Question

Solving Linear Equations via Matrix: I just don't know where to start? I know I can do it with a push in the right direction ;u;
Here's the problem:
4 12 | -1
6 4 | 4

Answer 1

-3/2 R1 + R2 -> R2 is your first step.

Answer 2

Basically, you can solve the system of linear equations using elimination. By doing -3/2 R1 + R2 -> R2 , you eliminate 6 in the second row.

Answer 3

So multiply the first row by -3/2 and add it to row 2?

Answer 4

Yes. :)

Answer 5

I got:
-6 -18 | 1 1/2
0 -14 | -5 1/2
Should I try making -18=0 next?

Answer 6

Ok I did it! end answers were -173/42 and -361/43 which may have been a bit outlandish? But it worked?

Answer 7

The operation -3/2 R1 + R2 -> R2 does not change R1.
You multiply every entry in row 1 by -3/2, then add the corresponding entry to row 2, the result should be written in row 2, i.e., the entries in row 1 does NOT change in this operation. Here is how you do it:
\[\left[\begin{matrix}4 & 12 & | & -1 \\6 & 4 & | &4\end{matrix}\right]\]\[\rightarrow\left[\begin{matrix}4 & 12 & | & -1 \\0 & -14 & | &\frac{11}{2}\end{matrix}\right]\]
Then, divide R2 by -14, you'll get
\[\rightarrow\left[\begin{matrix}4 & 12 & | & -1 \\0 & 1 & | &-\frac{11}{28}\end{matrix}\right]\]
Now, divide R1 by 4, you'll get
\[\rightarrow\left[\begin{matrix}1 & 3 & | & -\frac{1}{4} \\0 & 1 & | &-\frac{11}{28}\end{matrix}\right]\]
The last step is to eliminate the 1,2-entry, i.e. 3 in row 1, by carrying the operation:
-3R2 + R1 -> R1. I'll leave it to you.