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Question

Show that the matrix, A=10-2-2-12341 satisfies the equation, A3-A2-3A-I3=O. Hence, find A−1.

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Solution

We have, A= 1 0 -2-2 -1 2 3 4 1 A= 1 0 -2-2 -1 2 3 4 1 =1-9+0-2-8=-9+16 =7 Since, A0Hence, A-1 exists.Now,A2= 1 0 -2-2 -1 2 3 4 1 1 0 -2-2 -1 2 3 4 1= 1+0-6 0+0-8 -2+0-2-2+2+6 0+1+8 4-2+2 3-8+3 0-4+4 -6+8+1= -5 -8 -4 6 9 4-2 0 3A3=A2.A =-5 -8 -4 6 9 4-2 0 3 1 0 -2-2 -1 2 3 4 1 =-5+16-12 0+8-16 10-16-4 6-18+12 0-9+16 -12+18+4 -2+0+9 0+0+12 4+0+3 = -1 - 8 -10 0 7 10 7 12 7 Now, A3-A2-3A-I3 = -1 - 8 -10 0 7 10 7 12 7 --5 -8 -4 6 9 4-2 0 3-3 1 0 -2-2 -1 2 3 4 1 - 1 0 0 0 1 0 0 0 1 = -1+5-3-1 -8+8+0+0 -10+4+6-00-6+6-0 7-9+3-1 10-4-6-0 7+2-9-0 12+0-12-0 7-3-3-1 = 0 0 0 0 0 0 0 0 0 =OHence proved.Now, A3-A2-3A-I3=O (Null matrix)A-1A3-A2-3A-I3 =A-1O (Pre-multiplying by A-1)A2-A1-3I3=A-1 -5 -8 -4 6 9 4-2 0 3- 1 0 -2-2 -1 2 3 4 1-3 1 0 0 0 1 0 0 0 1 =A-1 -5-1-3 -8-0-0 -4+2+0 6+2+0 9+1-3 4-2 -2-3-0 0-4-0 3-1-3 = -9 -8 -2 8 7 2 -5 -4 -1=A-1A-1 = -9 -8 -2 8 7 2 -5 -4 -1

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