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प्रश्न
Solve the system of linear equations using the matrix method.
2x + 3y + 3z = 5
x − 2y + z = −4
3x − y − 2z = 3
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उत्तर
`[(2,3,3),(1,-2,1),(3,-1,-2)] [(x),(y),(z)] = [(5),(-4),(3)]` AX = B
A = `[(2,3,3),(1,-2,1),(3,-1,-2)]`, X = `[(x),(y),(z)]` or B = `[(5),(-4),(3)]`
Now, |A| = `|(2,3,3),(1,-2,1),(3,-1,-2)|`
= 2(4 + 1) − 3(−2 − 3) + 3(−1 + 6)
= 10 + 15 + 15
= 40 ≠ 0
∴ A−1 exists and hence the given equation has a unique solution.
A11 = `(-1)^(1 + 1) |(-2,1),(-1,-2)|`
= 4 + 1
= 5
A12 = `(-1)^(1 + 2) |(1,1),(3,-2)|`
= (−2 − 3)
= 5
A13 = `(-1)^(1 + 3) |(1,-2),(3,-1)|`
= −1 + 6
= 5
A21 = `(-1)^(2 + 1) |(3,3),(-1,-2)|`
= −(−6 + 3)
= 3
A22 = `(-1)^(2 + 2) |(2,3),(3,-2)|`
= −4 − 9
= −13
A23 = `(-1)^(2 + 3) |(2,3),(3,-1)|`
= −(−2 − 9)
= 11
A31 = `(-1)^ (3 + 1) |(3,3),(-2,1)|`
= 3 + 6
= 9
A32 = `(-1)^(3 + 2) |(2,3),(1,1)|`
= −(2 − 3)
= 1
A33 = `(-1)^(3 + 3) |(2,3),(1, -2)|`
= −4 − 3
= −7
∴ A−1 = `1/|A|` (Adj A)
= `1/40 [(5,5,5),(3,-13,11),(9,1,-7)]`
= `1/40 [(5,3,9),(5,-13,1),(5,11,-7)]`
X = A−1B
⇒ `[(x),(y),(z)] = 1/40 [(5,3,9),(5,-13,1),(5,11,-7)] [(5),(-4),(3)]`
= `1/40 [(25 - 12 + 27),(25 + 52 + 3),(25 - 44 - 21)]`
= `1/40 [(40),(80),(-40)]`
= `[(1),(2),(-1)]`
∴ x = 1, y = 2 and z = −1
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