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Question
If A = `[(1, 0, -1),(2, 1, 3 ),(0, 1, 1)]`, then verify that A2 + A = A(A + I), where I is 3 × 3 unit matrix.
Solution
We have, A = `[(1, 0, -1),(2, 1, 3 ),(0, 1, 1)]`
∴ A2 = A · A
= `[(1, 0, -1),(2, 1, 3),(0, 1, 1)] [(1, 0, -1),(2, 1, 3),(0, 1, 1)]`
= `[(1 + 0 + 0, 0 + 0 - 1, -1 + 0 - 1),(2 + 2 + 0, 0 + 1 + 3, -2 + 3 + 3),(0 + 2 + 0, 0 + 1 + 1, 0 + 3 + 1)]`
= `[(1, -1, -2),(4, 4, 4),(2, 2, 4)]`
∴ A2 + A = `[(1, -1, -2),(4, 4, 4),(2, 2, 4)] + [(1, 0, -4),(2, 1, 3),(0, 1, 1)]`
= `[(2, -1, -3),(6, 5, 7),(2, 3, 5)]` ......(i)
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, 2)]`
So, A(A + I) = `[(1, 0, -1),(2, 1, 3),(0, 1, 1)] [(2, 0, -1),(2, 2, 3),(0, 1, 2)]`
= `[(2 + 0 + 0, 0 + 0 - 1, -1 + 0 - 2),(4 + 2 + 0, 0 + 2 + 3, -2 + 3 + 6),(0 + 2 + 0, 0 + 2 + 1, 0 + 3 + 2)]`
= `[(2, -1, -3),(6, 5, 7),(2, 3, 5)]` .....(iii)
From (i) and (ii)
We get A2 + A = A(A + I)
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