Given, a matrix A is such that 3A3+2A2+5A+I=0 Consider, 3A3+2A2+5A+I=0 Pre - Multiply the above polynomial by A−1 ⇒A−1(3A3+2A2+5A+I)=A−1(0) ⇒3A−1A3+2A−1A2+5A−1A+A−1I=0 ⇒3A−1AA2+2A−1AA+5A−1A+A−1I=0 We know that A−1A=I and A−1I=A−1 ⇒3IA2+2IA+5I+A−1=0 ⇒3A2+2A+5+A−1=0 ⇒3A2+2A+5+A−1=0 ⇒A−1=−(3A2+2A+5)