00:01
One way to solve the system of linear equations is to write an augmented matrix.
00:06
And so we have in the first equation, 1x, 0y, 1z, 1w equals 4.
00:16
The second equation, we have no x, 1y, minus 1 z, no w equals negative 4.
00:24
In the third equation, we have 1x negative 2y, 3 z, 1w, 3 z, 1w.
00:33
Equals 12 and in the fourth and final equation we have 2x no y negative 2 z and 5 w equals negative 1 what we can do with this is take this to a calculator enter the matrix this would be a four row five column matrix and then you would input the elements so again in the first equation it was 1x no y 1 z one w you equals 4 and we have no x 1y negative 1 z no w equals negative 4 third equation 1 x negative 2 y equals 12 and fourth equation 2 x no y negative 2 z 5 w equals negative 1 once this is stored in the calculator then you can do row reduced row echelon form of that matrix and what what this tells us is there are infinitely many solutions because we don't have a unique solutions but we don't have the 4x4 identity on the left here.
01:48
So what we can say is from the first equation that would be x plus 1 .75 w equals 1 .75 w equals negative 1 .75 z minus 0 .75 w equals 2 .2 .2...