Matrices & Determinants
Matrix Multiplication & Powers
Grade 12
Question:
<p>If \( A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{bmatrix} \) and \(I\) is third order unit matrix, then</p>
<p>\( A^2 = \begin{bmatrix} 0 & 0 & 1 \\ p & q & r \\ pr & p+qr & q+r^2 \end{bmatrix} \)</p>
<p>\( A^3 = \begin{bmatrix} p & q & r \\ pr & p+qr & q+r^2 \\ pq+r^2p & p+q^2+qr^2 & p+2qr+r^3 \end{bmatrix} \)</p>
<p>\(A^3 = pI + qA + rA^2\)</p>
<p>\( A^3 = \begin{bmatrix} p & q & r \\ pr & 0 & q+r^2 \\ pq+r^2p & p+q^2+qr^2 & p+2qr+r^3 \end{bmatrix} \)</p>
Step-by-Step Solution
Key Concept: The matrix A is a companion matrix whose characteristic polynomial is determined by the last row coefficients. The minimal and characteristic polynomials relate directly to the entries p, q, r, making A satisfy its own characteristic equation (Cayley-Hamilton theorem).
<p><strong>Step 1:</strong> Compute $A^2$:</p><p>$$A^2 = A \cdot A = \begin{bmatrix} 0 & 0 & 1 \\ p & q & r \\ pr & p+qr & q+r^2 \end{bmatrix}$$</p><p><strong>Step 2:</strong> Compute $A^3$:</p><p>$$A^3 = A^2 \cdot A = \begin{bmatrix} p & q & r \\ pr & p+qr & q+r^2 \\ p^2+pqr & pq+q^2+qr^2 & pr+2qr+r^3 \end{bmatrix}$$</p><p><strong>Step 3:</strong> Find the characteristic polynomial using $\det(A - \lambda I)$:</p><p>$$\det(A - \lambda I) = -\lambda^3 + r\lambda^2 + q\lambda + p$$</p><p>By the Cayley-Hamilton theorem: $A^3 = rA^2 + qA + pI$</p><p><strong>Step 4:</strong> Verify the statements:</p><p>• From Cayley-Hamilton: $A^3 - rA^2 - qA - pI = 0$ ✓ (Always true by theorem)</p><p>• Rearranging: $A^3 = pI + qA + rA^2$ ✓ (Always true by theorem)</p><p>• Option (C) matches the Cayley-Hamilton result</p><p>∴ <strong>Answer: (A) and (C)</strong></p>
Correct Answer: A,C