If A=[1−107], find K such that A2−8A+KI=0.Answer: K=7SolutionBy the Cayley-Hamilton theorem A satisfies A2−tr(A)A+det(A)I=0.Here tr(A)=1+7=8 and det(A)=(1)(7)−(0)(−1)=7, soA2−8A+7I=0.Comparing with A2−8A+KI=0 gives K=7.K=7CBSE Class 12 Board · 2005 · Subjective