Matrices & Determinants
Types of Matrices
Grade 12
Question:
<p>If matrix \(A = [a_{ij}]_{3 \times 3}\), matrix \(B = [b_{ij}]_{3 \times 3}\), where \(a_{ij} + a_{ji} = 0\) and \(b_{ij} - b_{ji} = 0\) \(\forall\, i, j\), then \(A^4 B^3\) is:</p>
<p>(a) singular</p>
<p>(b) zero matrix</p>
<p>(c) symmetric</p>
<p>(d) skew symmetric</p>
Step-by-Step Solution
Key Concept: Matrix A is skew-symmetric (A^T = -A), so A² is symmetric and A⁴ = (A²)² is symmetric. Matrix B is symmetric (B^T = B), making B³ symmetric. The product of symmetric matrices in this order yields a symmetric result.
<p><strong>Step 1:</strong> Identify matrix types from given conditions.</p><p>From a<sub>ij</sub> + a<sub>ji</sub> = 0: We have a<sub>ij</sub> = -a<sub>ji</sub>, so A is <strong>skew-symmetric</strong> (A<sup>T</sup> = -A)</p><p>From b<sub>ij</sub> - b<sub>ji</sub> = 0: We have b<sub>ij</sub> = b<sub>ji</sub>, so B is <strong>symmetric</strong> (B<sup>T</sup> = B)</p><p><strong>Step 2:</strong> Determine the nature of A⁴.</p><p>Since A is skew-symmetric: A<sup>T</sup> = -A</p><p>(A²)<sup>T</sup> = (A<sup>T</sup>)² = (-A)² = A² → A² is symmetric</p><p>(A⁴)<sup>T</sup> = (A²)² is the product of A² with itself. Since A² is symmetric: (A⁴)<sup>T</sup> = A⁴ → <strong>A⁴ is symmetric</strong></p><p><strong>Step 3:</strong> Determine the nature of B³.</p><p>Since B is symmetric: B<sup>T</sup> = B</p><p>(B³)<sup>T</sup> = (B<sup>T</sup>)³ = B³ → <strong>B³ is symmetric</strong></p><p><strong>Step 4:</strong> Find the nature of A⁴B³.</p><p>(A⁴B³)<sup>T</sup> = (B³)<sup>T</sup>(A⁴)<sup>T</sup> = B³ · A⁴ = A⁴B³</p><p>Since A⁴ and B³ both are symmetric matrices, their product (when multiplied in this order) is <strong>symmetric</strong>.</p><p>∴ Answer: A (A⁴B³ is symmetric)</p>
Correct Answer: A