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Question
Solve the following system of equations by using inversion method
x + y = 1, y + z = `5/3`, z + x = `4/3`
Solution
Matrix form of the given system of equations is
`[(1, 1, 0),(0, 1, 1),(1, 0, 1)] [(x),(y),(z)] [(1),(5/3),(4/3)]`
This is of the form AX = B,
where A = `[(1, 1, 0),(0, 1, 1),(1, 0, 1)]` X = `[(x),(y),(z)]` B = `[(1),(5/3),(4/3)]`
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),(0, 1)| - 1|(0, 1),(1, 1)| + 0`
= 1(1 – 0) –1(0 – 1)
= 1 + 1
= 2 ≠ 0
∴ A−1 exists.
Consider, AA−1 = I
∴ `[(1, 1, 0),(0, 1, 1),(1, 0, 1)]` A−1 = `[(1, 0, 0),(0, 1, 0),(0, 0, 1)]`
Applying R3 → R3 – R1, we get
`[(1, 1, 0),(0, 1, 1),(1, -1, 1)]` A−1 = `[(1, 0, 0),(0, 1, 0),(-1, 0, 1)]`
Applying R2 → R2 − R3, we get
`[(1, 1, 0),(0, 2, 0),(0, -1, 1)]` A−1 = `[(1, 0, 0),(1, 1, -1),(-1, 0, 1)]`
Applying R1 → `(1/2)` R2, we get
`[(1, 1, 0),(0, 1, 0),(0, -1, 1)]` A−1 = `[(1, 0, 0),(1/2, 1/2, -1/2),(-1, 0, 1)]`
Applying R1 → R1 − R2, we get
`[(1, 1, 0),(0, 1, 0),(0, -1, 1)]` A−1 = `[(1/2, 1/2, 1/2),(1/2, 1/2, -1/2),(-1, 0, 1)]`
Applying R3 → R3 + R2, we get
`[(1, 1, 0),(0, 1, 0),(0, 0, 1)]` A−1 = `[(1/2, 1/2, 1/2),(1/2, 1/2, -1/2),(-1/2, 1/2, 1/2)]`
∴ A−1 = `[(1/2, 1/2, 1/2),(1/2, 1/2, -1/2),(-1/2, 1/2, 1/2)]`
∴ X = `[(1/2, 1/2, 1/2),(1/2, 1/2, -1/2),(-1/2, 1/2, 1/2)] [(1),(5/3),(4/3)]` .......[From (i)]
∴ `[(x),(y),(z)] = [(1/3),(2/3),(3/1)]`
∴ By equality of matrices, we get
x = `1/3` y = `2/3` and z = 1
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