Question

if A=$\left[\begin{array}{cc}2& 3\\ 5& -7\end{array}\right]then{\left(A^{{}^{\prime }}\right)}^2=$

• A. $\left[\begin{array}{ccc}5& -7& 12\\ 1& 4& 22\end{array}\right]$
• B. $\left[\begin{array}{cc}1& 17\\ 1& -4\\ 0& 2\end{array}\right]$
• C. $\left[\begin{array}{cc}-19& -25\\ -15& 64\end{array}\right]$
• D. $\left[\begin{array}{cc}19& -25\\ -15& 64\end{array}\right]$

Answer 1

The correct option is D $\left[\begin{array}{cc}19& -25\\ -15& 64\end{array}\right]$

Given,

$A=\left(\begin{array}{cc}2& 3\\ 5& -7\end{array}\right)$

$\Rightarrow A^{\prime }=\left(\begin{array}{cc}2& 5\\ 3& -7\end{array}\right)$

Now,$(A^{\prime }{)}^2$

$={\left(\begin{array}{cc}2& 5\\ 3& -7\end{array}\right)}^2$

$=\left(\begin{array}{cc}2& 5\\ 3& -7\end{array}\right)\left(\begin{array}{cc}2& 5\\ 3& -7\end{array}\right)$

$=\left(\begin{array}{cc}2\cdot 2+5\cdot 3& 2\cdot 5+5(-7)\\ 3\cdot 2+(-7)\cdot 3& 3\cdot 5+(-7)(-7)\end{array}\right)$

$=\left(\begin{array}{cc}19& -25\\ -15& 64\end{array}\right)$