If A = […] and A^3 = kA , then the value of k is

The correct answer is Option (2) → \(-9\)

Multiplying row by column,

    \( \displaystyle A^2 = \begin{bmatrix}0 - 1 - 4&0 + 0 - 4&0 + 2 + 0\\0 + 0 - 4&-1 + 0 - 4&-2 + 0 + 0\\0 + 2 + 0&-2 + 0 + 0&-4 - 4 + 0\end{bmatrix} = \begin{bmatrix}-5&-4&2\\-4&-5&-2\\2&-2&-8\end{bmatrix} \)

Since \(A^3 = A^2 \cdot A = kA\), it is enough to compare one non-zero element. The element in row 1 and column 2 of \(A^2 \cdot A\) uses row 1 of \(A^2\) and column 2 of A:

    \((-5)(1) + (-4)(0) + (2)(-2) = -9\)

The same element of kA is \(k \times 1 = k\). Hence

    \(k = -9\)

Check with row 1, column 3: \((-5)(2) + (-4)(2) + (2)(0) = -18 = -9 \times 2\).

In fact \(A^3 = -(1^2 + 2^2 + 2^2)A = -9A\), where 1, 2, 2 are the elements above the diagonal.