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Question

If A=010001pqr, and I is the identity matrix of order 3, show that A3 = pI + qA +rA2.

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Solution

Given: A=010001pqrNow,A2=AAA2=010001pqr010001pqrA2=0+0+00+0+00+1+00+0+p0+0+q0+0+r0+0+rpp+0+rq0+q+r2A2=001pqrrpp+rqq+r2A3=A2AA3=001pqrrpp+rqq+r2010001pqrA3=0+0+p0+0+q0+0+r0+0+rpp+0+rq0+q+r20+0+pq+r2prp+0+q2+r2q0+p+rq+rq+r3A3=pqrrpp+rqq+r2pq+r2prp+q2+r2qp+2rq+r3 ...1pI+qA+rA2pI+qA+rA2=p100010001+q010001pqr+r001pqrrpp+rqq+r2pI+qA+rA2=p000p000p+0q000qpqq2qr+00rrprqr2r2prp+r2qrq+r3pI+qA+rA2=p+0+00+q+00+0+r0+0+rpp+0+rq0+q+r20+pq+r2p0+q2+rp+r2qp+qr+qr+r3pI+qA+rA2=pqrrpp+rqq+r2pq+r2pq2+r2q+rpp+2qr+r3 ...2 A3=pI+qA+rA2 From eqs. (1) and (2)

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