Gaussian elimination with matrices

Question

Hi I'm doing Gaussian elimination to read the solutions off a system of linear equations. I've gotten this far but I don't know where to go from here. I'm assuming I've made a mistake but I've been through it again and I get the same conclusion. (https://i.stack.imgur.com/qv3Ch.jpg)

Answer 1

Just because there's a row of zeroes doesn't mean you've made a mistake. Here's a neato thing that's going to happen - free variables.

I don't know your skill level but I'm going to approach this in the linear algebra way. Here goes.

The matrix you currently have is

\begin{pmatrix} 1 & 0 & 1.5 & -3 & 0\\ 0 & 1 & -0.5 & 2 & 1\\ 0 & 0 & 1 & -7 & -5\\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}

with the row of zeroes at the bottom. Continue the pivoting process, beginning from row 3. That is, transform the matrix into one in which the first three columns correspond to columns of $I_4$, the identity matrix of size $4 \times 4$. This will give

\begin{pmatrix} 1 & 0 & 0 & 7.5 & 7.5\\ 0 & 1 & 0 & -1.5 & -1.5\\ 0 & 0 & 1 & -7 & -5\\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}

This corresponds to the following system:

\begin{cases} x_1 + 7.5x_4 = 7.5 \\ x_2 - 1.5x_4 = -1.5 \\ x_3 - 7x_4 = -5 \\ x_4 \ \text{free} \end{cases}

To say that $x_4$ is free means that it can assume any value I choose - it is a parameter. But why is it free? The fourth column has no pivot, and so I cannot express $x_4$ in terms of the other variables.

To generate a solution, solve each of the preceding three equations for $x_4$, and generate a line. This is easily done and I suggest you try it out for yourself (including the remaining steps in the pivot process). If you get stuck I'll give the result.

Answer 2

I haven't checked the calculations very carefully, but they seem to be alright.

Now in the next step "clear" the third column, by which I mean using elementary row operations make $A_{33} = 1$ and $A_{13} = A_{23} = A_{43} = 0$. Then you can write the corresponding system of equations, but you will have the last variable ($x_4$), as a free variable, so you should express $x_1,x_2,x_3$ by $x_4.$