Question

Let \(A = \begin{bmatrix}n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n\end{bmatrix}\) and \(B = \begin{bmatrix}0 & 0 & n \\ 0 & n & 0 \\ n & 0 & 0\end{bmatrix}\). Then \(A^2 + B^2 + AB =\)

• A. \(n^2 I\)
• B. \(n^2 B\)
• C. \(n^2 (2I + B)\)
• D. \(n^2 (I + B)\)

Hint:

Matrix Operations}
Diagonal matrices are easy to exponentiate
Permutation matrices cycle their entries when multiplied
Factor out common terms for simplified expressions

Answer 1

The Correct Option is C

Step 1: Compute \(A^2\): \[ A^2 = \begin{bmatrix}n^2 & 0 & 0 \\ 0 & n^2 & 0 \\ 0 & 0 & n^2\end{bmatrix} = n^2 I \] Step 2: Compute \(B^2\): \[ B^2 = \begin{bmatrix}n^2 & 0 & 0 \\ 0 & n^2 & 0 \\ 0 & 0 & n^2\end{bmatrix} = n^2 I \] Step 3: Compute \(AB\): \[ AB = \begin{bmatrix}0 & 0 & n^2 \\ 0 & n^2 & 0 \\ n^2 & 0 & 0\end{bmatrix} = n^2 B \] Step 4: Combine results: \[ A^2 + B^2 + AB = n^2 I + n^2 I + n^2 B = n^2 (2I + B) \]