# An exercise from a linear algebra book (hoffman and kunze)

I am not sure how to solve this exercise:

Find all solutions of \begin{align*}2x_1-3x_2-7x_3+5x_4+2x_5&=-2\\ x_1-2x_2-4x_3+3x_4+x_5&=-2\\ 2x_1\qquad-4x_3+2x_4+x_5&=3\\ x_1-5x_2-7x_3+6x_4+2x_5&=-7\;. \end{align*}

my row-reduced echelon matrix is:

$$\pmatrix{1&0&-2&1&0&1\\ 0&1&1&-1&0&2\\ 0&0&0&0&1&1\\ 0&0&0&0&0&0}$$

I don't see my mistake right now, I think I did something wrong xD
Thank you very much!

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Why do you think you did something wrong? – Gerry Myerson Aug 20 '12 at 23:27
Please double-check to make sure that I copied the equations correctly. – Brian M. Scott Aug 20 '12 at 23:28
You might want to bring this matrix into a canonical form, you are not done yet (and double check your work if you think you made a mistake along the way) – Belgi Aug 20 '12 at 23:33
@Belgi, the matrix is certainly in reduced row-echelon form, which is canonical. What did you have in mind? – Gerry Myerson Aug 20 '12 at 23:39
@GerryMyerson - my mistake. – Belgi Aug 20 '12 at 23:42

From the reduced form, you have the following equations: $$x_5 = 1, \ \ x_1-2 x_3 + x_4 = 1, \ \ x_2+x_3 - x_4 = 2.$$ Since this gives $x_1 = 1+2x_3-x_4$ and $x_2 =2-x_3+x_4$, we can choose $x_3,x_4$ arbitrarily and determine the corresponding $x_1,x_2$. Hence the solution set is given by: $$\begin{bmatrix} 1 \\ 2 \\ 0 \\ 0 \\ 1 \end{bmatrix}+ x_3 \begin{bmatrix} 2 \\ -1 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -1 \\ 1 \\ 0 \\ 1 \\ 0 \end{bmatrix},$$ with $x_3, x_4$ arbitrary.