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If `A=[{:(1,0,-1),(2,1,3),(0,1, 1):}]` then verify that `A^(2)+A=(A+I)` , where I is `3xx3` unit matrix. |
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Answer» We have, `A=[{:(1,0,-1),(2,1,3),(0,1,1):}]` `therefore A^(2)=A.A` `=[{:(1,0,-1),(2,1,3),(0,1,1):}][{:(1,0,-1),(2,1,3),(0,1,1):}]=[{:(1,-1,-2),(4,4,4),(2,2,4):}]` `therefore A^(2)+A=[{:(1,-1,-2),(4,4,4),(2,2,4):}]+[{:(1,0,-1),(2, 1,3),(0,1,1):}]` `=[{:(2,-1,-3),(6,5,7) ,(2,3,5):}]` Now, `A+I=[{:(1,0,-1),(2,1,3),(0,1,1):}]+[{:(1,0,0),(0,1,0),(0,0,1):}]=[{:(2,0,-1),(2,2,3),(0,1,3):}]` and `A(A+I)=[{:(1,0,-1),(2,1,3),(0,1,1):}][{:(2,0,-1),( 2,2,3),(0,1,2):}]=[{:(2,-1,-3),(6,5,7),(2,3,5):}]` Thus, we see that `A^(2)+A=A(A+I)` |
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