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Question
Solve system of linear equations, using matrix method.
2x + y + z = 1
x – 2y – z =` 3/2`
3y – 5z = 9
Solution
The given equation,
2x + y + z = 1
x - 2y - z = `3/2`
3y - 5z = 9
The equation can be written as a system so X = A-1 B
Where, A `= [(2,1,1),(1,-2,-1),(0,3,-5)], X = [(x),(y),(z)] and B = [(1),(3/2),(9)]`
`therefore abs A = [(2,1,1),(1,-2,-1),(0,3,-5)]`
`= 2 [10 + 3] - 1 [-5 + 0] + 1 [3 + 0]`
`= 2 xx 13 - 1 xx (-5) + 1 xx 3`
`= 26 + 5 + 3 = 34 ne 0`
The cofactors of the elements of matrix A are as follows:
`A_11 = (-1)^(1 + 1) abs ((-2,-1),(3,-5)) = (-1)^2 [10 + 3] = 1 xx 13 = 13`
`A_12 = (-1)^(1 + 2) abs ((1,-1),(0,-5)) = (- 1)^3 [-5 + 0] = -1 xx (-5) = 5`
`A_13 = (- 1)^(1 + 3) abs ((1,-2),(0,3)) = (-1)^4 [3 + 0] = 1 xx 3 = 3`
`A_21 = (-1)^(2 + 1) abs ((1,1),(3,-5)) = (-1)^3 [-5 -3] = -1 xx (- 8) = 8`
`A_22 = (-1)^(2+2) |(2,1), (0,-5)| = (-5-3) = 8`
`A_23 = (-1)^(2 + 3) abs ((2,1),(0,3)) = (-1)^5 [6 - 0] = -1 xx 6 = - 6`
`A_31 = (-1)^(3 + 1) abs ((1,1),(-2,-1)) = (-1)^4 [-1 + 2] = 1 xx 1 = 1`
`A_32 = (-1)^(3 + 2) abs ((2,1),(1,-1)) = (-1)^5 [-2 -1] = -1 xx (-3) = 3`
`A_33 = (-1)^(3 + 3) abs ((2,1),(1,-2)) = (-1)^6 [-4 -1] = 1 xx (-5) = - 5`
Hence, the matrix made up of the elements of the cofactors = `[(13,5,3),(8,-10,-6),(1,3,-5)]`
`therefore adj A = [(13,5,3),(8,-10,-6),(1,3,-5)] = [(13,8,1),(5,-10,3),(3,-6,-5)]`
`A^-1 = 1/abs A (adj A)`
`= 1/34 [(13,8,1),(5,-10,3),(3,-6,-5)]`
`therefore X = A^-1 B`
`= 1/34 [(13,8,1),(5,-10,3),(3,-6,-5)] [(1),(3/2),(9)]`
`= 1/34 [(13 + 12 + 9),(5 - 15 + 27),(3 - 9 - 45)]`
`= 1/34 [(34),(17),(-51)] = [(34/34),(17/34),((-51)/34)]`
`=> [(x),(y),(z)] = [(1),(1/2),(-3/2)]`
`=> x = 1, y = 1/2 and z = (-3)/2`
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