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I am preparing for my Linear Algebra paper which is coming up in 4 weeks so I am working through questions in each paper.

I am puzzled with one question I have come across and perhaps someone can help me see the light.

Express the following matrix equation

$$x_1 \begin{bmatrix} 2\\ 5\\ \end{bmatrix}+x_2 \begin{bmatrix} 3\\-4 \end{bmatrix} +x_3 \begin{bmatrix} -2\\2 \end{bmatrix} =\begin{bmatrix} 5\\6 \end{bmatrix}$$

into a system of linear equations and show $x_1 = 2, x_2 = 3, x_3 = 4$ is a solution to the matrix equation and to the resulting system of linear equations.

So having an understanding of matrix multiplication and linear combinations I figure I'll go ahead and find the reduced row echelon form of the matrix by applying a series of elementary row operations.

\begin{bmatrix}2&3&-2&5\\5&-4&2&6\end{bmatrix}

\begin{bmatrix}1&3/2&-1&5/2\\5&-4&2&6\end{bmatrix}

\begin{bmatrix}1&3/2&-1&5/2\\0&-23/2&7&-13/2\end{bmatrix}

\begin{bmatrix}1&3/2&-1&5/2\\0&1&-14/2&13/23\end{bmatrix}

\begin{bmatrix}1&0&-2/23&38/23\\0&1&-14/23&13/23\end{bmatrix}

I cannot see how I could get $x_1 = 2, x_2 = 3, x_3 = 4$ as solutions to this matrix.

Please could someone advise if I am overlooking something here?

Thanks

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    $\begingroup$ You can choose $x_3$ aribitary. Then, $x_1$ and $x_2$ are also determined. Set $x_3=4$ and solve the equations using this information. $\endgroup$ – Peter Apr 18 '16 at 19:14
  • $\begingroup$ Clearly, your problem is that the second row is supposed to have a $1$, not a $-1$. $\endgroup$ – Noble Mushtak Apr 18 '16 at 19:22
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    $\begingroup$ @Peter No, his reduced row echelon form is wrong. Look at the link I sent you. $\endgroup$ – Noble Mushtak Apr 18 '16 at 19:25
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    $\begingroup$ @NobleMushtak No that was a typo on my part which I have corrected. Thanks $\endgroup$ – Karma88 Apr 18 '16 at 19:25
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Hint : Your final equations are

$$x_1-\frac{2}{23}x_3=\frac{38}{23}$$

and

$$x_2-\frac{14}{23}x_3=\frac{13}{23}$$

Set $x_3=4$ and determine $x_1$ and $x_2$. You get one of infinite many solutions this way.

Choosing another real value for $x_3$, you get another solution this way.

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    $\begingroup$ This answers my question! Thank you. $\endgroup$ – Karma88 Apr 18 '16 at 19:26

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