(2)

$\text{Here }{A}^n=\left[\begin{array}{cc}2n+1& -4n\\ n& -2n+1\end{array}\right]$

$\Rightarrow A^{15}=\left[\begin{array}{cc}31& -60\\ 15& -29\end{array}\right]$

$\Rightarrow A^{15}+B=\left[\begin{array}{cc}2& -11\\ 2& -11\end{array}\right]$

$\text{Now }(A^{15}+B)\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}2& -11\\ 2& -11\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$

$\Rightarrow 2x-11y=0$

(4)

$B={(I+A)}^{-1},A+C=I$

$\Rightarrow B(I+A)=(I+A)B=I$

$\Rightarrow B+BA=B+AB$

$\Rightarrow B+B(I-C)=B+(I-C)B$

$\Rightarrow 2B-BC=2B-CB$

$\Rightarrow BC=CB$

$\therefore$ $CB\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{cc}1& -5\\ -1& 2\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}12\\ -6\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]={\left[\begin{array}{cc}1& -5\\ -1& 2\end{array}\right]}^{-1}\left[\begin{array}{c}12\\ -6\end{array}\right]=\frac{-1}{3}\left[\begin{array}{cc}2& 5\\ 1& 1\end{array}\right]\left[\begin{array}{c}32\\ -6\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}2\\ -2\end{array}\right]$

$\therefore$ $x_1+x_2=0$

(3)

Ans.     ${\mathit{A}}^{\mathbf{2}}\mathbf{-}{\mathit{B}}^{\mathbf{2}}\mathbf{+}\mathit{B}\mathit{A}\mathbf{-}\mathit{A}\mathit{B}$

Explanation:

$(A+B)(A-B)$$=A(A-B)+B(A-B)$

                           $=A^2-AB+BA-B^2$

(1)

Ans.   Skew-symmetric matrix

Explanation:

Since (AB' - BA')' = (AB')' - (BA')' 

                             = (BA' - AB') 

                             = - (AB' - BA')

(4)

Ans.    4 elements

Explanation:

Since 3 × 4 signifies that there are 3 rows and 4 columns, each row has 4 elements.

(4)

Ans.     Diagonal matrix

Explanation:

A matrix in which all elements are zero except the diagonal elements is called a diagonal matrix.

(2)

Ans.     6, 9, 6

Explanation:

The number of elements in $m\times n$ matrix is equal to $mn$.

$\therefore$   Elements in A = 3 × 2 = 6

$\therefore$   Elements in B = 3 × 3 = 9

$\therefore$   Elements in C = 2 × 3 = 6

(4)

Ans.    4 × 4

Explanation:

We know that if a matrix is of order $m\times n$, then it has $mn$ elements. Thus, to find all possible orders of a matrix with 8 elements, we will find all ordered pairs of natural numbers whose product is 8. Thus, all possible ordered pairs are (1, 8), (2, 4), (4, 2).

(4)

Ans.     512

Explanation:

Number of entries in 3 × 3 matrix is 9. Since each entry has 2 choices, namely 2 or 0, therefore, number of possible matrices 

$=\underset{9 \text{times}}{\underbrace{2\times 2\times 2\times \cdots \times 2}}=2^9=512$

(3)

Ans.    Unit matrix

Explanation:

A square matrix in which all diagonal elements are 1 and all other elements are 0 is called a unit matrix.

(2)

Ans.   not possible to find

Explanation:

Consider,  $\left[\begin{array}{cc}3x+7& 5\\ y+1& 2-3x\end{array}\right]=\left[\begin{array}{cc}0& y-2\\ 8& 4\end{array}\right]$

$\Rightarrow 3x+7=0,5=y-2,y+1=8,2-3x=4$

$\Rightarrow x=\frac{-7}{3},y=7;y=7,x=\frac{-2}{3}$

(4)

Ans.    B + D

Explanation:

Only B + D is defined because only matrices of the same order can be added.

(1)

Ans.   20

Explanation:

We have,

         $\left[\begin{array}{cc}1& 2\\ -2& -b\end{array}\right]+\left[\begin{array}{cc}a& 4\\ 3& 2\end{array}\right]=\left[\begin{array}{cc}5& 6\\ 1& 0\end{array}\right]$

$\Rightarrow$             $\left[\begin{array}{cc}a+1& 6\\ 1& 2-b\end{array}\right]=\left[\begin{array}{cc}5& 6\\ 1& 0\end{array}\right]$

$\Rightarrow$                             $a+1=5,2-b=0$

$\Rightarrow$                                    $a=4,b=2$

$\Rightarrow$                         $a^2+b^2=4^2+2^2=20$

(3)

Ans.   $\mathbf{-}\mathbf{6}\mathbf{,}\mathbf{-}\mathbf{4}\mathbf{,}\mathbf{-}\mathbf{9}$

Explanation:

                       $kA=\left[\begin{array}{cc}0& 3a\\ 2b& 24\end{array}\right]$

$\Rightarrow$      $k\left[\begin{array}{cc}0& 2\\ 3& -4\end{array}\right]=\left[\begin{array}{cc}0& 3a\\ 2b& 24\end{array}\right]$

$\Rightarrow$    $\left[\begin{array}{cc}0& 2k\\ 3k& -4k\end{array}\right]=\left[\begin{array}{cc}0& 3a\\ 2b& 24\end{array}\right]$

$\therefore$                 $-4k=24$

$\Rightarrow$                      $k=-6$

$\Rightarrow$                   $3a=2\times (-6)$

$\Rightarrow$                   $3a=-12$

$\Rightarrow$                      $a=-4$

and                    $3k=2b$

$\Rightarrow$         $3\times (-6)=2b$

$\Rightarrow$                      $b=-9$

(4)

Ans.    $\frac{\mathbf{1}}{\mathbf{2}}\mathbf{I}$

Explanation:

We have,

                $B=\left[\begin{array}{cc}-\frac{{\displaystyle 1}}{{\displaystyle \pi }}{\cos }^{-1}\left(\pi x\right)& \frac{{\displaystyle 1}}{{\displaystyle \pi }}{\tan }^{-1}\left(\frac{{\displaystyle x}}{{\displaystyle \pi }}\right)\\ \frac{{\displaystyle 1}}{{\displaystyle \pi }}{\sin }^{-1}\left(\frac{{\displaystyle x}}{{\displaystyle \pi }}\right)& -\frac{{\displaystyle 1}}{{\displaystyle \pi }}{\tan }^{-1}\left(\pi x\right)\end{array}\right]$

and         $A=\left[\begin{array}{cc}\frac{{\displaystyle 1}}{{\displaystyle \pi }}{\sin }^{-1}\left(\pi x\right)& \frac{{\displaystyle 1}}{{\displaystyle \pi }}{\tan }^{-1}\left(\frac{{\displaystyle x}}{{\displaystyle \pi }}\right)\\ \frac{{\displaystyle 1}}{{\displaystyle \pi }}{\sin }^{-1}\left(\frac{{\displaystyle x}}{{\displaystyle \pi }}\right)& \frac{{\displaystyle 1}}{{\displaystyle \pi }}{\cot }^{-1}\left(\pi x\right)\end{array}\right]$

$\therefore$  $A-B=\left[\begin{array}{cc}\frac{1}{\pi }\left({\sin }^{-1}\right(\pi x)+{\cos }^{-1}(\pi x\left)\right)& 0\\ 0& \frac{1}{\pi }\left({\cot }^{-1}\right(\pi x)+{\tan }^{-1}(\pi x\left)\right)\end{array}\right]$

                    $=\left[\begin{array}{cc}\frac{{\displaystyle 1}}{{\displaystyle \pi }}·\frac{{\displaystyle \pi }}{{\displaystyle 2}}& 0\\ 0& \frac{{\displaystyle 1}}{{\displaystyle \pi }}·\frac{{\displaystyle \pi }}{{\displaystyle 2}}\end{array}\right]=\frac{1}{2}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\frac{1}{2}I$

(4)

Ans.    $\mathbf{3}\mathbf{\times }\mathit{n}$

Explanation:

$A_{3\times m}$ and $B_{3\times n}$ are two matrices. If $m=n$, then A and B have the same order as $3\times n$ each, so the order of (5A − 2B) should also be $3\times n$.

(2)

Ans.    symmetric matrix

Explanation:

We have  $A=\left[\begin{array}{ccc}1& 2& 4\\ 2& 5& 6\\ 4& 6& 7\end{array}\right]$

$\Rightarrow$     $A\text{'}=\left[\begin{array}{ccc}1& 2& 4\\ 2& 5& 6\\ 4& 6& 7\end{array}\right]$

$\because$       $A\text{'}=A$

So, the given matrix is a symmetric matrix.

(3)

Ans.    $\mathit{m}\mathbf{=}\mathit{n}$

Explanation:

It is known that a given matrix is said to be a square matrix if the number of rows is equal to the number of columns.

$\therefore$  $\text{ If }m=n,\text{ then }{\left[a_{ij}\right]}_{m\times n}\text{ is a square matrix.}$

(1)

Ans.    $\mathit{x}\mathbf{=}\mathit{y}$

Explanation:

Let               $A=\left[\begin{array}{cc}5& x\\ y& 0\end{array}\right]$

For a symmetric matrix,

                    $A\text{'}=A$

$\therefore$        $\left[\begin{array}{cc}5& x\\ y& 0\end{array}\right]=\left[\begin{array}{cc}5& y\\ x& 0\end{array}\right]$

$\therefore$              $x=y$

(2)

Ans.    $\frac{\mathbf{\pi }}{\mathbf{3}}$

Explanation:

Let            $A=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$

$\therefore$       $A\text{'}=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]$

$\therefore$$A+A\text{'}=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]+\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]$

$\Rightarrow$$A+A\text{'}=\left[\begin{array}{cc}2\cos \alpha & 0\\ 0& 2\cos \alpha \end{array}\right]$

                    $I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\therefore$$\left[\begin{array}{cc}2\cos \alpha & 0\\ 0& 2\cos \alpha \end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\therefore$                   $2\cos \alpha =1$

$\Rightarrow$                     $\cos \alpha =\frac{1}{2}$

$\Rightarrow$                            $\alpha =\frac{\pi }{3}$

(3)

Ans.       it is not necessary that one of the matrices is a zero matrix

Explanation:

Let  $A=\left[\begin{array}{cc}0& -1\\ 0& 2\end{array}\right]$  and  $B=\left[\begin{array}{cc}3& 5\\ 0& 0\end{array}\right]$

Then,  $AB=0$  but $A$ and $B$ are non-zero.

(4)

Ans.   $\mathit{m}\mathbf{\times }\mathit{n}$

Explanation:

Let     $A={\left[a_{ij}\right]}_{m\times n}$
and    $B={\left[b_{ij}\right]}_{p\times q}$

$\therefore$$B\text{'}={\left[b_{ij}\right]}_{q\times p}$

Now, $AB\text{'}$ is defined, so $n=q$

and $B\text{'}A$ is also defined, so $p=m$

$\therefore$$\text{Order of }B\text{ is }m\times n$

(3)

Ans.        AB and BA both are defined

Explanation:

Let $A={\left[a_{ij}\right]}_{2\times 3}$ and $B={\left[b_{ij}\right]}_{3\times 2}$

Both AB and BA are defined.

(1)

Ans.        [28]

Explanation:

$AB+XY=\left[\begin{array}{ccc}2& -3& 4\end{array}\right]\left[\begin{array}{c}3\\ 2\\ 2\end{array}\right]+\left[\begin{array}{ccc}1& 2& 3\end{array}\right]\left[\begin{array}{c}2\\ 3\\ 4\end{array}\right]$

                 $=\left[8\right]+\left[20\right]$

                 $=\left[28\right]$

(1)

Ans.    $\left[\begin{array}{cc}{a}^2+b^2& 0\\ 0& a^2+b^2\end{array}\right]$

Explanation:

$\left[\begin{array}{cc}a& b\\ -b& a\end{array}\right]\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right]=\left[\begin{array}{cc}{a}^2+b^2& -ab+ba\\ -ba+ab& b^2+a^2\end{array}\right]$

                                $=\left[\begin{array}{cc}{a}^2+b^2& 0\\ 0& a^2+b^2\end{array}\right]$

(1)

Ans.     $\mathbf{I}$

Explanation:

We have,   $A={\left[a_{ij}\right]}_{2\times 2}$

where       $a_{ij}=$$\{\begin{array}{cc}1,& i\ne j\\ 0,& i=j\end{array}$ 

$\therefore$             $A=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$

and           $A^2=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I$

(1)

Ans.        $\left[\begin{array}{cc}1& n\\ 0& 1\end{array}\right]$

Explanation:

Let          $A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$

               $A^2=A\times A$

                     $=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$

                      $=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]$

                 $A^3=A^2\times A$

                      $=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$

                       $=\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]$

                         $\begin{array}{ccc}⋮& ⋮& ⋮\end{array}$

                 $A^n=\left[\begin{array}{cc}1& n\\ 0& 1\end{array}\right]$

(2)

Ans.       1

Explanation :

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

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

$\because$    $A^2=kA-2I$

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

on comparing $3k-2=1$

                $3k=3$$\Rightarrow k=1$

(3)

Ans.     (A), (B), (D) only

Explanation:

(A)             $A={\left[\begin{array}{cc}2& -3\\ 3& 5\end{array}\right]}_{2\times 2}$

(B)            $\left|A\right|$$=\left(2\right)\left(5\right)-(-3)\left(3\right)=10+9=19$

       $\because$     $\left|A\right|$$\ne 0$

        $\therefore$    $A^{-1} \text{exists}$

(C)    For a symmetric matrix,

                   $A=A^T$

         but     $\left[\begin{array}{cc}2& -3\\ 3& 5\end{array}\right]\ne \left[\begin{array}{cc}2& 3\\ -3& 5\end{array}\right]$             (False)

(D)    $\left|A\right|$$=\left[\begin{array}{cc}2& -3\\ 3& 5\end{array}\right]=19$

(E)    The matrix whose every element is zero is called a null matrix. Therefore, A is not a null matrix.

(4)

Ans.    41

Explanation:

For a skew symmetric matrix,

$\Rightarrow$                  $A^{\text{'}}=-A$

$\Rightarrow$    $\left[\begin{array}{ccc}x& 2& -1\\ y& u& 6\\ z& v& w\end{array}\right]=\left[\begin{array}{ccc}-x& -y& -z\\ -2& u& -v\\ 1& -6& -w\end{array}\right]$

Comparing the corresponding elements,

$\Rightarrow$    $y=-2,v=-6,z=1$

          $x=0,u=0,w=0$

$\therefore$  $x^2+y^2+z^2+u^2+v^2+w^2=0+4+1+0+36+0=41$