If A= […] then A^-1 is:

The correct answer is Option (3) → \(\displaystyle\frac{1}{32}\begin{bmatrix} -1 & -6 \\ 5 & -2 \end{bmatrix}\)

The determinant:

    \(|A| = (-2)(-1) - (6)(-5)\)
    \(= 2 + 30 = 32\)

For a 2 × 2 matrix, the adjoint is found by swapping the diagonal entries and changing the signs of the other two:

    \( \displaystyle \mathrm{adj}\,A = \begin{bmatrix} -1 & -6 \\ 5 & -2 \end{bmatrix} \)

Then \(A^{-1} = \frac{1}{|A|}\,\mathrm{adj}\,A\):

    \( \displaystyle A^{-1} = \frac{1}{32}\begin{bmatrix} -1 & -6 \\ 5 & -2 \end{bmatrix} \)

Shortcut: Since |A| = 32, only Options (2) and (3) can be right. Option (2) keeps the off-diagonal entries 6 and −5 with their original signs, which is wrong for an adjoint.