Question

If $A=\left[\begin{array}{cc}2& 3\\ 5& -2\end{array}\right]$ be such that $A^{-1}=kA,$ then $k$ is equal to

• A. $19$
• B. $\frac{1}{19}$
• C. $-19$
• D. $-\frac{1}{19}$

Answer 1

Answer: (B)

Given, $A=\left[\begin{array}{cc}2& 3\\ 5& -2\end{array}\right]$
$\therefore A^{-1}=\frac{1}{-4-15}\left[\begin{array}{cc}-2& -3\\ -5& 2\end{array}\right]$
$=\frac{-1}{19}\left[\begin{array}{cc}-2& -3\\ -5& 2\end{array}\right]=\frac{1}{19}\left[\begin{array}{cc}2& 3\\ 5& -2\end{array}\right]$
[multiplying $-1$ each element of a matrix]
$=\frac{1}{19}A$
Given, $A^{-1}=kA$
$\therefore k=\frac{1}{19}$