The inverse of the matrix \(\begin{bmatrix}3&5\\1&2\end{bmatrix}\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\begin{bmatrix}2&-5\\-1&3\end{bmatrix}\)
For \(A = \begin{bmatrix}a&b\\c&d\end{bmatrix}\) with \(ad - bc \ne 0\), the inverse is found by swapping a and d, changing the signs of b and c, and dividing by \(ad - bc\):
\( \displaystyle A^{-1} = \frac{1}{ad - bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix} \)
Here
\(ad - bc = 3(2) - 5(1) = 1\)
\( \displaystyle A^{-1} = \begin{bmatrix}2&-5\\-1&3\end{bmatrix} \)
Check: \(AA^{-1}\) has elements \(6 - 5 = 1\), \(-15 + 15 = 0\), \(2 - 2 = 0\), \(-5 + 6 = 1\), so \(AA^{-1} = I\).