There are two ways to do this by matrices.
Augmented matrix (Gauss-Jordan) and inverse
method. I picked the 1st method:
We abbreviate that system with this augmented matrix:
We want 0's where the 2 and 5 are.
and 1's where the 3 (upper left corner)
and -4 are.
We'll get the 0's first.
To get a 0 where the 5 is,
multiply row 1 by -5 and
row 2 by 3 and replace row 2.
Here's the work: -15 -10 | -15
15 -12 | 81
---------------
0 -22 | 66
Since it turns out that that all those numbers
can be divided through by -22, we will do that
too, getting
0 1 | -3
and replace row 2 by that:
Then we get a 0 where the 2 is by multiplying the
second row by -2 and adding it to the first row:
Here's the work: 3 2 | 3
0 -2 | 6
---------------
3 0 | 9
Since it turns out that that all those numbers
can be divided through by 3, we will do that
too, getting
1 0 | 3
and replace row 1 by that:
This augmented matrix is the abbreviation for
this system:
or x = 3
y = -3
Edwin
