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Question
If A = `[(2, -3, 5),(3, 2, -4),(1, 1, -2)]`, then find A–1. Using A–1, solve the following system of equations:
2x – 3y + 5z = 11
3x + 2y – 4z = –5
x + y – 2z = –3
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Solution
Given, A = `[(2, -3, 5),(3, 2, -4),(1, 1, -2)]`
|A| = `|(2, -3, 5),(3, 2, -4),(1, 1, -2)|`
= 2(–4 + 4) + 3(–6 + 4) + 5(3 – 2)
= 3(–2) + 5(1)
= –1 ≠ 0
∴ A–1 exists
Suppose Aij is the cofactor of element aij of A
Now, A11 = 0, A12 = –(–2) = 2, A13 = 1,
A21 = –1, A22 = –9, A23 = –5,
A31 = 2, A32 = 23,
A33 = 13
∴ Adj A = `[(0, -1, 2),(2, -9, 23),(1, -5, 13)]`
`A^-1 = 1/|A| xx Adj A`
= `[(0, 1, -2),(-2, 9, -23),(-1, 5, -13)]`
Also, we need to solve the following system of equations
2x – 3y + 5z = 11
3x + 2y – 4z = –5
x + y – 2z = –3
Here, A = `[(2, -3, 5),(3, 2, -4),(1, 1, -2)]` B = `[(11),(-5),(-3)]`, X = `[(x),(y),(z)]`
∴ X = A–1B
`[(x),(y),(z)] = [(0, 1, -2),(-2, -, -23),(-1, 5, -13)] [(11),(-5),(-3)]`
⇒ `[(x),(y),(z)] = [(0 xx 11 + 1 xx (-5) + (-2) xx (-3)),((-2) xx 11 + 9 xx (-5) + (-23) xx (-3)),((-1) xx 11 + 5 xx (-5) + (-13) xx (-3))] = [(1),(2),(3)]`
From equality of matrices,
x = 1, y = 2, z = 3
