<p>If <span>\(A = \begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix}\)</span> and <span>\(B = \begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix}\)</span> such that \(A^2 = B\) then</p>
Step-by-Step Solution
Key Concept: Compute A² by matrix multiplication and equate it element-wise to B to find α. The (1,1) entry gives α² = 1, while the (2,1) entry gives α + 1 = 5, which determines α uniquely.
<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>= <begin>bmatrix</begin> α² & 0 \\ α+1 & 1 <end>bmatrix</end></p><p><strong>Step 2:</strong> Equate A² = B element-wise.</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><strong>Step 3:</strong> From (1,1) entry: α² = 1 ⟹ α = ±1</p><p><strong>Step 4:</strong> From (2,1) entry: α + 1 = 5 ⟹ α = 4</p><p><strong>Step 5:</strong> Since α must satisfy both conditions simultaneously and they are contradictory, <strong>no real value of α satisfies A² = B</strong>.</p><p>∴ Answer: D (No solution exists)</p>
Correct Answer: D