If $$A \times \begin{bmatrix} 1 & 1\\ 0 & 2 \end{bmatrix} = [1 \ \ 2],$$ then A =

$$\left[1 \ \ \ \dfrac {1}{2} \right]$$

$$\left[\dfrac {1}{2} \ \ \ 1\right]$$

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

$$\begin{bmatrix} 2 & 1\\ 1 & 5 \end{bmatrix}$$

If $$A$$ is a non-singular matrix, then

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

$$A^{-1}$$ is symmetric if $$A$$ is symmeteric

$$A^{-1}$$ is skew-symmetric if $$A$$ is symmeteric

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

MCQ Questions for Class 12 Commerce Maths Matrices Quiz 4

$\mathrm{A}=\left[\begin{array}{ll} 0 & 1\\ 1 & 0 \end{array}\right]$ , then

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

0%
$I$
0%
$\mathrm{O}$
0%
$\mathrm{A}$
0%
$\mathrm{A}^{2}$

Explanation

Given, A

$=$

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

$A^2$

$=$

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

$A^5$

$=$

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

$A^5$

$=$

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

$=$

$A$

$\mathrm{A}=\left[\begin{array}{ll} 1 & 2\\ 0 & 3 \end{array}\right]$ and

$\mathrm{B}=[3 \space-1]$ , then

$\mathrm{B}\mathrm{A}=$

0%
$\left[\begin{array}{ll} 3 & 0\\ 0 & 3 \end{array}\right]$
0%
$[3 \ \ \ 0]$
0%
$[3 \ \ \ 3 ]$
0%
$[0 \ \ -3]$

Explanation

The value of

$\mathrm{BA}$

$=[3 \space -1]\begin{bmatrix}1 & 2\\0 & 3\end{bmatrix}$

$=[ 3\times 1-1\times 0 \ \ \space 3\times 2-1\times 3 ]$

$=[3 \ \ \ \space 3]$

$If \space A= \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix}$ , then A is

0%
a nilpotent matrix
0%
an involutory matrix
0%
a symmetric matrix
0%
an idempotent matrix

Explanation

Here The transpose of the matrix A is

$= \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix}$

$\therefore A = A^{T}$

And a symmetric matrix is a square matrix that is equal to its transpose.

Hence the answer is option C

If A =

$\begin{bmatrix} x & 1\\ 1 & 0 \end{bmatrix}$ and

$A^{2}$ is identity matrix, then

$x=$

0%
$1$
0%
$-1$
0%
$\pm 1$
0%
$0$

Explanation

Since,

$A=\begin{bmatrix} x &1 \\ 1 &0 \end{bmatrix}, A^2 =I$

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

$\Rightarrow \begin{bmatrix} x^{2}+1 &x \\ x &1 \end{bmatrix}= \begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}$

$\Rightarrow$ So

$x=0$

Option D is the correct answer.

$L=\left[\begin{array}{lll} 2 & 3 & 5\\ 4 & 1 & 2\\ 1 & 2 & 1 \end{array}\right] =P+Q$ ,

$P$ is a symmetric matrix,

${Q}$ is a skew-symmetric matrix then

${P}$ is equal to

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$\left[\begin{array}{lll} 3 & 5 & 6\\ 5 & 6 & 4\\ 9 & 4 & 3 \end{array}\right]$
0%
$\left[\begin{array}{lll} 2 & 3.5 & 3\\ 3.5 & 1 & 2\\ 3 & 2 & 1 \end{array}\right]$
0%
$\left[\begin{array}{lll} 6 & 5 & 4\\ 3 & 6 & 3\\ 5 & 2 & 5 \end{array}\right]$
0%
$\left[\begin{array}{lll} 6 & 5 & 4\\ 4 & 5 & 3\\ 3 & 4 & 3 \end{array}\right]$

Explanation

Given ,

$P+Q=\begin{bmatrix} 2 & 3 & 5\\ 4 & 1 & 2\\ 1 & 2 & 1 \end{bmatrix}$ -----(i)

ans

$P$ is symmetric and

$Q$ is skew symmetric matrix. So, by property of transpose (7),

$P=P^{T}$ and

$Q^{T}=-Q$

$\therefore$ equation (i),

$P^{T}-Q=\begin{bmatrix} 2 & 4 & 1\\ 3 & 1 & 2\\ 5 & 2 & 1 \end{bmatrix}$ ---------(ii)

Add (i) and (ii), we get

$P+P^{T}=\begin{bmatrix} 4 & 7 & 6\\ 7 & 2 & 4\\ 6 & 4 & 2 \end{bmatrix}, 2P=\begin{bmatrix} 4 & 7 & 6\\ 7 & 2 & 4\\ 6 & 4 & 2 \end{bmatrix}$

$\Rightarrow P=\begin{bmatrix} 2 & 3.5 & 3\\ 3.5 & 1 & 2\\ 3 & 2 & 1 \end{bmatrix}$

$\mathrm{A}=\left[\begin{array}{ll} 2 & -1\\ 3 & -2 \end{array}\right],$ then

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

0%
$I$
0%
$A$
0%
$-A$
0%
$A^{2}$

Explanation

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

$A^3=AI=A$

$A^5=A^3(A^2)=A^3I=A(A^2I)=A$

$I=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$ and E =

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

$\left ( 2I+3E \right )^{3}=$

0%
$8I+ 18E$
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$4I+ 36E$
0%
$8I +36E$
0%
$2I+ 3E$

Explanation

Consider,

$2I+3E$

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

$\Rightarrow \begin{bmatrix} 2 &0 \\ 0 &2 \end{bmatrix}+\begin{bmatrix} 0 &3 \\ 0 &0 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 2 &3 \\ 0 &2 \end{bmatrix}$
Now

$(2I+3E)^3$

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

$\Rightarrow \begin{bmatrix} 4 &6+6 \\ 0 &4 \end{bmatrix}\begin{bmatrix} 2 &3 \\ 0 &2 \end{bmatrix}$

$=\begin{bmatrix} 8 &12+24 \\ 0 &8 \end{bmatrix}=\begin{bmatrix} 8 &36 \\ 0 &8 \end{bmatrix}=8I+36E$

$\therefore(C)$

Let

$A$ and

$B$ be

$3\times3$ matrices such that

$A^{T}=-A, \, B^{T}=B$ , then matrix

$(\lambda AB+3BA)$ is a skew symmertric matrix for

0%
$\lambda =3$
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$\lambda =-3$
0%
$\lambda =3$ or $\lambda =-3$
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$\lambda =3$ and $\lambda =-3$

Explanation

If the matrix

$\lambda AB +3BA$ to be skew-symmetry,

Then, it must satisfy

$\left ( \lambda AB+3BA\right )^{T}=-\left ( \lambda AB+3BA \right )$

Now, considering

$\left ( \lambda AB+3BA\right )^{T}$

$=\lambda \left ( AB \right )^{T}+3\left ( BA \right )^{T}$

$=\lambda\ B^{T} A^{T}+3\ A^{T}B^{T}$
Substituting values of

$A^{T}=-A$ and

$B^{T}=B$ , we get

$=-\lambda BA-3AB = -(3AB+\lambda BA)$

From this, we get

$\lambda = 3$ clearly.

Hence, option A.

If in a square matrix

$A=\left[ { a }_{ ij } \right]$ , we find that

${ a }_{ ij }={ a }_{ ji }\quad \forall \quad i,j$ , then

$A$ is

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Symmetric
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Skew Symmetric
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Idempotent
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none of these

Explanation

Given square matrix

$A=\left\{ { a }_{ ij } \right\}$ and

${ a }_{ ij }={ a }_{ ji }$

We know that in a square matrix if

${ a }_{ ij }={ a }_{ ji }$ for all i and j, then the matrix is a symmetric matrix

So A is a symmetric matrix

$\mathrm{A}$ : If

$\mathrm{A}=\left\{\begin{array}{ll} 1 & -1\\ -1 & 1 \end{array}\right\}$ and

$\mathrm{B}=\left\{\begin{array}{ll} 2 & 2\\ 2 & 2 \end{array}\right\},$ then

$\mathrm{A}\mathrm{B}=0$

$\mathrm{R}$ : If

$\mathrm{A}\mathrm{B}=0\Rightarrow \mathrm{A}$ or

$\mathrm{B}$ need not be null matrices.

The correct answer is

0%
Both $A$ and $R$ are true, $R$ is correct explanation to $A$
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Both $A$ and $R$ are true but $R$ is not correct explanation to $A$
0%
$A$ is true, $R$ is false
0%
$A$ is false, $R$ is true

Explanation

$\mathrm{A}=\left\{\begin{array}{ll} 1 & -1\\ -1 & 1 \end{array}\right\} ;\mathrm{B}=\left\{\begin{array}{ll} 2 & 2\\ 2 & 2 \end{array}\right\}$
Then,

$\mathrm{AB}=\left\{\begin{array}{ll} 1\times 2-1\times 2 & 1\times 2-1\times 2\\ -1\times 2+1\times 2 & -1\times 2+1\times 2 \end{array}\right\} =O$
Hence both statement are correct and Reason is correct explanation of Assertion.

If P =

$\begin{bmatrix} 1\\ 3\\ 4\end{bmatrix}$ , Q =

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

0%
$\begin{bmatrix}2 & -1 & 5\\ 6& -3& 15\\ 8& -4 & 20\end{bmatrix}$
0%
$\begin{bmatrix}2 & -3 & 20\end{bmatrix}$
0%
$\begin{bmatrix}2\\ -3\\ 20\end{bmatrix}$
0%
$[19]$

Explanation

$\text{PQ}=\begin{bmatrix} 1\\3\\4\end{bmatrix}\begin{bmatrix} 2&-1&5\end{bmatrix}$

$\quad = \begin{bmatrix} 1\times 2&1\times(-1)&1\times 5\\3\times2&3\times(-1)&3\times 5 \\4\times 2&4\times(-1)&4\times 5\end{bmatrix}$

$\quad = \begin{bmatrix} 2 & -1 & 5\\ 6& -3& 15\\ 8& -4 & 20 \end{bmatrix}$

$\mathrm{A}=\left(\begin{array}{lll} x & 1 & 4\\ -1 & 0 & 7\\ -4 & -7 & 0 \end{array}\right)$ such that

$\mathrm{A}^{\mathrm{T}}=-\mathrm{A}$ , then

$\mathrm{x}=$

0%
$-1$
0%
$0$
0%
$1$
0%
$4$

Explanation

Given

$A=\begin{pmatrix} x & 1 & 4\\ -1 & 0 & 7\\ -4 & -7 & 0 \end{pmatrix}$
and

$A^{T}=-A$
So By property of traspose (I), and opertaion of matrixes (4),

$A^{T}=\begin{pmatrix} x & -1 & -4\\ 1 & 0 & -7\\ 4 & 7 & 0 \end{pmatrix}=-A=\begin{pmatrix} -x & -1 & -4\\ 1 & 0 & -7\\ 4 & 7 & 0 \end{pmatrix}$
So, By equality of matrixes,

$x=-x$

$\Rightarrow x=0$

$\mathrm{If}\mathrm{A}=\left[\begin{array}{lll} 2 & x-3 & x-2\\ 3 & -2 & -1\\ 4 & -1 & -5 \end{array}\right]$ is a symmetric matrix then

$\mathrm{x}$

0%
$0$
0%
$3$
0%
$6$
0%
$8$

Explanation

Given

$A=\begin{bmatrix} 2 & x-3 & x-2\\ 3 & -2 & -1\\ 4 & -1 & -5 \end{bmatrix}$ is a symmetric matrix.
So, by property of symmetric matrices

$A=A^{T}$

$\Rightarrow \begin{bmatrix} 2 & x-3 & x-2\\ 3 & -2 & -1\\ 4 & -1 & -5 \end{bmatrix}=\begin{bmatrix} 2 & 3 & 4\\ x-3 & -2 & -1\\ x-2 & -1 & -5 \end{bmatrix}$
So, By operation of matrices - equality,

$\Rightarrow x-3=3$

$\Rightarrow x=6$

$\begin{bmatrix} 10 & 20 & 30 \\ 20 & 45 & 80 \\ 30 & 80 & 171 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 4 & 1 \end{bmatrix}\begin{bmatrix} x & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix}$ then

${x}=$

0%
$10$
0%
$20$
0%
$30$
0%
$40$

Explanation

Given,

$\Rightarrow \begin{bmatrix} 10 & 20 & 30 \\ 20 & 45 & 80 \\ 30 & 80 & 171 \end{bmatrix}=\begin{bmatrix} x & 0 & 0 \\ 2x & 5 & 0 \\ 3x & 20 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 10 & 20 & 30 \\ 20 & 45 & 80 \\ 30 & 80 & 171 \end{bmatrix}=\begin{bmatrix} x & 2x & 3x \\ 2x & 4x+5 & 6x+20 \\ 3x & 6x+20 & 9x+81 \end{bmatrix}$

On comparing, we get

$x=10$

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

$A^{2}-A=$

0%
$\begin{bmatrix} 3 & 0 & 0\\ 0 & 1 & -1\\ 0 & 5 & 4 \end{bmatrix}$
0%
$\begin{bmatrix} 1 & 2 & -1\\ -3 & 1 & 1\\ 3 & 5 & -4 \end{bmatrix}$
0%
$\begin{bmatrix} 3 & 1 & -1\\ -3 & 1 & -1\\ 3 & 5 & 4 \end{bmatrix}$
0%
$\begin{bmatrix} 3 & -1 & 1\\ 3 & -1 & 1\\ 3 & 5 & 4 \end{bmatrix}$

Explanation

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

${ A }^{ 2 }=\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & -1 \\ 3 & -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & -1 \\ 3 & -1 & 1 \end{bmatrix}=\begin{bmatrix} 4 & 3 & 0 \\ -3 & 2 & -2 \\ 6 & 4 & 5 \end{bmatrix}$

Now,

${ A }^{ 2 }-A=\begin{bmatrix} 4 & 3 & 0 \\ -3 & 2 & -2 \\ 6 & 4 & 5 \end{bmatrix}-\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & -1 \\ 3 & -1 & 1 \end{bmatrix}$

$=\begin{bmatrix} 3 & 1 & -1 \\ -3 & 1 & -1 \\ 3 & 5 & 4 \end{bmatrix}$

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

$\mathrm{C}=\left[\begin{array}{l} 1\\ 2\\ -1 \end{array}\right]$ and

$(3\mathrm{B}-2\mathrm{A})\mathrm{C}+2\mathrm{X}=0$ then

$\mathrm{X}$ is equal to

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$\frac{1}{2}} \begin{bmatrix} 3\\ 13 \end{bmatrix$
0%
$\frac{1}{2}}\begin{bmatrix} 3\\ -13 \end{bmatrix$
0%
$\frac{1}{2}} \begin{bmatrix} -3\\ 13 \end{bmatrix$
0%
$\begin{bmatrix} 3\\ -13 \end{bmatrix}$

Explanation

Given,

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

Now,

$(3{ B }-2{ A }){ C }+2{ X}=0$

$\Rightarrow \left( \begin{bmatrix} 6 & 0 & -3 \\ 9 & 3 & 12 \end{bmatrix}-\begin{bmatrix} 8 & 2 & 0 \\ 2 & -4 & 4 \end{bmatrix} \right) \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}+2X=0$

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

$\begin{bmatrix} -3 \\ 13 \end{bmatrix}+2X=0$

$\Rightarrow 2X=\begin{bmatrix} 3 \\ -13 \end{bmatrix}$

$\Rightarrow X=\displaystyle \frac{1}{2}\begin{bmatrix} 3 \\ -13 \end{bmatrix}$

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

$\mathrm{A}^{2}=\lambda I$ then

$\lambda=$

0%
$0$
0%
$1$
0%
$\dfrac{1}{2}$
0%
$-2$

Explanation

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

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

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

But given

${ A }^{ 2 }=\lambda I$

$\Rightarrow \lambda=0$

A :

$\begin{vmatrix} 0 &p-e & e-r\\ e-p& 0 &r-p \\ r-e& p-r & 0 \end{vmatrix}$ =0
R : The determinant of a skew symmetric matrix is zero
The correct answer is

0%
Both A and R are true R is correct explanation to A
0%
Both A and R are true but R is not correct explanation to A
0%
A is true R is false
0%
A is false R is true

Explanation

Let

$A=\begin{bmatrix}0 &p-e & e-r\\ e-p& 0 &r-p \\ r-e& p-r & 0\end{bmatrix}$

Here,

$A^{T}=\begin{bmatrix} 0 & e-p & r-e \\ p-e & 0 & p-r \\ e-r & r-p & 0 \end{bmatrix}$

$\Rightarrow A^{T}=-A$

Hence, A is skew-symmetric matrix of order 3.
Determinant of a skew symmetric matrix of odd order is 0.

$A=\left[\begin{array}{lll} 2 & 2 & 1\\ 1 & 2 & 1\\ 3 & 4 & 2 \end{array}\right]$ then (

$\mathrm{A}-\mathrm{I}$ )

$(\mathrm{A}-2I)=$

0%
$\left[\begin{array}{lll} 5 & 6 & -2\\ 1 & 0 & 7\\ 0 & 1 & 1 \end{array}\right]$
0%
$\left[\begin{array}{lll} 5 & 6 & 3\\ 4 & 6 & 2\\ 7 & 10 & 7 \end{array}\right]$
0%
$\left[\begin{array}{lll} 1 & 0 & 7\\ 7 & 10 & 7\\ 4 & 6 & 2 \end{array}\right]$
0%
$\left[\begin{array}{lll} -1 & 1 & 2\\ 3 & 4 & 1\\ 1 & 1 & 2 \end{array}\right]$

Explanation

$A-I=\left[\begin{array}{} 1 & 2 & 1\\ 1 & 1 & 1\\ 3 & 4 & 1 \end{array}\right]$

$A-2I=\left[\begin{array}{} 0 & 2 & 1\\ 1 & 0 & 1\\ 3 & 4 & 0 \end{array}\right]$

$\therefore (A-I)(A-2I)=\left[\begin{array}{} 1 & 2 & 1\\ 1 & 1 & 1\\ 3 & 4 & 1 \end{array}\right]\left[\begin{array}{} 0 & 2 & 1\\ 1 & 0 & 1\\ 3 & 4 & 0 \end{array}\right]=\left[\begin{array}{} 5 & 6 & 3\\ 4 & 6 & 2\\ 7 & 10 & 7 \end{array}\right]$

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

$B=\begin{bmatrix} 10 & -4 & -1 \\ -11 & 5 & 0 \\ 9 & -5 & 1 \end{bmatrix}$ then

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$\mathrm{A}\mathrm{B}=\mathrm{B}\mathrm{A}$
0%
$\mathrm{A}\mathrm{B}=-\mathrm{A}\mathrm{B}$
0%
$\mathrm{A}\mathrm{B}=2\mathrm{B}\mathrm{A}$
0%
$\mathrm{A}\mathrm{B}=3\mathrm{B}\mathrm{A}$

Explanation

Given,

$A=\begin{bmatrix} 1 & 2 & 1 \\ 3 & 4 & 2 \\ 1 & 3 & 2 \end{bmatrix}, B=\begin{bmatrix} 10 & -4 & -1 \\ -11 & 5 & 0 \\ 9 & -5 & 1 \end{bmatrix}$

$AB=\begin{bmatrix} 1 & 2 & 1 \\ 3 & 4 & 2 \\ 1 & 3 & 2 \end{bmatrix}\begin{bmatrix} 10 & -4 & -1 \\ -11 & 5 & 0 \\ 9 & -5 & 1 \end{bmatrix}$

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

$BA=\begin{bmatrix} 10 & -4 & -1 \\ -11 & 5 & 0 \\ 9 & -5 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 1 \\ 3 & 4 & 2 \\ 1 & 3 & 2 \end{bmatrix}$

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

Hence,

$AB=BA$

$A=\begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}$ then

$A^{3}-35A=$

0%
$\mathrm{A}$
0%
$2\mathrm{A}$
0%
$3\mathrm{A}$
0%
$4\mathrm{A}$

Explanation

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

${ A }^{ 2 }=\begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}\begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}=\begin{bmatrix} 12 & 12 & 12 \\ 12 & 12 & 12 \\ 12 & 12 & 12 \end{bmatrix}$

${ A }^{ 3 }=\begin{bmatrix} 12 & 12 & 12 \\ 12 & 12 & 12 \\ 12 & 12 & 12 \end{bmatrix}\begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}=\begin{bmatrix} 72 & 72 & 72 \\ 72 & 72 & 72 \\ 72 & 72 & 72 \end{bmatrix}$

${ A }^{ 3 }-35A=\begin{bmatrix} 72 & 72 & 72 \\ 72 & 72 & 72 \\ 72 & 72 & 72 \end{bmatrix}-35\begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}$

$=\begin{bmatrix} 72 & 72 & 72 \\ 72 & 72 & 72 \\ 72 & 72 & 72 \end{bmatrix}-\begin{bmatrix} 70 & 70 & 70 \\ 70 & 70 & 70 \\ 70 & 70 & 70 \end{bmatrix}$

$=\begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}$

$A=\left[\begin{array}{ll} 2 & 1\\ 3 & 0 \end{array}\right]$ then

$\mathrm{A}^{2}+2\mathrm{A}+I=$

0%
$\left[\begin{array}{ll} 12 & 4\\ 12 & 4 \end{array}\right]$
0%
$\left[\begin{array}{ll} 12 & -4\\ 4 & 12 \end{array}\right]$
0%
$\left[\begin{array}{ll} 4 & 12\\ 12 & 4 \end{array}\right]$
0%
$\left[\begin{array}{ll} 4 & 12\\ -12 & -4 \end{array}\right]$

Explanation

$A^2+2A+I=(A+I)^2$

Now,

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

$(A+I)^2=A^2+2A+I=\begin{bmatrix} 3 & 1\\ 3 & 1 \end{bmatrix}\begin{bmatrix} 3 & 1\\ 3 & 1 \end{bmatrix}=\begin{bmatrix} 12 & 4\\ 12 & 4 \end{bmatrix}$

Let

$\left[\begin{array}{ll} 2 & -2\\-2 & \ 5 \end{array}\right]=\left[\begin{array}{ll} 1 & 0\\ -1 & 1 \end{array}\right]\left[\begin{array}{ll} 2 & 0\\ 0 & x \end{array}\right]\left[\begin{array}{ll} 1 & -1\\ 0 & 1 \end{array}\right]$ , then the value of

$x$ is

0%
$-3$
0%
$-2$
0%
$0$
0%
$3$

Explanation

Consider,

$\begin{bmatrix} 1 & 0\\ -1 & 1 \end{bmatrix}\begin{bmatrix} 2 & 0\\ 0 & x \end{bmatrix}\begin{bmatrix} 1 & -1\\ 0 & 1 \end{bmatrix}$

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

By given, we have

$\begin{bmatrix} 2 & -2\\ -2 & 2+x \end{bmatrix}=\begin{bmatrix} 2 & -2\\ -2 & 5 \end{bmatrix}$

By equality of above matrices, we have

$2+x=5$

$\therefore x =3$

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

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$A^{T}$
0%
$A^{-1}$
0%
$-A$
0%
$-A^{-1}$

If A and B are two matrices such that

$AB = B$ and

$BA = A$ , then

$A^2 + B^2$ is equal to

0%
$2AB$
0%
$2BA$
0%
$A + B$
0%
$AB$

Explanation

$A^{2}+B^{2}=AA+BB$

$=A(BA)+B(AB)$ (Given

$AB=B \, and \, BA =A$ )

$=(AB)A+(BA)B$

$=BA+AB$

$=A+B$

$A \times \begin{bmatrix} 1 & 1\\ 0 & 2 \end{bmatrix} = [1 \ \ 2],$ then A =

0%
$\left[\dfrac {1}{2} \ \ \ 1\right]$
0%
$\begin{bmatrix} 1 & 2\\ 1 & 0 \end{bmatrix}$
0%
$\begin{bmatrix} 2 & 1\\ 1 & 5 \end{bmatrix}$
0%
$\left[1 \ \ \ \dfrac {1}{2} \right]$

$A$ is a non-singular matrix, then

0%
$A^{-1}$ is symmetric if $A$ is symmeteric
0%
$A^{-1}$ is skew-symmetric if $A$ is symmeteric
0%
$\left| { A }^{ -1 } \right| =\left| A \right|$
0%
$\left| { A }^{ -1 } \right| ={ \left| A \right| }^{ -1 }$

Explanation

Since A is a non singular matrix

$\left| A \right| \neq 0$ , thus

${ A }^{ -1 }$ exists.

Now

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

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

$\Rightarrow { \left( { A }^{ -1 } \right) }^{ -1 }{ A }^{ -1 }={ I }={ \left( A \right) }^{ -1 }{ \left( { A }^{ -1 } \right) }^{ ' }\left( \therefore { A }^{ ' }=A \right)$

$\Rightarrow { \left( { A }^{ -1 } \right) }={ \left( { A }^{ -1 } \right) }^{ ' }$ . So

${ A }^{ -1 }$ is symmetric

Also since

$\left| A \right| \neq 0$

${ A }^{ -1 }$ exists such that

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

$\left| A \right| \left| { A }^{ -1 } \right| =1\quad \therefore \left| AB \right| =\left| A \right| \left| B \right|$

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

$\begin{bmatrix}a\ \ b\end{bmatrix}$ x

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

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$\begin{bmatrix}ax+ay +bx+by \end{bmatrix}$
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$\begin{bmatrix}ax \\ by \end{bmatrix}$
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$\begin{bmatrix}ax +by \end{bmatrix}$
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$\begin{bmatrix}ax\ \ by \end{bmatrix}$

$\left[3x^{2}+10xy+5y^{2} \right]=\begin{bmatrix}x & y \end{bmatrix}A\begin{bmatrix} x\\ y\end{bmatrix}$ , and

${A}$ is a symmetric matrix then

$\mathrm{A}=$

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$\left[\begin{array}{ll} 3 & 10\\ 10 & 5 \end{array}\right]$
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$\left[\begin{array}{ll} 10 & 3\\ 5 & 10 \end{array}\right]$
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$\left[\begin{array}{ll} +3 & -5\\ -5 & +5 \end{array}\right]$
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$\left[\begin{array}{ll} 3 & 5\\ 5 & 5 \end{array}\right]$

Explanation

Given

$\begin{bmatrix} 3{ x }^{ 2 }+10xy & +5{ y }^{ 2 } \end{bmatrix}=\begin{bmatrix} x & y \end{bmatrix}A\begin{bmatrix} x \\ y \end{bmatrix}$

We will check using options
If option A is the matrix A, then consider RHS

$\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} 3 & 10 \\ 10 & 5 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}$

$=\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} 3x+10y \\ 10x+5y \end{bmatrix}$

$=\begin{bmatrix} 3{ x }^{ 2 }+20xy & +5{ y }^{ 2 } \end{bmatrix}$
which is not equal to given LHS

Now, option B, so let

$A=\begin{bmatrix} 10 & 3 \\ 5 & 10 \end{bmatrix}$

$\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} 10 & 3 \\ 5 & 10 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}$

$=\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} 10x+3y \\ 5x+10y \end{bmatrix}$

$=\begin{bmatrix} 3{ x }^{ 2 }+8xy & +5{ y }^{ 2 } \end{bmatrix}$
which is not equal to given LHS

Now, we will try option C
Let

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

$\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} 3 & -5 \\ -5 & 5 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}$

$=\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} 3x-5y \\ -5x+5y \end{bmatrix}$

$=\begin{bmatrix} 3{ x }^{ 2 }-10xy & +5{ y }^{ 2 } \end{bmatrix}$
which is not equal to given LHS

Now, lastly we will try option D.
Let

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

$\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} 3 & 5 \\ 5 & 5 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}$

$=\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} 3x+5y \\ 5x+5y \end{bmatrix}$

$=\begin{bmatrix} 3{ x }^{ 2 }+10xy & +5{ y }^{ 2 } \end{bmatrix}$
which is equal to given LHS

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

$A^2 + B^2 =$

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$2 I$
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$4 I$
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$\begin{bmatrix} 7 & 0\\ 0 & 7 \end{bmatrix}$
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$\begin{bmatrix} 1 & -1\\ 0 & 5 \end{bmatrix}$

CBSE Questions for Class 12 Commerce Maths Matrices Quiz 4 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Matrices Quiz 4 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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