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I'm trying to solve the following matrix and I'm getting a contradiction yet I know there is a solution as I've graphed it and the book has the same answer. What am I doing wrong?

$$\left[\begin{array}{rrr|r}1&1&1&3\\2&-1&1&0\\-3&5&7&7\end{array}\right]$$ Now the book gets (1,2,0) as an answer. And here is my attempt.

Replace R2 with -2R1+R2 and I also replace R3 with 3R1+R3: $$\left[\begin{array}{rrr|r}1&1&1&3\\0&3&1&6\\0&8&10&16\end{array}\right]$$

Then I multiply R2 by 1/3 as to get a 1 in the leading term: $$\left[\begin{array}{rrr|r}1&1&1&3\\0&1&1/3&2\\0&8&10&16\end{array}\right]$$

Now I want to eliminate the leading term of R3 so I replace it with -8R2+R3 and get: $$\left[\begin{array}{rrr|r}1&1&1&3\\0&1&1/3&2\\0&0&22/3&0\end{array}\right]$$

And so 22/3 = 0 is a contradiction. What the heck am I doing wrong?

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There is not any contradiction, we simply have $z=0$ and indeed the solution is $(1,2,0)$.

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  • $\begingroup$ Ahh...right because that term 22/3z = 0 and so z = 0. Ugh...Thanks! $\endgroup$ – maybedave Oct 12 '19 at 22:02
  • $\begingroup$ @maybedave Yes of course! The problem there was if we had for example something like $$\left[\begin{array}{rrr|r}1&1&1&3\\0&1&1/3&2\\0&0&0&22/3\end{array}\right]$$ $\endgroup$ – user Oct 12 '19 at 22:03
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Remember what the augmented matrix is shorthand notation for. You're trying to solve $Ax=b,$ where $$A=\begin{bmatrix} 1 & 1 & 1\\ 2 & -1 & 1\\ -3 & 5 & 7\end{bmatrix},$$ $$x=\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix},$$ and $$b=\begin{bmatrix}3 \\ 0 \\ 7\end{bmatrix},$$ and you're getting that $\frac{22}{3}x_3=0.$ So, you're not getting that $22/3=0,$ but rather that $x_3=0,$ which is perfectly fine.

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