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प्रश्न
If A = `[(1,-1),(2,3)]` show that A2 - 4A + 5I2 = 0 and also find A-1.
उत्तर
Given A = `[(1,-1),(2,3)] |"A"| = |(1,-1),(2,3)|` = 3 + 2 = 5 ≠ 0
∴ A-1 exists.
LHS = A2 - 4A + 5I
Premultiply by A-1 we get
LHS = `"A"^-1. "A"^2 - 4"A"^-1."A" + 5"A"^-1."I"`
`= ("A"^-1 "A")."A" - 4"I" + 5"A"^-1` ...[∵ A-1A = I]
= IA - 4I + 5A-1
= A - 4I + 5A-1
Now `"A"^-1 = 1/|"A"| "adj A"`
`= 1/5 [(3,1),(-2,1)]`
∴ LHS = `[(1,-1),(2,3)] - 4[(+1,0),(0,1)] + 5 . 1/5 [(3,1),(-2,1)]`
`= [(1,-1),(2,3)] + [(-4,0),(0,-4)] + [(3,1),(-2,1)]`
`= [(1-4+3,-1+0+1),(2+0-2,3-4+1)] = [(0,0),(0,0)]`
= RHS
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