The system is:
$x_1 + 2x_2 - x_3 = 2, $
$2x_1 - x_2 + x_3 = 1 $,
$-x_1 + 4x_2 -2x_3 = a $
To begin solving the system, I did Row 2 - 2(Row 1). Row 3 + Row 1, then Row 3 + Row 2.
This left me with
\begin{bmatrix}1&2&-1&2\\0&-5&3&-3\\0&0&0&a-1\end{bmatrix}
If all I am trying to find is if the system is inconsistent, do I need to continue reducing it completely to reduced echelon form? Or is it that since I solved the row with a, I can stop there leaving the system as consistent when a = 1.
$\endgroup$ 01 Answer
$\begingroup$Generally you don't need to continue reducing to reduced echelon form since you can see that for the system to be consistent you need $a-1=0$ so $a=1$. But there is a a mistake.
Row 3 + Row 1 gives
$$\begin{bmatrix}1 & 2 & -1 & 2\\0 & -5 & 3 & -3\\0 & 6 & -3 & a+2\end{bmatrix}$$
so Row 3 + Row 2 will give
$$\begin{bmatrix}1 & 2 & -1 & 2\\0 & -5 & 3 & -3\\0 & 1 & 0 & a-1\end{bmatrix}$$
Then you can continue to do Row 2 + 5$\times$Row3 to give$$\begin{bmatrix}1 & 2 & -1 & 2\\0 & 0 & 3 & 5a-8\\0 & 1 & 0 & a-1\end{bmatrix}$$
and similarly Row 1 -2$\times$Row 3
$$\begin{bmatrix}1 & 0 & -1 & 4-2a\\0 & 0 & 3 & 5a-8\\0 & 1 & 0 & a-1\end{bmatrix}$$
and finally Row 1 +$\frac{1}{3}\times$ Row 2 gives$$\begin{bmatrix}1 & 0 & 0 & \frac{4}{3}-\frac{a}{3}\\0 & 0 & 3 & 5a-8\\0 & 1 & 0 & a-1\end{bmatrix}$$
So you can see that the solutions are $x_{1}=\frac{4}{3}-\frac{a}{3},x_{2}=a-1$ and $x_{3}=\frac{5a}{3}-\frac{8}{3}$. Thus the system is consistent for all a.
Indeed we can check:
- $x_{1}+2x_{2}-x_{3}=\frac{4}{3}-\frac{a}{3}+2a-2-\frac{5a}{3}+\frac{8}{3}=2$
- $2x_{1}-x_{2}+x_{3}=\frac{8}{3}-\frac{2a}{3}-a+1+\frac{5a}{3}-\frac{8}{3}=1$
- $-x_{1}+4x_{2}-2x_{3}=-\frac{4}{3}+\frac{a}{3}+4a-4-\frac{10a}{3}+\frac{16}{3}=a$
as required.
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