Question

How do you solve `5x - 5y + 10z = -11`, `10x + 5y - 5z = 1` and `15x - 15y -10z = -1` using matrices?

Answer 1

The solution is `((x),(y),(z))=((-2/5),(1/5),(-4/5))`

Explanation:

Perform the Gauss-Jordan Elimination on the augmented matrix

`A=((5,-5,10,|,-11),(10,5,-5,|,1),(15,-15,-10,|,-1))`

Make the pivot in the first column and first row

`R1larr((R1)/5)`

`((1,-1,2,|,-2.2),(10,5,-5,|,1),(15,-15,-10,|,-1))`

Eliminate the first column

`R2larr(R2-10R1)` and `R3larr(R3-15R1)`

`((1,-1,2,|,-2.2),(0,15,-25,|,23),(0,0,-40,|,32))`

Make the pivot in the second column and second row

`R2larr((R2)/15)`

`((1,-1,2,|,-2.2),(0,1,-5/3,|,23/15),(0,0,-40,|,32))`

Eliminate the second column

`R1larr(R1+R2)`

`((1,0,1/3,|,-2/3),(0,1,-5/3,|,23/15),(0,0,-40,|,32))`

Make the pivot in the third column and third row

`R3larr((R3)/-40)`

`((1,0,1/3,|,-2/3),(0,1,-5/3,|,23/15),(0,0,1,|,-4/5))`

Eliminate the third column

`R1larr(R1-1/3R3)`, and `R2larr(R2+5/3R3)`,

`((1,0,0,|,-2/5),(0,1,0,|,1/5),(0,0,1,|,-4/5))`

The solution is

`((x),(y),(z))=((-2/5),(1/5),(-4/5))`