Matrices & Determinants
Matrix multiplication
Grade 12
Question:
<p>Let \(A = \begin{bmatrix} a & b \\ b & a \end{bmatrix}\) and \(A^2 = \begin{bmatrix} \alpha & \beta \\ \beta & \alpha \end{bmatrix}\). Then which of the following is correct?</p>
<p>\(\alpha = a^2 + b^2,\ \beta = ab\)</p>
<p>\(\alpha = a^2 + b^2,\ \beta = 2ab\)</p>
<p>\(\alpha = 2ab,\ \beta = a^2 + b^2\)</p>
<p>\(\alpha = a^2 - b^2,\ \beta = 2ab\)</p>
Step-by-Step Solution
Key Concept: When a symmetric matrix has the form with equal diagonal and equal off-diagonal elements, squaring it preserves this structure. Computing A² directly: the diagonal elements become a² + b² and off-diagonals become 2ab, revealing the relationship between (a,b) and (α,β).
<p><strong>Step 1:</strong> Compute A²</p><p>A² = <begin>bmatrix</begin> a & b \\ b & a <end>bmatrix</end> × <begin>bmatrix</begin> a & b \\ b & a <end>bmatrix</end> = <begin>bmatrix</begin> a² + b² & ab + ab \\ ab + ab & b² + a² <end>bmatrix</end> = <begin>bmatrix</begin> a² + b² & 2ab \\ 2ab & a² + b² <end>bmatrix</end></p><p><strong>Step 2:</strong> Compare with given form</p><p>From A² = <begin>bmatrix</begin> α & β \\ β & α <end>bmatrix</end>, we have:</p><p>• α = a² + b²</p><p>• β = 2ab</p><p><strong>Step 3:</strong> Derive constraint relation</p><p>α² - β² = (a² + b²)² - (2ab)² = a⁴ + 2a²b² + b⁴ - 4a²b² = a⁴ - 2a²b² + b⁴ = (a² - b²)²</p><p>Therefore: <strong>α² - β² = (a² - b²)² ≥ 0</strong> (always true for real a,b)</p><p>∴ Answer: B</p>
Correct Answer: B