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Question

If A=2-11-12-11-12. Verify that A3-6A2+9A-4I=O and hence find A−1.

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Solution

A = 2 -1 1-1 2 -1 1 -1 2 A= 2 -1 1-1 2 -1 1 -1 2=2×4-1+1-2+1+11-2 =6-1-1=4 Since, A0 Hence, A-1 exists.Now,A2= 2 -1 1-1 2 -1 1 -1 2 2 -1 1-1 2 -1 1 -1 2 = 4+1+1 -2-2-1 2+1+2-2-2-1 1+4+1 -1-2-2 2+1+2 -1-2-2 1+1+4 = 6 -5 5-5 6 -5 5 -5 6 A3 =A2.A= 6 -5 5-5 6 -5 5 -5 6 2 -1 1-1 2 -1 1 -1 2 = 12+5+5 -6-10-5 6+5+10-10-6-5 5+12+5 -5-6-10 10+5+6 -5-10-6 5+5+12 = 22 -21 21-21 22 -21 21 -21 22 Now, A3-6A2+9A-4I= 22 -21 21-21 22 -21 21 -21 22-6 6 -5 5-5 6 -5 5 -5 6+9 2 -1 1-1 2 -1 1 -1 2-4 1 0 0 0 1 0 0 0 1 = 22-36+18-4 -21+30-9-0 21-30+9-0-21+30-9-0 22-36+18-4 -21+30-9-0 21-30+9-0 -21+30-9-0 22-36+18-4 = 0 0 0 0 0 0 0 0 0 =O [Null matrix]Hence proved.Now, A3-6A2+9A-4I=OA-1×A3-6A2+9A-4I=A-1×O A2-6A+9I-4A-1 =O4A-1 =A2-6A+9I4A-1 = 6 -5 5-5 6 -5 5 -5 6 -6 2 -1 1-1 2 -1 1 -1 2+9 1 0 0 0 1 0 0 0 14A-1 = 6-12+9 -5+6+0 5-6+0-5+6+0 6-12+9 -5+6+0 5-6+0 -5+6+0 6-12+9 4A-1 = 3 1 -1 1 3 1 -1 1 3A-1=14 3 1 -1 1 3 1 -1 1 3

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