Back substitution for an augmented matrix

Question

So let's say I have an augmented matrix $ \left[ \begin{array}{ccc|c} 1&1&-1&-2\\ 2&-1&3&14\\ -1&-2&1&3 \end{array} \right] $ and I have to solve for x such that $\begin{bmatrix}1&1&-1\\2&-1&3\\-1&-2&3\end{bmatrix}$ x $\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$ = $\begin{bmatrix}-2\\14\\3\end{bmatrix}$.

I started by using Gaussian elimination by putting the matrix into upper triangular form. $ \left[ \begin{array}{ccc|c} 1&1&-1&-2\\ 0&-3&5&18\\ 0&0&-5/3&-5 \end{array} \right] $ Then in order to solve for x I need to do back substitution but I'm not sure how to do that. The answer for this example is:

x$_3$ = $-5\over-5/3$ = 3

x$_2$ = $18 - (5)(3)\over-3$ = -1

x$_1$ = $-2 - (1)(-1) - (-1)(3)\over1$ = 2

I understand the solution for x$_3$ but what's going on for x$_2$ and x$_1$ and what is the value that's being multiplied in those equations?

Answer 1

The second equation says

$$-3x_2 +5x_3=18$$

Hence $$x_2=\frac{18-5x_3}{-3}$$

Notice that we have solved $x_3$ earlier.

The first equations say $$x_1+x_2-x_3=-2$$

Hence $$x_1=\frac{-2-(1)(x_2)+(1)x_3}{1}$$

Noticed that we have solved $x_2$ and $x_3$ earlier.

Answer 2

Note that for $x_2$, from the second equation of RREF, we have

$$-3x_2+5x_3=18 \implies -3x_2=18-5x_3$$

and from here we can solve.

For $x_1$ we can find a similar expression with $x_2$ and $x_3$ at the RHS.

Answer 3

It's much simpler than that: just proceed till you obtain the R.R.E.F. The solution is the last column, since the augmented matrix has maximal rank (I suppose the upper triangular form you obtain is correct): \begin{align} &\begin{bmatrix} 1&1&-1&\!\!\!|\!\!&-2\\ 0&-3&5&\!\!\!|\!\!&18\\ 0&0&-5/3&\!\!\!|\!\!&-5 \end{bmatrix}\rightsquigarrow \begin{bmatrix} 1&1&-1&\!\!\!|\!\!&-2\\ 0&1&-5/3&\!\!\!|\!\!&-6\\ 0&0&1&\!\!|\!\!\!&\phantom{-}3 \end{bmatrix}\rightsquigarrow \begin{bmatrix} 1&1&0&\!\!|\!\!&\phantom{+}1\\ 0&1& 0&\!\!|\!\!&-1\\ 0&0&1&\!\!|\!\!&\phantom{-}3 \end{bmatrix} \\[1ex]\rightsquigarrow &\begin{bmatrix} 1&0&0&\!\!|\!\!&\phantom{+}\color{red}2\\ 0&1& 0&\!\!|\!\!&\color{red}{-1} \\ 0&0&1&\!\!|\!\!&\phantom{-}\color{red}3 \end{bmatrix}. \end{align}