Question
Given \[ A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{bmatrix} \] and \(I\) is the identity matrix of order 3. Prove that: \[ A^3 = pI + qA + rA^2 \]
Step 1: Compute \(A^2 = A \cdot A\)
\[ A^2 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{bmatrix} \] \[ = \begin{bmatrix} 0 & 0 & 1 \\ p & q & r \\ rp & p + rq & q + r^2 \end{bmatrix} \]Step 2: Compute \(A^3 = A^2 \cdot A\)
\[ A^3 = \begin{bmatrix} 0 & 0 & 1 \\ p & q & r \\ rp & p + rq & q + r^2 \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{bmatrix} \] \[ = \begin{bmatrix} p & q & r \\ rp & p + rq & q + r^2 \\ p(q + r^2) & rp + q^2 + r^2q & p + 2rq + r^3 \end{bmatrix} \]Step 3: Compute \(pI + qA + rA^2\)
\[ pI = \begin{bmatrix} p & 0 & 0 \\ 0 & p & 0 \\ 0 & 0 & p \end{bmatrix} \] \[ qA = \begin{bmatrix} 0 & q & 0 \\ 0 & 0 & q \\ pq & q^2 & qr \end{bmatrix} \] \[ rA^2 = \begin{bmatrix} 0 & 0 & r \\ rp & rq & r^2 \\ r^2p & rp + r^2q & rq + r^3 \end{bmatrix} \] Adding all: \[ pI + qA + rA^2 = \begin{bmatrix} p & q & r \\ rp & p + rq & q + r^2 \\ p(q + r^2) & rp + q^2 + r^2q & p + 2rq + r^3 \end{bmatrix} \]Final Result
\[ A^3 = \begin{bmatrix} p & q & r \\ rp & p + rq & q + r^2 \\ p(q + r^2) & rp + q^2 + r^2q & p + 2rq + r^3 \end{bmatrix} \] \[ pI + qA + rA^2 = \begin{bmatrix} p & q & r \\ rp & p + rq & q + r^2 \\ p(q + r^2) & rp + q^2 + r^2q & p + 2rq + r^3 \end{bmatrix} \]Hence Proved: \( A^3 = pI + qA + rA^2 \)