<p>Let \(A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{bmatrix}\). Then \(A^2 = AA\). Which of the following are correct about \(A^2\)?</p>
<p>(a) \(A^2\) can be computed as \(AA\)</p>
<p>(b) \(A^2\) is a \(3 \times 3\) matrix</p>
<p>(c) The result depends on \(p, q, r\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Computing A² requires multiplying the matrix by itself; the key insight is that the third row of A determines the structure of A², and recognizing the cyclic shift pattern in the first two rows reveals that A² will have zeros in specific positions.
<p><strong>Step 1:</strong> Compute A² by multiplying A × A:</p><p>A² = <span style='font-family: monospace;'>[0 1 0] × [0 1 0] [0 0 1]</span></p><p><span style='font-family: monospace;'>[0 0 1] [0 0 1] = [p q r]</span></p><p><span style='font-family: monospace;'>[p q r] [p q r] [pq q²+rp qr+r²]</span></p><p><strong>Step 2:</strong> Compute each entry systematically:</p><p>Row 1: (0,1,0)·cols = [0, 0, 1]</p><p>Row 2: (0,0,1)·cols = [p, q, r]</p><p>Row 3: (p,q,r)·cols</p><ul><li>(3,1): 0·p + 0·0 + p·p = p²</li><li>(3,2): 0·1 + 0·0 + p·q = pq (also = 0·0 + 0·1 + 1·q = q)</li><li>(3,3): 0·0 + 0·1 + p·r + q·0 + r·r = pr + r²</li></ul><p><strong>Step 3:</strong> A² = <span style='font-family: monospace;'>[0 0 1]</span></p><p><span style='font-family: monospace;'>[p q r]</span></p><p><span style='font-family: monospace;'>[p² pq pr+r²]</span></p><p>∴ The structure shows A is a shift matrix with characteristic cyclic behavior; A³ produces the identity multiplied by the third row pattern.</p>
Correct Answer: abc