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Question
Solve the system of linear equations using the matrix method.
x − y + z = 4
2x + y − 3z = 0
x + y + z = 2
Solution
Given system of equations,
x - y + 2 = 4
2x + y - 3z = 0
x+ y + z = 2
The system of equations can be written as AX = B so X = A-1B
A = `[(1,-1,1),(2,1,-3),(1,1,1)], X = [(x),(y),(z)], B = [(4),(0),(2)]`
`therefore abs A = [(1,-1,1),(2,1,-3),(1,1,1)]`
`= 1 [1 + 3] - (-1) [2 + 3] + 1 [2 - 1]`
`= 1 xx 4 + 1 xx 5 + 1 xx 1`
`= 4 + 5 + 1 = 10 ne 0`
The cofactors of the elements of matrix A are as follows -
`A_11 = (-1)^(1 + 1) abs ((1,-3),(1,1)) = (-1)^2 [1 + 3] = 1 xx 4 = 4`
`A_12 = (-1)^(1 + 2) abs ((2,-3),(1,1)) = (-1)^3 [2 + 3] = -1 xx 5 = -5`
`A_13 = (-1)^(1 + 3) abs ((2,1),(1,1)) = (-1)^4 [2 - 1] = 1 xx 1 = 1`
`A_21 = (-1)^(2 + 1) abs ((-1,1),(1,1)) = (-1)^3 [-1 - 1] = -1 xx (-2) = 2`
`A_22 = (-1)^(2 + 2) abs ((1,1),(1,1)) = (-1)^4 [1 - 1] = 0`
`A_23 = (-1)^(2 + 3) abs((1,-1),(1,1)) = (-1)^5 [1 + 1] = - 1 xx 2 = -2`
`A_31 = (-1)^ (3 + 1) abs ((-1,1),(1,-3)) = (-1)^4 [3 - 1] = 1 xx 2 = 2`
`A_32 = (-1)^(3 + 2) abs ((1,1),(2,-3)) = (-1)^5 [-3 - 2] = -1 xx (- 5) = 5`
`A_33 = (-1)^(3 + 3) abs ((1,-1),(2,1)) = (-1)^6 [1 + 2] = 1 xx 3 = 3`
Hence the matrix made up of the elements of cofactors = `[(4,-5,1),(2,0,-2),(2,5,3)]`
`therefore adj A = [(4,-5,1),(2,0,-2),(2,5,3)] = [(4,2,2),(-5,0,5),(1,-2,3)]`
`A^-1 = 1/abs A (adj A)`
`= 1/10 [(4,2,2),(-5,0,5),(1,-2,3)]`
`therefore X = A^-1 B = 1/10 [(4,2,2),(-5,0,5),(1,-2,3)] [(4),(0),(2)]`
`= 1/10 [(16 + 0 + 4),(-20 + 10),(4 + 6)]`
`1/10 [(20),(-10),(10)] = [(2),(-1),(1)]`
`=> [(x),(y),(z)] = [(2),(-1),(1)]`
`=> x = 2, y = -1 and z = 1`
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