I figure it never hurts getting as much practice as possible solving systems of linear equations, so let's solve this one.
What I'm going to do is I'm going to solve it using an augmented matrix, and I'm going to put it in reduced row echelon form.
Each pivot entry in each successive row is to the right of the pivot entry before it.
My pivot entries are the only entries in their columns. So let's go back from the augmented matrix world and kind of put back our variables there.
So let me replace the third row with the third row minus 2 times the second row. 3 minus 2 times 2, that's 3 minus 4, or minus 1.
2 minus 2 times 1, that's 2 minus 2, that's 0.
This method can also be used to find the rank of a matrix, to calculate the determinant of a matrix, and to calculate the inverse of an invertible square matrix.
The method is named after Carl Friedrich Gauss (1777–1855), although it was known to Chinese mathematicians as early as 179 AD.
For example, in the following sequence of row operations (where multiple elementary operations might be done at each step), the third and fourth matrices are the ones in row echelon form, and the final matrix is the unique reduced row echelon form.
Using row operations to convert a matrix into reduced row echelon form is sometimes called Gauss–Jordan elimination.