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प्रश्न
Solve by matrix inversion method:
x – y + 2z = 3; 2x + z = 1; 3x + 2y + z = 4
उत्तर
The given system can be written as
`[(1,-1,2),(2,0,1),(3,2,1)][(x),(y),(z)] = [(3),(1),(4)]`
AX = B
Where A = `[(1,-1,2),(2,0,1),(3,2,1)]`, X = `[(x),(y),(z)]` and B = `[(3),(1),(4)]`
|A| = `|(1,-1,2),(2,0,1),(3,2,1)|`
= 1(0 – 2) – (-1)(2 – 3) + 2(4 – 0)
= -2 – (-1)(-1) + 2(4)
= -2 – 1 + 8
= 5
[Aij] = `[(-2,-(-1),4),(-|(-1,2),(2,1)|,|(1,2),(3,1)|,-|(1,-1),(3,2)|),(|(-1,2),(0,1)|,-|(1,2),(2,1)|,|(1,-1),(2,0)|)]`
`= [(-2,1,4),(-(-1-4),(1-6),-(2+3)),((-1-0),-(1-4),(0+2))]`
`=> [(-2,1,4),(5,-5,-5),(-1,3,2)]`
adj A = `["A"_"ij"]^"T" = [(-2,5,-1),(1,-5,3),(4,-5,2)]`
`"A"^-1 = 1/|"A"|`(adj A)
`= 1/(5)[(-2,5,-1),(1,-5,3),(4,-5,2)]`
X = A-1B
`= 1/(5)[(-2,5,-1),(1,-5,3),(4,-5,2)][(3),(1),(4)]`
`=> 1/(5)[(-6+5-4),(3-5+12),(12-5+8)]`
`=> 1/(5)[(-5),(10),(15)]`
`[(x),(y),(z)] = [(-1),(2),(3)]`
∴ x = -1, y = 2, z = 3.
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