Matrices & Determinants
Square root of a matrix
Grade 12
Question:
<p>If \(\begin{bmatrix}\alpha & \beta\\ \gamma & -\alpha\end{bmatrix}\) is to be the square root of two-rowed unit matrix, then \(\alpha,\ \beta,\) and \(\gamma\) should satisfy the relation</p>
<p>(1) \(1 - \alpha^2 + \beta\gamma = 0\)</p>
<p>(2) \(\alpha^2 + \beta\gamma - 1 = 0\)</p>
<p>(3) \(1 + \alpha^2 + \beta\gamma = 0\)</p>
<p>(4) \(1 - \alpha^2 - \beta\gamma = 0\)</p>
Step-by-Step Solution
Key Concept: If matrix M is a square root of the identity matrix I, then M² = I. Multiply the given matrix by itself and equate the resulting matrix to I to find the constraint relations.
<p><strong>Step 1:</strong> Let M = ⎡α β⎤. For M to be a square root of I₂, we need M² = I₂.</p><p><strong>Step 2:</strong> Compute M²:</p><p>M² = ⎡α β⎤ · ⎡α β⎤ = ⎡α² + βγ αβ - αβ⎤ = ⎡α² + βγ β(α - α)⎤</p><p> ⎣γ -α⎦ ⎣γ -α⎦ ⎣γα - γα γβ + α²⎦ ⎣0 γβ + α²⎦</p><p><strong>Step 3:</strong> Set M² = ⎡1 0⎤:</p><p> ⎣0 1⎦</p><p>From (1,1) entry: α² + βγ = 1</p><p>From (1,2) entry: β(α - α) = 0 ✓ (automatically satisfied)</p><p>From (2,2) entry: γβ + α² = 1</p><p>From (2,1) entry: γα - γα = 0 ✓ (automatically satisfied)</p><p><strong>Step 4:</strong> Both non-trivial equations give the same constraint:</p><p><strong>∴ Answer: α² + βγ = 1</strong></p>
Correct Answer: A