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Question
Solve the following equations by using inversion method.
x + y + z = −1, x − y + z = 2 and x + y − z = 3
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Solution
Matrix form of the given system of equations is,
`[(1, 1, 1),(1, -1, 1),(1, 1, -1)] [(x),(y),(z)] = [(-1),(2),(3)]`
This is of the form AX = B,
where, A = `[(1, 1, 1),(1, -1, 1),(1, 1, -1)], "X" = [(x),(y),(z)], "B" = [(-1),(2),(3)]`
Pre-multiplying AX = B by A−1, we get
A−1(AX) = A−1B
∴ (A−1A)X = A−1B
∴ IX = A−1B
∴ X = A−1B .......(i)
To determine X, we have to find A−1
|A| = `|(1, 1, 1),(1, -1, 1),(1, 1, -1)|`
= 1(1 − 1) − 1(−1 − 1) + 1(1 + 1)
= 2 + 2
= 4 ≠ 0
∴ A−1 exists.
A11 = (−1)1+1 M11 = `|(-1, 1),(1, -1)|` = 1 − 1 = 0
A12 = (−1)1+2 M12 = `-|(1, 1),(1, -1)|` = −(−1 − 1) = 2
A13 = (−1)1+3 M13 = `|(1, 1),(1, -1)|` = 1 + 1 = 2
A21 = (−1)2+1 M21 = `-|(1, 1),(1, -1)|` = − (−1 − 1) = 2
A22 = (−1)2+2 M22 = `|(1, 1),(1, -1)|` = −1 − 1 = −2
A23 = (−1)2+3 M23 = `-|(1, 1),(1, 1)|` = − (1 − 1) = 0
A31 = (−1)3+1 M31 = `|(1, 1),(-1, 1)|` = 1 + 1 = 2
A32 = (−1)3+2 M32 = `-|(1, 1),(1, 1)|` = −(1 − 1) = 0
A33 = (−1)3+3 M33 = `|(1, 1),(1, -1)|` = −1 − 1 = −2
Hence, the matrix of cofactors is
`|("A"_11, "A"_12, "A"_13),("A"_21, "A"_22, "A"_23),("A"_31, "A"_32, "A"_33)| = [(0, 2, 2),(2, -2, 0),(2, 0, -2)]`
∴ adj A = `[(0, 2, 2),(2, -2, 0),(2, 0, -2)]`
= `2[(0, 1, 1),(1, -1, 0),(1, 0, -1)]`
∴ A−1 = `1/|"A"|` (adj A)
= `1/4 xx 2[(0, 1, 1),(1, -1, 0),(1, 0, -1)]`
∴ A−1 = `1/2[(0, 1, 1),(1, -1, 0),(1, 0, -1)]`
∴ X = `1/2[(0, 1, 1),(1, -1, 0),(1, 0, -1)] [(-1),(2),(3)]` .......[From (i)]
= `1/2[(5),(-3),(-4)]`
∴ `[(x),(y),(z)] = [(5/2),(-3/2),(-2)]`
By equality of matrices, we get
x = `5/2`, y = `-3/2`, z = −2
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