Question

Using matrices, solve the following system of equations :

2x - 3y + 5z = 11

3x + 2y - 4z = -5

x + y - 2z = -3

Answer 1

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