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

$$A = \begin{bmatrix}x & y & z\end{bmatrix}$$, $$B = \begin{bmatrix} a  & h & g \\ h & b & f \\ g & f & c\end{bmatrix}$$, $$C = \begin{bmatrix}x \\ y \\ z\end{bmatrix}$$
$$AB=\begin{bmatrix}ax+hy+gz & hx+by+z & gx+fy+cz\end{bmatrix}$$
$$ABC=\begin{bmatrix}ax+hy+gz & hx+by+z & gx+fy+cz\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}$$

Given, 

Comparing LHS and RHS gives us 

$$\displaystyle \left[ \begin{matrix} 1 & x & 1 \end{matrix} \right] \begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{bmatrix}\: \begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix}=O$$
$$\Rightarrow \begin{bmatrix} 16+2x & \quad 6+5x & \quad 4+x \end{bmatrix}\begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix}=O\\ \Rightarrow { x }^{ 2 }+16x+28\quad =\quad 0\\ \therefore \quad x=-2\quad or\quad x=-4$$

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

Order of $$A$$ is $$3\times2$$
Order of $$B$$ is $$3\times3$$

Order of $$A$$ and $$B$$ are not same. So , $$A+B$$ does not exists.

Again , $$AB$$ also does not exists as the number of columns of $$A$$ are not equal to the number of rows of $$ B.$$

Here, $$BA$$ exists as the number of columns of $$B$$ is equal to the number of rows of $$A$$.

$$\displaystyle A = \begin{bmatrix} 0 & c & -b \\ -c & 0 & a \\ b & -a & 0 \end{bmatrix}$$ and $$\displaystyle B = \begin{bmatrix} a^2 & ab & ac \\ ba & b^2 & bc \\ ca & cb & c^2 \end{bmatrix}$$

Now, $$AB=\begin{bmatrix} 0 & c & -b \\ -c & 0 & a \\ b & -a & 0 \end{bmatrix}\begin{bmatrix} a^{ 2 } & ab & ac \\ ba & b^{ 2 } & bc \\ ca & cb & c^{ 2 } \end{bmatrix}$$

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

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

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

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

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

$$\Rightarrow A^{2}=-I$$

$$A = \begin{bmatrix}4 & x+2 \\ 2x-3 & x+1\end{bmatrix}$$
$$A^{T}=\begin{bmatrix}4 & 2x-3 \\ x+2 & x+1\end{bmatrix}$$
$$A$$ is symmetric.
$$\therefore\quad A=A^{T}$$
$$\Rightarrow 2x-3=x+2$$
$$\therefore  x=5$$
Hence, option B.

Given, $$\displaystyle A = \begin{bmatrix} 4 & -1 & -4 \\ 3 & 0 & -4 \\ 3 & -1 & -3 \end{bmatrix}$$

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

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

$$\displaystyle A \times  \begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}$$

We will check using options
Option A
Consider, $$A\begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix}$$
         $$=\begin{bmatrix} 2 & 4 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix}$$

    $$=\begin{bmatrix} 6 & 12 \\ 0 & -6 \end{bmatrix}$$
    $$\ne RHS$$

Hence, option $$A$$ cannot be matrix $$A$$.

Option B, 
Consider, $$A\begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix}$$
        $$\begin{bmatrix} -1 & 1 \\ 4 & 2 \end{bmatrix}\begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix}$$

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

        $$\ne RHS$$
Hence, option $$B$$ cannot be matrix $$A$$.

Option C,
Consider, $$A\begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix}$$

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

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

      $$= RHS$$
Hence, option C is correct for $$A$$.

Given : $$\displaystyle A=\begin{bmatrix}0 &c  &-b \\-c  &0  &a \\b  &-a  &0 \end{bmatrix}$$ $$\displaystyle B=\begin{bmatrix}a^{2} &ab  &ac \\ab  &b^{2}  &bc \\ac  &bc  &c^{2} \end{bmatrix}$$

Given

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

$$\therefore

A^2 = A.A = \begin{pmatrix}1& 2 & 2 \\ 2 & 1 & 2 \\ 2

& 2 & 1\end{pmatrix}\times\begin{pmatrix}1& 2 & 2 \\ 2

& 1 & 2 \\ 2 & 2 & 1\end{pmatrix}$$

$$\quad      

                            = \begin{pmatrix}1+4+4 & 2+2+4 &

2+4+2 \\ 2+2+4 & 4+1+4 & 4+2+2 \\ 2+4+2 & 4+2+2 &

4+4+1 \end{pmatrix}$$ 

$$\quad                                

  = \begin{pmatrix}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8

& 9\end{pmatrix}$$

$$\therefore \quad A^2 - 4A 

= \begin{pmatrix}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8

& 9\end{pmatrix} - 4\begin{pmatrix}1& 2 & 2 \\ 2 & 1

& 2 \\ 2 & 2 & 1\end{pmatrix}$$

$$\quad                              

                       = \begin{pmatrix}9&8&8 \\ 8&9&8

\\ 8&8&9\end{pmatrix} - \begin{pmatrix}4&8&8 \\

8&4&4 \\ 4&4&8\end{pmatrix} =

\begin{pmatrix}5&0&0 \\ 0&5&0 \\

0&0&5\end{pmatrix}=5I$$

$$\displaystyle A = \begin{bmatrix}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{bmatrix}$$
$$\displaystyle A^2=\begin{bmatrix}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{bmatrix}\begin{bmatrix}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$$
$$\therefore A^2=I$$
Hence, option D.

Consider, $$(A-A')'=A'-(A')'$$
               $$=A'-A$$
               $$=-(A-A')$$
$$\Rightarrow (A-A')'=-(A-A')$$
Hence, $$A-A'$$ is skew-symmetric

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

$$AB=\begin{bmatrix} 2 & 1 & 4 \\ -4 & -2 & -8 \\ -6 & -3 & -12 \end{bmatrix}$$

$$(AB)A=\begin{bmatrix} 2 & 1 & 4 \\ -4 & -2 & -8 \\ -6 & -3 & -12 \end{bmatrix}\begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix}$$

$$\Rightarrow (AB)A=\begin{bmatrix} 12 \\ -24 \\ -36 \end{bmatrix}=-12\begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix}=-12A$$

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

$$\Rightarrow \quad Write \space A A^{-1}= I$$

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

$$\quad \begin{matrix}R_{21}(-2)\\ \mbox{~}\\ R_{31}(1)\end{matrix}\begin{bmatrix}1 & 0 & 5 \\ 2 & 3 & 1 \\ -1 & 1 & 1\end{bmatrix}A^{-1} = \begin{bmatrix}1& 0 & 0 \\ -2 & 1 & 0 \\ 1 & 0 & 1\end{bmatrix}$$

$$\quad \begin{matrix}R_2(-1) \\ \mbox{~} \\ R_3(1/3)\end{matrix}\begin{bmatrix}1 & 2 & 5 \\ 0 & 1 & 9 \\ 0 & 1 & 2\end{bmatrix}A^{-1} = \begin{bmatrix}1& 0 & 0 \\ 2 & -1 & 0 \\ \displaystyle\frac{1}{3} & 0 & \displaystyle\frac{1}{3}\end{bmatrix}$$

$$\quad \begin{matrix}R_{12}(-2) \\ \mbox{~} \\ R_{32}(-1)\end{matrix}\begin{bmatrix}1 & 0 & -13 \\ 0 & 1 & 9 \\ 0 & 0 & -7\end{bmatrix} A^{-1}= \begin{bmatrix}-3 & 2 & 0 \\ 2 & -1 & 0 \\ -\displaystyle\frac{5}{3} & 1 & \displaystyle\frac{1}{3}\end{bmatrix}$$

$$\quad \begin{matrix}R_3(-1/7)\\ \mbox{~}\end{matrix}\begin{bmatrix}1 & 0 & -13 \\ 0 & 1 & 9 \\ 0 & 0 & 1\end{bmatrix}A^{-1} = \begin{bmatrix}-3 & 2 & 0 \\ 2 & -1 & 0 \\ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$$

$$\quad \begin{matrix}R_{13}(13) \\ \mbox{~} \\ R_{23}(-9)\end{matrix}\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}A^{-1} = \begin{bmatrix}\displaystyle\frac{2}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{13}{21} \\ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\\ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$$

Hence, $$\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{2}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{13}{21} \\ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\\ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$$

$$\displaystyle A=\begin{pmatrix}0 &a  &b \\-a  &0  &c \\-b  &-c  &0 \end{pmatrix}$$
$$\displaystyle \therefore  A^{T}=\begin{pmatrix}0 &-a  &-b \\a  &0  &-c \\b  &c  &0 \end{pmatrix}$$
$$\displaystyle A^{T}=-A$$
$$\therefore $$ Assertion (A) & Reason (R) both are true & Reason(R) is correct explanation of Assertion (A).

$$\displaystyle { A }^{ 2 }=\begin{bmatrix} 1 & \frac { 1 }{ 2 }  \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & \frac { 1 }{ 2 }  \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\\ { A }^{ 4 }=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}\\ { A }^{ 8 }=\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 4 \\ 0 & 1 \end{bmatrix}$$

Let $$A=\begin{bmatrix} a & \alpha  & \beta  \\ -\alpha  & b & \gamma  \\ -\beta  & -\gamma  & c \end{bmatrix}\quad \Rightarrow \quad -A=\begin{bmatrix} -a & -\alpha  & -\beta  \\ \alpha  & -b & -\gamma  \\ \beta  & \gamma  & -c \end{bmatrix}$$

$$\displaystyle A^{2}=AA=\begin{pmatrix}a &b \\b  &a \end{pmatrix}\begin{pmatrix}a &b \\b  &a \end{pmatrix}$$ $$\displaystyle =\begin{pmatrix}a^{2}+b^{2} &2ab \\2ab  &a^{2}+b^{2} \end{pmatrix}$$ $$\displaystyle =
\begin{pmatrix} \alpha &\beta \\\beta  &\alpha \end{pmatrix}$$

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

Given $$A$$ be a symmetric matrix such that $$A^5 =0$$ and $$B=I +A + A^2 +A^3 +A^4$$
$$B^{-1}=(I+A+A^2+A^3+A^4)^{-1}=I'+A^{-1}+(A^{-1})^2+(A^{-1})^3+(A^{-1})^4=I +A + A^2 +A^3 +A^4=B$$
$$\therefore B$$ is symmetric.
$$BA=(I +A + A^2 +A^3 +A^4)A=A + A^2 +A^3 +A^4+A^5=B-I$$
$$\Rightarrow BA=B-I$$
$$\Rightarrow B(I-A)=I \Rightarrow B^{-1}=I-A$$
$$B^{-1}$$ exists.
$$\therefore B$$ is non-singualr.
Hence, options A and C.

$$A =\begin{bmatrix}\alpha &\beta  \\\gamma  &-\alpha  \end{bmatrix}$$
$$A^2=\begin{bmatrix}\alpha^2+\beta \gamma  &\alpha \beta-\alpha \beta  \\\alpha \gamma - \alpha \gamma&\beta \gamma +\alpha^2  \end{bmatrix}$$
but, $$A^2=I$$
$$\Rightarrow \alpha ^2+\beta \gamma = 1$$
$$\Rightarrow 1 -\alpha ^2-\beta \gamma=0$$
Hence, option B is correct. 

$$A =\begin{bmatrix} ab&b^2 \\-a^2 &-ab \end{bmatrix}$$
$$A^2=\begin{bmatrix} ab&b^2 \\-a^2 &-ab \end{bmatrix}\begin{bmatrix} ab&b^2 \\-a^2 &-ab \end{bmatrix}$$
$$=\begin{bmatrix} a^{ 2 }b^{ 2 }-a^{ 2 }b^{ 2 } & ab^{ 3 }-ab^{ 3 } \\ -a^{ 3 }b+a^{ 3 }b & -a^{ 2 }b^{ 2 }+a^{ 2 }b^{ 2 } \end{bmatrix}=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$
$$\therefore  A^2=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} =O$$
Hence, option A.

$$\left[ \begin{matrix} 3\quad  & -2 \\ 3 & \quad 0 \\ 2 & \quad 4 \end{matrix} \right] \begin{bmatrix} y\quad  & y \\ x\quad  & x \end{bmatrix}=\left[ \begin{matrix} 3\quad  & 3 \\ 3y\quad  & 3y \\ 10\quad  & 10 \end{matrix} \right] \\ \Rightarrow \left[ \begin{matrix} 3y-2x\quad  & 3y-2x \\ 3y\quad  & 3y \\ 2y+4x\quad  & 2y+4x \end{matrix} \right] =\left[ \begin{matrix} 3 & \quad 3 \\ 3y\quad  & 3y \\ 10\quad  & 10 \end{matrix} \right] $$
$$\therefore 3y-2x=3$$ and $$2y+4x=10$$
$$\displaystyle \Rightarrow x=\frac { 3 }{ 2 } ,y=2$$

If $$A$$ is a square matrix, then and if $$A'$$ represents its transpose, then
$$A+A'$$ is symmetric and $$A-A'$$ is skew symmetric.
Hence matrix A can be written as
$$A=(\dfrac{A+A'}{2})+(\dfrac{A-A'}{2})$$

Therefore of all the above matrix,
$$A-A'$$ is not symmetric.

All positive odd integral power of a skew-symmetric matrix are skew-symmetric and positive even integral power of a skew-symmetric matrix are symmetric.