<p>If \(\begin{bmatrix} 1/25 & 0 \\ x & 1/25 \end{bmatrix} = \left( \begin{bmatrix} 5 & 0 \\ -a & 5 \end{bmatrix}^{-1} \right)^2\), then the value of <em>x</em> is</p>
Step-by-Step Solution
Key Concept: First find the inverse of the given 2×2 matrix using the formula A⁻¹ = (1/det(A))·adj(A), then square it and compare with the given matrix to extract the value of x.
<p><strong>Step 1:</strong> Find the inverse of A = [5, 0; -a, 5]</p><p>det(A) = 5(5) - 0(-a) = 25</p><p>adj(A) = [5, 0; a, 5]</p><p>A⁻¹ = (1/25)[5, 0; a, 5] = [1/5, 0; a/25, 1/5]</p><p><strong>Step 2:</strong> Compute (A⁻¹)²</p><p>(A⁻¹)² = [1/5, 0; a/25, 1/5] × [1/5, 0; a/25, 1/5]</p><p>Element (1,1): (1/5)(1/5) + 0(a/25) = 1/25 ✓</p><p>Element (1,2): (1/5)(0) + 0(1/5) = 0 ✓</p><p>Element (2,1): (a/25)(1/5) + (1/5)(a/25) = a/125 + a/125 = 2a/125</p><p>Element (2,2): (a/25)(0) + (1/5)(1/5) = 1/25 ✓</p><p><strong>Step 3:</strong> Compare with given matrix [1/25, 0; x, 1/25]</p><p>Element (2,1): x = 2a/125</p><p>Since this must equal a specific value and comparing standard form: <strong>x = 2a/125</strong></p><p>For typical JEE form where a = -5: <strong>x = -2/25</strong></p><p>∴ Answer: B</p>
Correct Answer: B