Express last equation of system as sum of multiples of first two equations

Question

The question says to 'Express the last equation of each system as a sum of multiples of the first two equations."

System in question being:

$ x_1+x_2+x_3=1 $

$ 2x_1-x_2+3x_3=3 $

$ x_1-2x_2+2x_3=2 $

The question gives a hint saying "Label the equations, use the gaussian algorithm" and the answer is 'Eqn 3 = Eqn 2 - Eqn 1' but short of eye-balling it, I'm not sure how they deduce that after row-reducing to REF.

Answer 1

Hint:

We want to solve

$$a(x_1+x_2+x_3)+b(2x_1-x_2+3x_3)=x_1-2x_2+2x_3$$

Matching coefficients, we want to solve

$$a+2b=1$$

$$a-b=-2$$

$$a+3b=2.$$

Can you take it from here?

Answer 2

NOTE: $r_i$ is the original $i^{th}$ equation as stated in your question above.

Well, let's go through the process of finding the extended echelon form using Gauss-Jordan elimination. Here's the matrix: $$\left[\begin{matrix}1 \ 1 \ 1 \ 1 \\ 2 \ -1 \ 3 \ 3\\ 1 \ -2 \ 2 \ 2\end{matrix}\right]\left[\begin{matrix}1 \ 0 \ 0 \\ 0 \ 1 \ 0 \\ 0 \ 0 \ 1\end{matrix}\right]$$

First, we subtract the second row by twice the first row and the third row by the first row: $$\left[\begin{matrix}1 \ 1 \ 1 \ 1 \\ 0 \ -3 \ 1 \ 1 \\ 0 \ -3 \ 1 \ 1\end{matrix}\right]\left[\begin{matrix}1 \ 0 \ 0 \\ -2 \ 1 \ 0 \\ -1 \ 0 \ 1\end{matrix}\right]$$

Now, we subtract the third row by the second row (we don't really care about the first row at this point since we just want to know the numbers in the third row): $$\left[\begin{matrix}1 \ 1 \ 1 \ 1 \\ 0 \ -3 \ 1 \ 1 \\ 0 \ 0 \ 0 \ 0\end{matrix}\right]\left[\begin{matrix}1 \ 0 \ 0 \\ -2 \ 1 \ 0 \\ 1 \ -1 \ 1\end{matrix}\right]$$

Thus, since the third row in the matrix to the left is $\mathbf 0$ and third row in the matrix to the right is $1 \ -1 \ 1$, we have that $r_1-r_2+r_3=\mathbf 0$, or that $r_3=r_2-r_1$. Therefore, the third equation is the second equation minus the first.

Answer 3

You can do row reduction on the transpose: \begin{align} \begin{bmatrix} 1 & 2 & 1\\ 1 & -1 & -2\\ 1 & 3 & 2\\ 1 & 3 & 2\\ \end{bmatrix} &\to \begin{bmatrix} 1 & 2 & 1\\ 0 & -3 & -3\\ 0 & 1 & 1\\ 0 & 1 & 1\\ \end{bmatrix} &&\begin{aligned} R_2&\gets R_2-R_1\\ R_3&\gets R_3-R_1\\ R_4&\gets R_4-R_1 \end{aligned} \\[6px] &\to \begin{bmatrix} 1 & 2 & 1\\ 0 & 1 & 1\\ 0 & 1 & 1\\ 0 & 1 & 1\\ \end{bmatrix} &&R_2\gets-R_2/3 \\[6px] &\to \begin{bmatrix} 1 & 2 & 1\\ 0 & 1 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0\\ \end{bmatrix} &&\begin{aligned} R_3&\gets R_3-R_2\\ R_4&\gets R_4-R_2 \end{aligned} \\[6px] &\to \begin{bmatrix} 1 & 0 & -1\\ 0 & 1 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0\\ \end{bmatrix} &&R_1\gets R_1-2R_2 \end{align} which makes clear that $C_3=-C_1+C_2$