If A=22101−1130, find A2−5A+4I and hence find a matrix X such that A2−5A+4I+X=0.Answer: X=11513−4310−2SolutionFirst compute A2=A⋅A=590−1−2−125−2.ThenA2−5A+4I=590−1−2−125−2−1010505−55150+400040004=−1−1−5−1−34−3−102.Since A2−5A+4I+X=0, we have X=−(A2−5A+4I):X=11513−4310−2.CBSE Class 12 Board · 2015 · Subjective