<p>If <span>\(A \times \begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix} = 6I\)</span> where \(I\) is a unit matrix of order \(2 \times 2\), then \(A\) is</p>
<p>(a) \(\begin{bmatrix} 2 & 4 \\ 1 & -1 \end{bmatrix}\)</p>
<p>(b) \(\begin{bmatrix} -1 & 1 \\ 4 & 2 \end{bmatrix}\)</p>
<p>(c) \(\begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix}\)</p>
<p>(d) none of these</p>
Step-by-Step Solution
Key Concept: Isolate matrix A by multiplying both sides by the inverse of the given matrix on the right: A = 6I × [given matrix]⁻¹. The determinant of the 2×2 matrix is needed to find its inverse efficiently.
<p><strong>Step 1:</strong> Given: A × B = 6I, where B = [1, -2; 1, 4]</p><p><strong>Step 2:</strong> Multiply both sides by B⁻¹ on the right: A = 6I × B⁻¹ = 6B⁻¹</p><p><strong>Step 3:</strong> Find B⁻¹. For B = [1, -2; 1, 4], det(B) = 1(4) - (-2)(1) = 4 + 2 = 6</p><p><strong>Step 4:</strong> B⁻¹ = (1/6)[4, 2; -1, 1]</p><p><strong>Step 5:</strong> A = 6 × (1/6)[4, 2; -1, 1] = [4, 2; -1, 1]</p><p><strong>Verification:</strong> [4, 2; -1, 1] × [1, -2; 1, 4] = [4+2, -8+8; -1+1, 2+4] = [6, 0; 0, 6] = 6I ✓</p><p>∴ Answer: C</p>
Correct Answer: C