How do you solve #-x + y+z = -4# and #-x + 5y-19z= -28# and #6x-2y-26z = 0# using matrices?

First construct an augmented matrix with the coefficients of #bb(x,y,z)# on the left and the constants on the right in the augmented part of the matrix. Due to formatting restrictions I will use bold type for the augmented section instead of the usual vertical bar that separates the two.

#[(1,-1,-1,bb(4)),(1,-5,19,bb(28)),(6,-2,-26,bb(0))]#

Notice the first two equations have been multiplied by #-1# to remove the negations from the first entries, this tends to make things easier.

We now use row operations to get the matrix in an upper triangular form.

The notation for row operation will be:

#R3=R3+2R1#

This means row 3 is row 3 plus 3 x row 1 added to it.

#R3=R3-6R1#

#[(1,-1,-1,bb(4)),(1,-5,19,bb(28)),(0,4,-20,bb(-24))]#

#R2=R2-R1#

#[(1,-1,-1,bb(4)),(0,-4,20,bb(24)),(0,4,-20,bb(-24))]#

#R3=R3+R2#

#[(1,-1,-1,bb(4)),(0,-4,20,bb(24)),(0,0,0,bb(0))]#

Notice the row of all zero entries. This is caused by the fact that two of the equations are linearly dependent. Linear dependency occurs when two or more equation are just multiples of other equations.

Example:

#x+y=5#

#2x+2y=10#

These are linearly dependent, since equation 2 is just equation 1 multiplied by 2. The consequence would be, when we try to solve them we will end up with only one equation.

Since we have lost an equation, we can only solve this by assigning an arbitrary value to one of the variables and finding the other variables in terms of it. This leads to an infinite number of solutions.

Using back substitution and arbitrary #z#:

#y=(24-20z)/-4=>y=5z-6#

#x=4+z+5z-6=>x=6z-2#

So solutions are:

#(6z-2,5z-6,z)#