If \[ A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -3 \end{bmatrix} \] then $ A^2 + B^2 = \underline{\hspace{2cm}} $

Question

For the matrices $ A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} $ and $ B = \begin{bmatrix} -29 & 49 \\ -13 & 18 \end{bmatrix} $, if $ (A^{15} + B) \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix} $, then among the following which one is true?}

Answer 1

Step 1: Understanding the Concept:
When dealing with diagonal matrices, the square of the matrix is simply the square of each diagonal element.
Step 2: Formula Application:
$A^2 = \text{diag}(2^2, 3^2, 4^2) = \text{diag}(4, 9, 16)$ $B^2 = \text{diag}(1^2, (-2)^2, (-3)^2) = \text{diag}(1, 4, 9)$
Step 3: Explanation:
Adding the corresponding elements: $A^2 + B^2 = \text{diag}(4+1, 9+4, 16+9) = \text{diag}(5, 13, 25)$.
Step 4: Final Answer:
The result is option (c).

If \[ A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -3 \end{bmatrix} \] then \( A^2 + B^2 = \underline{\hspace{2cm}} \)

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The Correct Option is C

Solution and Explanation

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