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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, 2021 at 18:47

1 Answer 1

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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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