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Question: If A = \(\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{bmatrix}\) then A<sup>3</sup> – rA...

If A = [010001pqr]\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{bmatrix} then A3 – rA2 – qA =

A

p I

B

q I

C

r I

D

None of these

Answer

p I

Explanation

Solution

A2 = AA = [010001pqr]\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{bmatrix} × [010001pqr]\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{bmatrix}= [001pqrprp+qrq+r2]\begin{bmatrix} 0 & 0 & 1 \\ p & q & r \\ pr & p + qr & q + r^{2} \end{bmatrix}Again A3 = A2A

=[001pqrprp+qrq+r2]\begin{bmatrix} 0 & 0 & 1 \\ p & q & r \\ pr & p + qr & q + r^{2} \end{bmatrix} × [010001pqr]\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{bmatrix}

= [pqrprp+qrq+r2pq+r2ppr+q2+qr2p+2qr+r3]\begin{bmatrix} p & q & r \\ pr & p + qr & q + r^{2} \\ pq + r^{2}p & pr + q^{2} + qr^{2} & p + 2qr + r^{3} \end{bmatrix}

= [p000p000p]\begin{bmatrix} p & 0 & 0 \\ 0 & p & 0 \\ 0 & 0 & p \end{bmatrix} + [0q000qpqq2qr]\begin{bmatrix} 0 & q & 0 \\ 0 & 0 & q \\ pq & q^{2} & qr \end{bmatrix} + [00rprqrr2pr2pr+qr2qr+r3]\begin{bmatrix} 0 & 0 & r \\ pr & qr & r^{2} \\ pr^{2} & pr + qr^{2} & qr + r^{3} \end{bmatrix}

= p [100010001]\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} + q [010001pqr]\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{bmatrix}

+ r [001pqrprp+qrq+r2]\begin{bmatrix} 0 & 0 & 1 \\ p & q & r \\ pr & p + qr & q + r^{2} \end{bmatrix}

= pI + qA + rA2

\ a3 – rA2 – qA = pI.