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$$A= \begin{pmatrix} 1 & 1 & 0 \\ 1 & 2 & 1 \\ 1 & 3 & 2 \\ \end{pmatrix} $$

If $b = [ 2, 4 ,6 ]^T$ , how many solutions are there to the system $Ax = b$?

My guess is infinitely many because the echelon form produces last row to 0's but not sure

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    Yes, you are correct. – 2017-02-05

1 Answers 1

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Did you agument your matrix as follows:

\begin{bmatrix} 1 & 1 & 0 & 2\\ 1 & 2 & 1 & 4\\ 1 & 3 & 2 & 6\\ \end{bmatrix}

For your row reduced echelon form:

\begin{bmatrix} 1 & 0 & -1 & 0\\ 0 & 1 & 1 & 2\\ 0 & 0 & 0 & 0\\ \end{bmatrix}

The last row shows us we have a free variable. Call it $z$. Then our second row shows us that $y+z=2$ and our first row shows us that $x-z=0$. Or, $y=2-z$ and $x=z$.

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    I slapped an x column with random values such as all 1's and then I got the last row as 0's. Does that mean that $x^3$ and $x^4$ are just some value T? meaning infinite many solutions? I'm not sure if that is the right answer – 2017-02-05
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    Consider the reduced form. Because of that last row of 0s we see that $z=z$. This means that $y=2-z$ and $x=z$. – 2017-02-05