0 & -1\\0 & -1 &0\\ -1 &0 &0\end{bmatrix}$

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

$A^{100}$ is equal to.

0%
$2^{100}A$
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$2^{99}A$
0%
$100A$
0%
$299A$

Explanation

Given,

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

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

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

$A^4=A^2\cdot A^2=2A\cdot 2A$

$=4A^2=4\times 2A$

$=8A=2^3A$
Similarly,

$A^8=2^7A$

$\therefore A^{100}=2^{99}A$

$\begin{bmatrix} 1 & 2 \\ 3 & -5 \end{bmatrix}$ , then

${A}^{-1}$ is equal to

0%
$\begin{bmatrix} \cfrac { 5 }{ 11 } & \cfrac { 2 }{ 11 } \\ \cfrac { 3 }{ 11 } & -\cfrac { 1 }{ 11 } \end{bmatrix}$
0%
$\begin{bmatrix} -\cfrac { 5 }{ 11 } & -\cfrac { 2 }{ 11 } \\ -\cfrac { 3 }{ 11 } & -\cfrac { 1 }{ 11 } \end{bmatrix}$
0%
$\begin{bmatrix} \cfrac { 5 }{ 11 } & \cfrac { 2 }{ 11 } \\ \cfrac { 3 }{ 11 } & \cfrac { 1 }{ 11 } \end{bmatrix}$
0%
$\begin{bmatrix} 5 & 2 \\ 3 & -1 \end{bmatrix}$

Explanation

Since

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

$\therefore \left| A \right| =\begin{bmatrix} 1 & 2 \\ 3 & -5 \end{bmatrix}=-5-6=-11$

and

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

$\therefore { A }^{ -1 }=\cfrac { 1 }{ \left| A \right| } adj(A)$

$=-\cfrac { 1 }{ 11 } \begin{bmatrix} -5 & -2 \\ -3 & 1 \end{bmatrix}=\cfrac { 1 }{ 11 } \begin{bmatrix} 5 & 2 \\ 3 & -1 \end{bmatrix}$

$\quad =\begin{bmatrix} \cfrac { 5 }{ 11 } & \cfrac { 2 }{ 11 } \\ \cfrac { 3 }{ 11 } & -\cfrac { 1 }{ 11 } \end{bmatrix}$

Let

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

$10B=\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}$ , if

$B$ is the inverse of matrix

$A$ , then

$\alpha$ is

0%
$-2$
0%
$1$
0%
$2$
0%
$5$

Explanation

Since,

$B$ is the inverse of

$A$ .
ie,

$B=10{ A }^{ -1 }$

$\therefore \left( 10 \right) { A }^{ -1 }=\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}$

$\therefore \left( 10 \right) { A }^{ -1 }\cdot A=\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}A$

$\Rightarrow 10I=\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}\begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{bmatrix}=\begin{bmatrix} 10 & 0 & 0 \\ -5+\alpha & 5+\alpha & -5+\alpha \\ 0 & 0 & 10 \end{bmatrix}$

$\Rightarrow 5+\alpha =10$

$\Rightarrow \alpha =5$

Inverse of the matrix

$\begin{bmatrix} \cos 2\theta & -\sin 2\theta\\ \sin 2\theta & \cos 2\theta\end{bmatrix}$ is.

0%
$\begin{bmatrix} \cos 2\theta & -\sin 2\theta\\ \sin 2\theta & \cos 2\theta\end{bmatrix}$
0%
$\begin{bmatrix} \cos 2\theta & \sin 2\theta\\ \sin 2\theta & -\cos 2\theta\end{bmatrix}$
0%
$\begin{bmatrix} \cos 2\theta & \sin 2\theta\\ \sin 2\theta & \cos 2\theta\end{bmatrix}$
0%
$\begin{bmatrix} \cos 2\theta & \sin 2\theta\\ -\sin 2\theta & \cos 2\theta\end{bmatrix}$

Explanation

Let

$A=\begin{bmatrix} \cos 2\theta & -\sin 2\theta\\ \sin 2\theta & \cos 2\theta\end{bmatrix}$

$\therefore |A|=\cos^2 2\theta+\sin^2 2\theta=1$
and

$adj(A)=\begin{bmatrix}\cos 2\theta &\sin 2\theta\\ -\sin 2\theta & \cos 2\theta\end{bmatrix}$

$\therefore A^{-1}=\displaystyle\frac{1}{1}\begin{bmatrix}\cos 2\theta & \sin 2\theta\\ -\sin 2\theta &\cos 2\theta\end{bmatrix}$

$=\begin{bmatrix}\cos 2\theta & \sin 2\theta\\ -\sin 2\theta & \cos 2\theta\end{bmatrix}$

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

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

${ \left( AB \right) }^{ T }$ is equal to

0%
$\begin{bmatrix} -3 & -2 \\ 10 & 7 \end{bmatrix}$
0%
$\begin{bmatrix} -3 & 10 \\ -2 & 7 \end{bmatrix}$
0%
$\begin{bmatrix} -3 & 7 \\ 10 & 2 \end{bmatrix}$
0%
None of these

Explanation

Given,

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

$\therefore AB=\begin{bmatrix} 2-6+1 & 1-4+1 \\ 4+3+3 & 2+2+3 \end{bmatrix}=\begin{bmatrix} -3 & -2 \\ 10 & 7 \end{bmatrix}$
Now,

${ \left( AB \right) }^{ T }=\begin{bmatrix} -3 & 10 \\ -2 & 7 \end{bmatrix}$

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

$\begin{bmatrix} x\\ y\\z\end{bmatrix}=\begin{bmatrix} 0 \\ 3\\4\end{bmatrix}$ , then

$\begin{bmatrix} x\\ y\\z\end{bmatrix}$ is equal to.

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

Explanation

Given,

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

$\therefore x+y+z=0$ ,

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

$x+3y=z=4$
On solving these equations, we get

$x=1, y=2$ and

$z=-3$

$\therefore \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\2\\-3\end{bmatrix}$

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

$A^2$ is equal to

0%
Null matrix
0%
Unit matrix
0%
$\begin{bmatrix} 1&0\\0&1 \end{bmatrix}$
0%
$\begin{bmatrix} 0&0\\0&1 \end{bmatrix}$

Explanation

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

$A.A.=A^{2}= \begin{bmatrix} -2&4\\-1&2 \end{bmatrix} \begin{bmatrix} -2&4\\-1&2 \end{bmatrix} = \begin{bmatrix} 4-4\,\,-2\times 4+4\times 2\\-1x-2+2x-1\,\,-1\times 4+2\times 2 \end{bmatrix}$

$A^{2}=\begin{bmatrix} 0&0\\0&0 \end{bmatrix} = O =$ null matrix.

$A=[x\, y\, z], B = \begin{bmatrix}a&h&g\\h&b&f\\g&f&c\end{bmatrix}$ and

$C=\begin{bmatrix}x\\y\\z\end{bmatrix}$
Then,

$ABC = 0$ , if

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$[ax^2+by^2+cz^2+2gxy+2fyz+2czx]=0$
0%
$[ax^2+cy^2+bz^2+xy+yz+zx]=0$
0%
$[ax^2+by^2+cz^2+2hxy+2by+2cz]=0$
0%
$[ax^2+by^2+cz^2+2gzx+2fyz+2hxy]=0$

Explanation

$A = [x\,y\,z] B = \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix} c = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$

$A.B = [x\,y\,z] \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix}$

$= [ax+hy+gz\,\,\,\,hx+by+fz\,\,\,\,gx+fy+cz]$

$A.B.C = (AB)\times \begin{bmatrix} x \\ y \\ z\end{bmatrix} = [ax^{2}+hyx+gyx+hxy+by^{2}+fzy+gxz+fyz+cz^{2}]$

$ABC = [ax^{2}+by^{2}+cz^{2}+2hxy+2gzx+2fzy]$

$ABC = 0$

when,

$ax^{2}+by^{2}+cz^{2}+2hxy+2gzx+2fzy = 0$

$A$ and

$B$ are any two matrices, then

0%
$AB=BA$
0%
$AB=I$
0%
$AB=0$
0%
$AB$ may or may not be defined

Explanation

$\textbf{Step 1: Finding the order}$

$\text{Matrix multiplication is only defined for two matrices whose order is of the form}$

$a\times b \text{ and }b\times c \text{ respectively}$

$\text{Since we cant say anything about the order of the given matrices it may or may not be defined}$

$\textbf{Hence, AB may or may not be defined}$

$P=\begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix}$ , then

${P}^{5}$ is equal to

0%
$P$
0%
$2P$
0%
$-P$
0%
$-2P$

Explanation

Given

$P=\begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix}$

${P}^{2}=P.P=\begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix}\begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix}$

$=\begin{pmatrix} 4+2-4 & -4-6+8 & -8-8+12 \\ -2-3+4 & 2+9-8 & 4+12-12 \\ 2+2-3 & -2-6+6 & -4-8+9 \end{pmatrix}$

$=\begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix}=P$

$\therefore$

${P}^{4}={P}^{2}=P$

$\Rightarrow$

${P}^{5}={P}^{2}=P$

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

$I$ is the unit matrix of order

$2$ , then

$A^2$ equals

0%
$4A - 3I$
0%
$3A - 4I$
0%
$A - I$
0%
$A + I$

Explanation

$A=\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\\ { A }^{ 2 }=A\times A=\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\times \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\\ \quad \quad =\begin{bmatrix} 4+1 & -2-2 \\ -2-2 & 1+4 \end{bmatrix}=\begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix}$

From the given option to get first entry i.e

$5$ of

$A^2$ , only option

$(A)$ is satisfied. Or we can check as below

$4A-3I=\begin{bmatrix} 8 & -4 \\ -4 & 8 \end{bmatrix}-3\times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\\ \quad \quad \quad \quad =\begin{bmatrix} 8 & -4 \\ -4 & 8 \end{bmatrix}-\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}\\ \quad \quad \quad \quad =\begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix}={ A }^{ 2 }$

$A$ is a matrix such that

$\begin{pmatrix}2&1 \\3&2 \end{pmatrix} A(1 \space \,1)=\begin{pmatrix}1&1 \\0&0 \end{pmatrix}$ then

$A=$

0%
$\begin{pmatrix}1&1 \\0&1 \end{pmatrix}$
0%
$(2 \,1)$
0%
$\begin{pmatrix}1&0 \\-1&1 \end{pmatrix}$
0%
$\begin{pmatrix}2 \\-3 \end{pmatrix}$

Explanation

$\begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}A \begin{pmatrix} 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}$

To maintain the rules of matrix multiplication,

$A$ has to be of the order

$2 \times 1$

$\begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \begin{pmatrix} 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}$

$\Rightarrow \begin{pmatrix} 2x + y \\ 3x + 2y \end{pmatrix} \begin{pmatrix} 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}$

$\Rightarrow \begin{pmatrix} 2x + y & 2x + y \\ 3x + 2y & 3x + 2y \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}$

$\Rightarrow 2x + y = 1$ and

$3x + 2y = 0$

$\Rightarrow 4x + 2y - (3x + 2y) = 2 - 0$

$\Rightarrow x = 2$ and

$y = -3$

For the matrices

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

$B=\begin{bmatrix} 2 & 2 \\ -1 & 4 \end{bmatrix}$
What is

$AB$ ?

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$\begin{bmatrix} 3 & 5 \\ -1 & 3 \end{bmatrix}$
0%
$\begin{bmatrix} 1 & -1 \\ -1 & 5 \end{bmatrix}$
0%
$\begin{bmatrix} -1 & 14 \\ 1 & -4 \end{bmatrix}$
0%
$\begin{bmatrix} 2 & 4 \\ -1 & -7 \end{bmatrix}$
0%
none of the above

Explanation

$A$ and

$B$ are given by

$A=\left[ \begin{matrix} 1 & 3 \\ 0 & -1 \end{matrix} \right] , B=\left[ \begin{matrix} 2 & 2 \\ -1 & 4 \end{matrix} \right]$

Thus,

$AB=\left[ \begin{matrix} (1\times 2)+(3\times -1) & (1\times 2)+(3\times 4) \\ (0\times 2)+(-1\times -1) & (0\times 2)+(-1\times 4) \end{matrix} \right]$

$=\left[ \begin{matrix} 2-3 & 2+12 \\ 0+1 & 0-4 \end{matrix} \right]$

$=\left[ \begin{matrix} -1 & 14 \\ 1 & -4 \end{matrix} \right]$

The symmetric part of the matrix A=

$\begin{bmatrix} 1 &2 &4 \\ 6 & 8 & 2\\ 2 & -2 &7 \end{bmatrix}$ .

0%
$\begin{bmatrix} 1 &4 &3 \\ 2 & 8 & 0\\ 3 & 0 &7 \end{bmatrix}$
0%
$\begin{bmatrix} 1 &4 &3 \\ 4 & 8 & 0\\ 3 & 0 &7 \end{bmatrix}$
0%
$\begin{bmatrix} 0 &-2 &-1 \\ -2 & 0 & -2\\ -1 & -2 &0 \end{bmatrix}$
0%
$\begin{bmatrix} 0 &-2 &-1 \\ 2 & 0 & 2\\ -1 & 2 &0 \end{bmatrix}$

Explanation

Symmetric part of any matrix A is given by

$\dfrac{A+A'}{2}$ , where

$A'$ transpose of A

So symmetric part of given matrix

$\begin{bmatrix}1&2&4\\6&8&2\\2&-2&7 \end{bmatrix}$

$= \dfrac{1}{2}\bigg ($

$\begin{bmatrix}1&2&4\\6&8&2\\2&-2&7 \end{bmatrix}$ +

$\begin{bmatrix}1&6&2\\2&8&-2\\4&2&7 \end{bmatrix}$

$\bigg)$

$=\dfrac{1}{2}$

$\begin{bmatrix}2&8&6\\8&16&0\\6&0&14 \end{bmatrix}$

$= \begin{bmatrix}1&4&3\\4&8&0\\3&0&7 \end{bmatrix}$

$A = \begin{bmatrix} 0& 1\\ 1 & 0\end{bmatrix}$ , then

$A^{2}$ is equal to ______

0%
$\begin{bmatrix} 0& 1\\ 1 & 0\end{bmatrix}$
0%
$\begin{bmatrix} 1& 0\\ 1 & 0\end{bmatrix}$
0%
$\begin{bmatrix} 1& 0\\ 0 & 1\end{bmatrix}$
0%
$\begin{bmatrix} 0& 1\\ 0 & 1\end{bmatrix}$

Explanation

We have,

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

Hence

$A^2=AA=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix} 0&1\\1&0\end{bmatrix}$

$\quad = \begin{bmatrix}0\times 0+1\times 1&0\times 1+1\times 0\\1\times 0+0\times 1 &1\times 1+0\times 0\end{bmatrix}$

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

$\begin{bmatrix} 3& 2\\ -1 & -2\\ 4 & 5\end{bmatrix} \begin{bmatrix} 0& a\\ b & c \end{bmatrix} = \begin{bmatrix}-4 &9 \\ 4 & -7\\ -10 & 19\end{bmatrix}$
Find

$a, b$ and

$c$ .

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

Explanation

The product is

$\begin{bmatrix} 3 & 2 \\ -1 & -2 \\ 4 & 5 \end{bmatrix}\begin{bmatrix} 0 & a \\ b & c \end{bmatrix}=$

$\begin{bmatrix} 2b & 3a+2c \\ -2b & -2c-a \\ 5b & 5c+4a \end{bmatrix}$

By given, we have

$\begin{bmatrix} 2b & 3a+2c \\ -2b & -2c-a \\ 5b & 5c+4a \end{bmatrix}$

$=\begin{bmatrix} -4 & 9 \\ 4 & -7 \\ -10 & 19 \end{bmatrix}$

By comparing, we get

$b=-2 , a = 1 , c = 3$

$P=\begin{pmatrix}2&-2&-4 \\-1&3&4\\1&-2&-3\end{pmatrix}$ then

$P^5$ equals

0%
$P$
0%
$2P$
0%
$-P$
0%
$-2P$

Explanation

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

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

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

Now,

${ P }^{ 2 }=P\quad { P }^{ 5 }={ P }^{ 2 }\times { P }^{ 3 }\Rightarrow { P }^{ 5 }={ P }^{ 4 }$

$\left( \because { P }^{ 2 }=P \right) \Rightarrow { P }^{ 5 }={ P }^{ 2 }\times { P }^{ 2 }$

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

$\Rightarrow { P }^{ 5 }={ P }^{ 2 }=P$

$\therefore { P }^{ 5 }=P$

$\therefore$ Option A is correct

For a matrix

$A \begin{pmatrix} 1& 0 & 0\\ 2 & 1 & 0\\ 3 & 2 & 1\end{pmatrix}$ , if

$U_{1}, U_{2}$ and

$U_{3}$ are

$3\times 1$ column matrices satisfying

$AU_{1} = \begin{pmatrix}1\\ 0 \\ 0 \end{pmatrix}, AU_{2} \begin{pmatrix}2\\3 \\ 0 \end{pmatrix}, AU_{3} = \begin{pmatrix}2\\ 3\\ 1 \end{pmatrix}$ and

$U$ is

$3\times 3$ matrix whose columns are

$U_{1}, U_{2}$ and

$U_{3}$
Then sum of the elements of

$U^{-1}$ is

0%
$6$
0%
$0 (zero)$
0%
$1$
0%
$2/3$

Explanation

Let

$U_{i} = \begin{pmatrix}a_{i}\\b_{i} \\c_{i} \end{pmatrix} i = 1, 2, 3$

$AU_{1} = \begin{pmatrix}a_{1}\\2a_{1} + b_{1} \\ 3a_{1} + 2b_{1} + c_{1} \end{pmatrix} = \begin{pmatrix}1\\0 \\ 0 \end{pmatrix}$ , So

$a_{1} = 1, b_{1} = -2, c_{1} = 1$

$AU_{2} = \begin{pmatrix}a_{2}\\2a_{2} + b^{2} \\ 3a_{2} + 2b_{2} + c_{2} \end{pmatrix} = \begin{pmatrix}2\\ 3\\ 0 \end{pmatrix}$ ,
So,

$a_{2} = 2, b_{2} = -1, c_{2} = -4$ . Similarly,

$a_{3} = 2, b_{3} = -1, c_{3} = -3$
So,

$U = \begin{pmatrix} 1& 2 & 2\\ -2 & -1 & -1\\ 1 & -4 & -3\end{pmatrix}$ . So, sum of elements of

$U^{-1}$ is zero.

CBSE Questions for Class 12 Commerce Maths Matrices Quiz 7 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Matrices Quiz 7 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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