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Question
Using matrices, solve the following system of linear equations :
x + 2y − 3z = −4
2x + 3y + 2z = 2
3x − 3y − 4z = 11
Solution
The system of equations can be written in the form AX = B, where
A `= [(1,2,-3),(2,3,2),(3,-3,-4)],` X`=[("x"),("y"),("z")]` and B =`[(-4),(2),(11)]`
|A| = 1 (-12+6) - 2 (-8 - 6) - 3 (-6 - 9) = 67 ≠ 0
Therefore, A is non singular and so its inverse exists.
A11 = -6, A12 = 14, A13 = -15
A21 = 17, A22 = 5, A23 = 9
A31 = 13, A32 = -8, A33 = -1
Therefore, `"A"^-1 = 1/67[(-6,17,13),(14,5,-8),(-15,9,-1)]`
So X = A-1 B `=1/67[(-6,17,13),(14,5,-8),(-15,9,-1)][(-4),(2),(11)]`
i.e. `[("x"),("y"),("z")]=1/67[(201),(-134),(67)]=[(3),(-2),(1)]`
Hence, x = 3, y = -2 and z = 1
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