Question

List I	List II
A. $\left|\begin{array}{ccc}-2& 2& 2\\ 2& -2& 2\\ 2& 2& -2\end{array}\right|$	I. $\left(\begin{array}{cc}2& 5\\ 5& 8\end{array}\right)$
B. A= $\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right]$ then $A^{-1}$ =	II. $A^T$
C. A= $\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$ then $A+A^T$ =	III. $\left[\begin{array}{cc}5& 4\\ -4& 1\end{array}\right]$
D. A=$\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$ , $A^2=A+B$ then $B$	IV. 32

Match the following with A, B, C, D

• A. IV, II, I, III
• B. II, IV, III, IV
• C. III, II, IV, I
• D. I, II, III, IV

Answer 1

Answer: (A)

IV, II, I, III

A) $\left|\begin{array}{ccc}-2& 2& 2\\ 2& -2& 2\\ 2& 2& -2\end{array}\right|$
Here, $|A|=0-2(-8)+2\left(8\right)=32$
B) $A=\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right]$
Here, $|A|=-1$
$adjA=C^T={\left[\begin{array}{ccc}0& -1& 0\\ -1& 0& 0\\ 0& 0& -1\end{array}\right]}^T$
$\Rightarrow adjA=\left[\begin{array}{ccc}0& -1& 0\\ -1& 0& 0\\ 0& 0& -1\end{array}\right]$
$\Rightarrow A^{-1}=\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right]$
$\Rightarrow A^{-1}=A=A^T$
C) $A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$
$A^T=\left[\begin{array}{cc}1& 3\\ 2& 4\end{array}\right]$
$A+A^T=\left[\begin{array}{cc}2& 5\\ 5& 8\end{array}\right]$
D) Given $A=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$
$A^2=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$
$\Rightarrow A^2=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]$
Also, $A^2=A+B$
$\Rightarrow A^2-A=B$
$B=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]-\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$
$\Rightarrow B=\left[\begin{array}{cc}5& 4\\ -4& 1\end{array}\right]$