If \(A = \begin{bmatrix}2&0&0\\0&-1&0\\0&0&3\end{bmatrix}\), then \(A^3\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\begin{bmatrix}8&0&0\\0&-1&0\\0&0&27\end{bmatrix}\)
A is a diagonal matrix. Multiplying two diagonal matrices of the same order gives the diagonal matrix of the products of the diagonal elements:
\( \displaystyle A^2 = \begin{bmatrix}4&0&0\\0&1&0\\0&0&9\end{bmatrix} \)
\( \displaystyle A^3 = A^2A = \begin{bmatrix}8&0&0\\0&-1&0\\0&0&27\end{bmatrix} \)
So each diagonal element is simply cubed: \(2^3 = 8\), \((-1)^3 = -1\), \(3^3 = 27\).