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प्रश्न
Solve the following :
If A = `[(2, -1),(-1, 2)]`, then show that A2 – 4A + 3I = 0.
योग
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उत्तर
A2 – 4A + 3I
= A.A – 4A + 3I
= `[(2, -1),(-1, 2)] [(2, -1),(-1, 2)] -4[(2, -1),(-1, 2)] + 3[(1, 0),(0, 1)]`
= `[(4 + 1, -2 -2),(-2 - 2, 1 + 4)] - [(8, -4),(-4, 8)] + [(3, 0),(0, 3)]`
= `[(5, -4),(-4, 5)] - [(8, -4),(-4, 8)] + [(3, 0),(0, 3)]`
= `[(5 - 8 + 3, -4 + 4 + 0),(-4 + 4 + 0, 5 - 8 + 3)]`
= `[(0, 0),(0, 0)]`
= 0.
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