Question

How do you solve `24x + 24y + 16z = 106`, `8x + 24y + 40z = 135` and `24x + 16y - 8z = 42` using matrices?

Answer 1

`x=11/8`, `y=13/8` and `z=17/8`

Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

`A=((24,24,16,|,106),(8,24,40,|,135),(24,16,-8,|,42))`

I have written the equations not in the sequence as in the question in order to get `1` as pivot.

Perform the folowing operations on the rows of the matrix

`R1larrR1-3R2` ; `R3larrR3-3R2`

`A=((0,-48,-104,|,-299),(8,24,40,|,135),(0,-56,-128,|,-363))`

`R1larr(R1)/(-48)`

`A=((0,1,13/6,|,299/48),(8,24,40,|,135),(0,-56,-128,|,-363))`

`R2larrR1-24R1` ; `R3larrR3+56R1`

`A=((0,1,13/6,|,299/48),(8,0,-12,|,-29/2),(0,0,-20/3,|,-85/6))`

`R3larr(R3)*(-3/20)`

`A=((0,1,13/6,|,299/48),(8,0,-12,|,-29/2),(0,0,1,|,17/8))`

`R1larrR1-13/6R3` ; `R2larrR2+12R3`

`A=((0,1,0,|,13/8),(8,0,0,|,11),(0,0,1,|,17/8))`

`R2larr(R2)/8`

`A=((0,1,0,|,13/8),(1,0,0,|,11/8),(0,0,1,|,17/8))`

Thus, solution of equation system is `x=11/8`, `y=13/8` and `z=17/8`