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Question
Using matrices, solve the following system of equations :
2x - 3y + 5z = 11
3x + 2y - 4z = -5
x + y - 2z = -3
Solution
Using this let us solve the system of given equation
2x - 3y + 5z = 11
3x + 2y - 4z = -5
x + y - 2z = -3
This can be written in the form AX = B
`[(2,-3,5),(3,2,-4),(1,1,-2)][(x),(y),(z)] = [(11),(-5),(-3)]`
where `A = [(2,-3,5),(3,2,-4),(1,1,-2)]X = [(x),(y),(z)]B = [(11),(-5),(-3)]`
we know `A^(-1) = 1/|A| (adj A)`
`|A| = 2(2xx-2-1xx-4) - (-3)(3xx-2-1xx -4) + 5(3xx1-2xx1)`
`= 2(-4+4) + 3(-6+4) + 5(3-2)`
`= 0 - 6+ 5 = -1 !=0`
Hence it is a non- singular matrix
Therefore `A^(-1)` exists
Let us findthe (adjA) by by finding the minors and cofactors
`M_11 = |(2,-4),(1,-2)| = -4+4 = 0`
`M_12 = |(3,-4),(1,-2)| = -6+4 = -2`
`M_13 = |(3,2),(1,1)| = 3-2 = 1`
`M_21 = |(-3,5),(1,-2)| = 6-5 = 1`
`M_22 = |(2,5),(1, -2)| = -4-5 = -9`
`M_23 = |(2,-3),(1,1)| = 2+3 = 5`
`M_31 = |(-3,5),(2,-4)| = 12 - 10 = 2`
`M_32 = |(2,5),(3,-4)| = -8-158 = -23`
`M_33 = |(2,-3),(3,2)| = 4 + 9 = 13`
`A_11 = 0 A_12 = 2 A_13 = 1`
`A_21 = -1 A_22 = -9 A_23 = -5`
`A_31 = 2 A_32 = 23 A_33 = 13`
`A^(-1) = 1/(-1) [(0,-1,2),(2,-9,23),(1,-5,13)] = [(0,1,-2),(-2,9,-23),(-1,5,-13)]`
We know AX = B, then X = A–1 B
Therefore `[(x),(y),(z)] = [(0,1,-2),(-2,9,-23),(-1,5,-13)][(11),(-5),(-3)]`
Matrix multiplication can be done by multiplying the rows of matrix A with the column of matrix B.
Therefore `[(x),(y),(z)] = [(0,-5,+6),(-22,-45,+69),(-11,-25, +39)]= [(1),(2),(3)]`
Hence x = 1, y = 2 and z = 3
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