<p>Let \(A\) be a matrix such that \(A\begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}\) is a scalar matrix and \(|3A| = 108\). Then \(A^2\) equals</p>
<p>\(\begin{bmatrix} 4 & -32 \\ 0 & 36 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 36 & 0 \\ -32 & 4 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 4 & 0 \\ -32 & 36 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 36 & -32 \\ 0 & 4 \end{bmatrix}\)</p>
Step-by-Step Solution
Key Concept: If AB is scalar, then A and B must commute and have special structure. Use the scalar matrix condition to determine A's form, then apply the determinant condition |3A| = 108 to find the scalar value.
<p><strong>Step 1:</strong> Let A = [a b; c d]. Since A·[1 2; 0 3] = k·I (scalar matrix), we have:</p><p>[a b; c d]·[1 2; 0 3] = [a 2a+3b; c 2c+3d] = [k 0; 0 k]</p><p><strong>Step 2:</strong> From the scalar matrix condition:</p><p>• a = k, c = 0</p><p>• 2a + 3b = 0 ⟹ b = -2k/3</p><p>• 2c + 3d = k ⟹ d = k/3</p><p>So A = [k, -2k/3; 0, k/3]</p><p><strong>Step 3:</strong> Apply |3A| = 108:</p><p>|3A| = 3²|A| = 9|A| = 108 ⟹ |A| = 12</p><p><strong>Step 4:</strong> Calculate |A| = k·(k/3) - 0 = k²/3 = 12 ⟹ k² = 36 ⟹ k = 6</p><p>Therefore A = [6, -4; 0, 2]</p><p><strong>Step 5:</strong> Compute A²:</p><p>A² = [6, -4; 0, 2]·[6, -4; 0, 2] = [36, -32; 0, 4]</p><p>∴ Answer: D</p>
Correct Answer: D