Solving linear system after Gaussian elimination

Question

If I had an initial $2 \times 3$ matrix $A$ and a $2 \times 1$ constant vector $b$, such that I have $A$ and $b$ of $Ax=b$:

$$A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}$$ $$b = \begin{pmatrix} 7 \\ 8 \end{pmatrix}$$

Such that in augmented form it appears as such;

$$\left(\begin{array}{ccc|c} 1 & 2 & 3 & 7 \\ 4 & 5 & 6 & 8 \end{array}\right)$$

And valid operations were performed such that it was in $r.e.f.$ (the values here are just examples):

$$\left(\begin{array}{ccc|c} 1 & -3 & 2 & 11 \\ 0 & 1 & -9 & 2 \end{array}\right)$$

How does one solve this system, if there is not a row that only contain:

$$\left(\begin{array}{ccc|c} 0 & 0 & 1 & N \end{array}\right)$$

...so that you could back solve the entire system?

Answer 1

Lets take a step back from the linear algebra to more basic algebra.

$x -3y +2z = 11\\ 0x +1y -9z = 2$

Since you have two equations and 3 unknowns there will be a "line of solutions."

You can set $z = 0$ and find a pair of $x,y$ that will solve the equation.

i.e.

$x - 3y = 11\\ 0x +1y = 2$

$(17,2,0)$ is a solution.

Then set $z = 1$

$x - 3y +2= 11\\ 0x +1y -9= 2$

$(42, 11, 1)$ is also a solution

the full line of solutions is $(17,2,0) + (25,9,1) t$

It is worth noting that:

$\begin{bmatrix} 1&-3&2\\0&1&-9\end{bmatrix}\begin{bmatrix} 17\\2\\0\end{bmatrix} = \begin{bmatrix} 11\\2\end{bmatrix}$

and $\begin{bmatrix} 1&-3&2\\0&1&-9\end{bmatrix}\begin{bmatrix} 29\\9\\1\end{bmatrix} = \bf 0$

Answer 2

In your example, you wind up with a system of two equations $y-9z=2$ and $x-3y+2z=11$... so $y=2+9z$ and, substituting into the second, $x-3(2+9z)+2z=11$... So $x=17+25z$...

Thus we get solutions of the form $(17+25z,2+9z,z)$, where $z$ can be anything at all... This is a $1$-dimensional space...