If $\begin{bmatrix}a&b^3\\2&0\end{bmatrix}=\begin{bmatrix}1&8\\2&0\end{bmatrix}$, then $\begin{bmatrix}a&b\\2&0\end{bmatrix}^{-1}=$
Answer & explanation
Correct answer: option 4
We have,
$\begin{bmatrix}a&b^3\\2&0\end{bmatrix}=\begin{bmatrix}1&8\\2&0\end{bmatrix}⇒a=1, b^3 =8⇒ a=1,b=2$.
$∴\begin{bmatrix}a&b\\2&0\end{bmatrix}=\begin{bmatrix}1&2\\2&0\end{bmatrix}$
Hence, $\begin{bmatrix}a&b\\2&0\end{bmatrix}^{-1}=-\frac{1}{4}\begin{bmatrix}0&-2\\-2&1\end{bmatrix}=\begin{bmatrix}0&1/2\\1/2&-1/4\end{bmatrix}$