<p>If \( A = \begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix} \), then the value of \( \alpha \) for which \( A^2 = B \) is</p>
Step-by-Step Solution
Key Concept: Compute A² by matrix multiplication and equate it element-by-element to B. The (1,1) entry gives α², while the (2,1) entry provides the constraint that must be satisfied.
<p><strong>Step 1:</strong> Compute A²</p><p>A² = <begin>bmatrix</begin> α & 0 \\ 1 & 1 <end>bmatrix</end> × <begin>bmatrix</begin> α & 0 \\ 1 & 1 <end>bmatrix</end></p><p>A² = <begin>bmatrix</begin> α·α + 0·1 & α·0 + 0·1 \\ 1·α + 1·1 & 1·0 + 1·1 <end>bmatrix</end> = <begin>bmatrix</begin> α² & 0 \\ α+1 & 1 <end>bmatrix</end></p><p><strong>Step 2:</strong> Equate A² = B</p><p><begin>bmatrix</begin> α² & 0 \\ α+1 & 1 <end>bmatrix</end> = <begin>bmatrix</begin> 1 & 0 \\ 5 & 1 <end>bmatrix</end></p><p>From (1,1) element: α² = 1 ⟹ α = ±1</p><p>From (2,1) element: α + 1 = 5 ⟹ α = 4</p><p><strong>Step 3:</strong> Resolve contradiction</p><p>Since α = ±1 contradicts α = 4, there is <strong>no real value</strong> of α satisfying A² = B simultaneously for all elements.</p><p>∴ Answer: D (No solution exists)</p>
Correct Answer: D