Question

How to solve this system of equation using matrices? Please don't make it too complicated because I still have some more to solve like this. I'm gonna use your answer as my guide :)

Answer 1

`(1,-4,-2)`

Explanation:

First form an augmented matrix with the coefficients on the left and the constants on the right in the augmented column. It isn't possible to format an augmented matrix, so I will use bolden text in the last column to represent the augmented part.

Matrix:

`[(1,1,1,bb(-5)),(1,-1,3,bb(-1)),(4,1,1,bb(-2))]`

We need to convert this to an upper triangular matrix using row operations. It should look like the matrix below.

`[(a,b,c,bb(d)),(0,f,g,bb(h)),(0,0,k,bb(m))]`

The notation for the row operations will be:

`R2=R2+3R`

This means, row 2 is row 2 + 3 times row 1.

`R3=R3-4R1`

`[(1,1,1,bb(-5)),(1,-1,3,bb(-1)),(0,-3,-3,bb(18))]`

`R2=R2-R1`

`[(1,1,1,bb(-5)),(0,-2,2,bb(4)),(0,-3,-3,bb(18))]`

`R2=R2-R3`

`[(1,1,1,bb(-5)),(0,1,5,bb(-14)),(0,-3,-3,bb(18))]`

`R3=R3+3R2`

`[(1,1,1,bb(-5)),(0,1,5,bb(-14)),(0,0,12,bb(-24))]`

`R3=1/12R3`

`[(1,1,1,bb(-5)),(0,1,5,bb(-14)),(0,0,1,bb(-2))]`

From this we can see:

`z=-2`

Now use back substitution in 2nd row:

`y+5(-2)=-14=>y=-4`

In 1st row:

`x-4-2=-5=>x=1`

So solutions in form `(x,y,z)` are:

`(1,-4,-2)`