Correct answer is (a) \(\begin{bmatrix}1&5\\0&8\end{bmatrix}\)
Best explanation: Consider the matrix A=\(\begin{bmatrix}1&5\\0&8\end{bmatrix}\)
Using the elementary column operations, we write A=AI
\(\begin{bmatrix}1&5\\0&8\end{bmatrix}\)=A\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)
Applying C2→C2-5C1
\(\begin{bmatrix}1&0\\0&8\end{bmatrix}\)=A\(\begin{bmatrix}1&-5\\0&1\end{bmatrix}\)
Applying C2→C2/8
\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)=A\(\begin{bmatrix}1&-\frac{5}{8}\\0&\frac{1}{8}\end{bmatrix}\)
A^-1=\(\begin{bmatrix}1&-\frac{5}{8}\\0&\frac{1}{8}\end{bmatrix}\).