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Question
Solve the following system of equations by matrix method:
3x + 4y + 7z = 14
2x − y + 3z = 4
x + 2y − 3z = 0
Solution
The given system can be written in matrix form as:
`[(3, 4, 7),(2, -1, 3),(1, 2, -3)][(x),(y),(z)] = [(14),(4),(0)]` or A X = B
A = `[(3, 4, 7),(2, -1, 3),(1, 2, -3)], X = [(x),(y),(z)] and B = [(14),(4),(0)]`
Now,
|A| = `3|(-1, 3),(2, -3)| -4|(2, 3),(1, -3)| + 7|(2, 3),(2, -3)|`
= 3(3 − 6) − 4(−6 − 3) + 7(4 + 1)
= −9 + 36 + 35
= 62
So, the above system has a unique solution, given by
X = A−1 B
Cofactors of A are:
C11 = (−1)1 + 1 3 − 6 = −3
C21 = (−1)2 + 1 − 12 − 14 = 26
C31 = (−1)3 + 1 12 + 7 = 19
C12 = (−1)1 + 2 − 6 − 3 = 9
C22 = (−1)2 + 1 − 3 − 7 = −10
C32 = (−1)3 + 1 9 − 14 = 5
C13 = (−1)1 + 2 4 + 1 = 5
C23 = (−1)2 + 1 6 − 4 = −2
C33 = (−1)3 + 1 − 3 − 8 = −11
adj A = `[(-3, 9, 5),(26, -5, -2),(19, 5, -11)]^T`
= `[(-3, 26, 19),(9, -16, 5),(5, -2, -11)]`
A−1 = `1/(|A|)`adj A
Now, X = A−1 B = `1/62[(-3, 26, 19),(9, -16, 5),(5, -2, -11)][(14),(4),(0)]`
X = `1/62[(-42 + 104 + 0),(126 - 64 + 0),(70 - 8 + 0)]`
X = `1/62[(-42 + 104 + 0),(126 - 64 + 0),(70 - 8 + 0)]`
X = `1/62 [(62),(62),(62)]`
C = `[(1),(1),(1)]`
Hence, X = 1, Y = 1 and Z = 1
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