If A = […] and A2 − 10A = C then C is equal to

The correct answer is Option (2) → \(\begin{bmatrix} -8 & -8 \\ -2 & -9 \end{bmatrix}\)

By matrix multiplication (row by column):

    \( A^2 = \begin{bmatrix} 6 & -8 \\ -2 & 5 \end{bmatrix}\begin{bmatrix} 6 & -8 \\ -2 & 5 \end{bmatrix} \)

    \( = \begin{bmatrix} 36 + 16 & -48 - 40 \\ -12 - 10 & 16 + 25 \end{bmatrix} \)

    \( = \begin{bmatrix} 52 & -88 \\ -22 & 41 \end{bmatrix} \)

    \( 10A = \begin{bmatrix} 60 & -80 \\ -20 & 50 \end{bmatrix} \)

So

    \( C = A^2 - 10A = \begin{bmatrix} 52 - 60 & -88 + 80 \\ -22 + 20 & 41 - 50 \end{bmatrix} \)

    \( = \begin{bmatrix} -8 & -8 \\ -2 & -9 \end{bmatrix} \)

Shortcut: By the Cayley-Hamilton theorem, a 2 × 2 matrix satisfies \(A^2 - (\mathrm{tr}\,A)A + |A|\,I = O\). Here tr A = 6 + 5 = 11 and |A| = 30 − 16 = 14, so

    \(A^2 = 11A - 14I\)
    \(A^2 - 10A = A - 14I\)

    \( = \begin{bmatrix} 6 - 14 & -8 \\ -2 & 5 - 14 \end{bmatrix} = \begin{bmatrix} -8 & -8 \\ -2 & -9 \end{bmatrix} \)