Question

If $\left[\begin{array}{cc}x& y^3\\ 2& 0\end{array}\right]=\left[\begin{array}{cc}1& 8\\ 2& 0\end{array}\right]$, then ${\left[\begin{array}{cc}x& y\\ 2& 0\end{array}\right]}^{-1}$ is equal to

• A. $-\frac{1}{4}$$\left[\begin{array}{cc}0& -2\\ -2& 1\end{array}\right]$
• B. $\frac{2}{4}$$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
• C. $\frac{1}{4}$$\left[\begin{array}{cc}0& -8\\ -2& 1\end{array}\right]$
• D. $\frac{1}{4}\left[\begin{array}{cc}1& 4\\ 7& 2\end{array}\right]$
  

Answer 1

The correct option is A $-\frac{1}{4}$$\left[\begin{array}{cc}0& -2\\ -2& 1\end{array}\right]$