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प्रश्न
Solve the system of linear equations using the matrix method.
x − y + 2z = 7
3x + 4y − 5z = −5
2x − y + 3z = 12
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उत्तर
`[(1,-1,2),(3,4,-5),(2,-1,3)] [(x),(y),(z)] = [(7),(-5),(12)]` AX = B
A = `[(1,-1,2),(3,4,-5),(2,-1,3)]` X = `[(x),(y),(z)]` or B = `[(7),(-5),(12)]`
Now, |A| = `|(1,-1,2),(3,4,-5),(2,-1,3)|`
= 1(12 − 5) + 1(9 + 10) + 2(−3 − 8)
= 7 + 19 − 22
= 4 ≠ 0
⇒ A−1 exists and hence the given equation has a unique solution.
A11 = `(-1)^(1 + 1) |(4,-5),(-1,3)|`
= 12 − 5
= 7
A12 = `(-1)^(1 + 2) |(3,-5),(2,3)|`
= −(9 + 10)
= −19
A13 = `(-1)^(1 + 3) |(3,4),(2,-1)|`
= −3 − 8
= −11
A21 = `(-1)^(2 + 1) |(-1,2),(-1,3)|`
= −(−3 + 2)
= 1
A22 = `(-1)^(2 + 2) |(1,2),(1,3)|`
= 3 − 4
= −1
A23 = `(-1)^(2 + 3) |(1,-1),(2,-1)|`
= −1(−1 + 2)
= −1
A31 = `(-1)^ (3 + 1) |(-1,2),(4,-5)|`
= 5 − 8
= −3
A32 = `(-1)^(3 + 2) |(1,2),(3,-5)|`
= −(−5 − 6)
= 11
A33 = `(-1)^(3 + 3) |(1,-1),(3, 4)|`
= 4 + 3
= 7
∴ A−1 = `1/|A|` (Adj A)
= `1/4 [(7,-19,-11),(1,1,-1),(-3,11,7)]`
= `1/4 [(7,1,-3),(-19,-1,11),(-11,-1,7)]`
X = A−1B
⇒ `[(x),(y),(z)] = 1/4 [(7,1,-3),(-19,-1,11),(-11,-1,7)] [(7),(-5),(12)]`
= `1/4 [(49 - 5 - 36),(-133 + 5 + 132),(-77 + 5 + 84)]`
= `1/4 [(8),(4),(12)]`
= `[(2),(1),(3)]`
So, x = 2, y = 1 and z = 3
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