Question

Let A be an invertible matrix and suppose that the inverse of A is $\left[\begin{array}{cc}-1& 2\\ 4& -7\end{array}\right]$. Then the matrix A is

• A. $\left[\begin{array}{cc}1& \frac{2}{7}\\ \frac{4}{7}& \frac{1}{7}\end{array}\right]$
• B. $\left[\begin{array}{cc}7& 2\\ 4& 1\end{array}\right]$
• C. $\left[\begin{array}{cc}1& \frac{-4}{7}\\ \frac{-2}{7}& \frac{1}{7}\end{array}\right]$
• D. $\left[\begin{array}{cc}7& 4\\ 2& 1\end{array}\right]$

Answer 1

The correct option is B $\left[\begin{array}{cc}7& 2\\ 4& 1\end{array}\right]$

$A^{-1}=\left[\begin{array}{cc}-1& 2\\ 4& -7\end{array}\right]$
$So(A^{-1}{)}^{-1}=\frac{1}{(-1)}\left[\begin{array}{cc}-7& -2\\ -4& -1\end{array}\right]$
$[\because \left|A^{-1}\right|=\left|\begin{array}{cc}-1& 2\\ 4& -7\end{array}\right|=7-8=-1]$
$\Rightarrow A=\left[\begin{array}{cc}7& 2\\ 4& 1\end{array}\right]$