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. \( n(nI + nB + B) \)
• B. \( n(2nI + B) \)
• C. \( n^2(2I + B) \)
• D. \( n(nI + nA + B) \)

Hint:

Matrix Operations}
Diagonal matrices (like A) are straightforward to square - just square each diagonal element
Permutation matrices (like B) cycle their entries when multiplied
The identity matrix \( I \) acts as 1 in matrix multiplication
Always verify matrix dimensions before multiplication

Answer 1

The Correct Option is C

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