Question

If $A=\left[\begin{array}{rr}-1 & 2 \\ 2 & -1\end{array}\right]$ and $B=\left[\begin{array}{l}3 \\ 1\end{array}\right]$ AX=B, then X is equal to

(A) $\left[\begin{array}{ll}0 & 7\end{array}\right]$

(B) $\frac{1}{3}\left[\begin{array}{l}5 \\ 7\end{array}\right]$

(C) $\frac{1}{3}\left[\begin{array}{ll}5 & 7]\end{array}\right.$

(D) $\left[\begin{array}{l}5 \\ 7\end{array}\right]$

Answer 1

$\begin{array}{l}\text {} A^{-1}=-\frac{1}{3}\left[\begin{array}{ll}-1 & -2 \\-2 & -1\end{array}\right]=\frac{1}{3}\left[\begin{array}{ll}1 & 2 \\2 & 1\end{array}\right] \\\therefore \quad X=A^{-1} B=\frac{1}{3}\left[\begin{array}{ll}1 & 2 \\2 & 1\end{array}\right]\left[\begin{array}{l}3 \\1\end{array}\right]=\frac{1}{3}\left[\begin{array}{l}5 \\7\end{array}\right]\end{array}$

So, The correct option is (B).