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Can somebody help me find the simplest way to use elementary operations of multiplying one row by another and scalar multiplication of a row to find the solution to the system of equations?

$$2x_1+4x_2-x_3=7$$ $$x_1+x_2-x_3=0$$ $$3x_1-2x_2+3x_3=8$$

Every time I solve this it feels like it could have been done simpler... Thanks!


My attempt:

I first exchanged row 1 and row 2 to get:

$$x_1+x_2-x_3=0$$ $$2x_1+4x_2-x_3=7$$ $$3x_1-2x_2+3x_3=8$$

Then I did $-2R_1+R_2 \rightarrow R_2$ and $-3R_1+R_3 \rightarrow R_3$ to get:

$$x_1+x_2-x_3=0$$ $$0x_1+2x_2+x_3=7$$ $$0x_1-5x_2+6x_3=8$$

But now I can't use row 2 to eliminate the $x_2$ in row 3 without introducing fractions. I realize i could use back substitution, but I want to use row operations. Was there a different way I could have carried out the row operations to make this nicer? Thank you!

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  • $\begingroup$ It is all right if your equation coefficients are fractions. $\endgroup$ – Saeed Jan 15 at 18:47
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What you can do is multiply rows by nonzero constants. For instance $5R_2 \to R_2$ and $2R_3 \to R_3$. Then you can cancel the $x_2$ term in the last equation without getting a fraction in the coefficients. Of course, if there are fractions in the right-hand side, we cannot avoid those.

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We are given $$\begin{pmatrix} 2 & 4 & -1\\ 1 & 1 & -1\\ 3 & -2 & 3 \end{pmatrix} \begin{pmatrix} x_1\\ x_2\\ x_3 \end{pmatrix}= \begin{pmatrix} 7\\ 0\\ 8 \end{pmatrix}$$

Take $R_1 - 2R_2 \to R_2$ and $-3R_1+2R_3 \to R_3$ to form $$\begin{pmatrix} 2 & 4 & -1\\ 0 & 2 & 1\\ 0 & -16 & 9 \end{pmatrix} \begin{pmatrix} x_1\\ x_2\\ x_3 \end{pmatrix}= \begin{pmatrix} 7\\ 7\\ -5 \end{pmatrix}$$ Then let $8R_2 + R_3 \to R_3$ to obtain $$\begin{pmatrix} 2 & 4 & -1\\ 0 & 2 & 1\\ 0 & 0 & 17 \end{pmatrix} \begin{pmatrix} x_1\\ x_2\\ x_3 \end{pmatrix}= \begin{pmatrix} 7\\ 7\\ 51 \end{pmatrix}$$ We then find that $x_1=1,x_2=2,x_3=3$.

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  • $\begingroup$ Note that $-3 R_1 + 2 R_3 \to R_3$ is not an elementary row operation $\endgroup$ – Dunham Jan 16 at 0:14

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