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Question
Solve the system of linear equations using the matrix method.
x − y + 2z = 7
3x + 4y − 5z = −5
2x − y + 3z = 12
Solution
`[(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, `abs "A" = [(1,-1,2),(3,4,-5),(2,-1,3)]`
`= 1 (12 - 5) + 1 (9 + 10) + 2 (-3 - 8)`
`= 7 + 19 - 22 = 4 ne 0`
`=> A^-1` exists and hence the given equation has a unique solution.
`A_11 = (-1)^(1 + 1) abs ((4,-5),(-1,3)) = 12-5 = 7`
`A_12 = (-1)^(1 + 2) abs ((3,-5),(2,3)) = -(9 + 10) = -19`
`A_13 = (-1)^(1 + 3) abs ((3,4),(2,-1)) = -3 - 8 = -11`
`A_21 = (-1)^(2 + 1) abs ((-1,2),(-1,3)) = -(-3 + 2) = 1`
`A_22 = (-1)^(2 + 2) abs ((1,2),(1,3)) = 3 - 4 = -1`
`A_23 = (-1)^(2 + 3) abs((1,-1),(2,-1)) = -1(-1 + 2) = -1`
`A_31 = (-1)^ (3 + 1) abs ((-1,2),(4,-5)) = 5 - 8 = -3`
`A_32 = (-1)^(3 + 2) abs ((1,2),(3,-5)) = -(-5 - 6) = 11`
`A_33 = (-1)^(3 + 3) abs ((1,-1),(3, 4)) = 4 + 3 = 7`
`therefore A^-1 = 1/abs 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^-1 B`
`=> [(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, or z = 3.`
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