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 :)

Question

Elias Ackerman · Accepted Answer

#(1,-4,-2)#

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)#