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:

If a matrix is a scalar multiple of the identity matrix, such as \(A=nI\), then multiplication becomes easier because \(AI=IA=A\).

Answer 1

The Correct Option is B

Step 1: Express matrix \(A\) in terms of identity matrix.
Given, \[ A=\begin{bmatrix} n & 0 & 0\\ 0 & n & 0\\ 0 & 0 & n \end{bmatrix} \] So, \[ A=nI \]

Step 2: Find \(A^2\).
\[ A^2=(nI)^2 \] \[ A^2=n^2I \]

Step 3: Find \(B^2\).
Given, \[ B=\begin{bmatrix} 0 & 0 & n\\ 0 & n & 0\\ n & 0 & 0 \end{bmatrix} \] Multiplying \(B\) by itself, \[ B^2= \begin{bmatrix} 0 & 0 & n\\ 0 & n & 0\\ n & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & n\\ 0 & n & 0\\ n & 0 & 0 \end{bmatrix} \] \[ = \begin{bmatrix} n^2 & 0 & 0\\ 0 & n^2 & 0\\ 0 & 0 & n^2 \end{bmatrix} \] Therefore, \[ B^2=n^2I \]

Step 4: Find \(AB\).
Since \[ A=nI \] we get \[ AB=(nI)B \] \[ AB=nB \]

Step 5: Add \(A^2+B^2+AB\).
\[ A^2+B^2+AB=n^2I+n^2I+nB \] \[ =2n^2I+nB \] Taking \(n\) common, \[ = n(2nI+B) \]

Step 6: Final conclusion.
Therefore, \[ \boxed{n(2nI+B)} \]