find a basis and the dimensions of the solution space w

Question

$$x+2y-2z+2s-t=0$$

$$x+2y-z+3s-2t=0$$

$$2x+4y-7z+s+t=0$$

I need to find the basis and dimensions. I'm not sure how to do it. The book I have doesn't have a very good example.

I end up with: $$ \left( \begin{array}{ccc} 1 & 2 & 0 & 4 & -3 \\ 0 & 0 &1 & 1 &-1 \\ 0 &0& 0&0 &0\\ \end{array} \right)$$

I got this as my row reduced form, the here I'm not sure where to go to get the basis. I get that the dimension will be three

Here is my work:

R2-R1 = 0 0 1 1 -1 0

R3-R1 = 0 0 -3 -3 3 0

R3/3 = 0 0 -1 -1 1 0

-R3-R2 = 0 0 0 0 0 0

R1-R2 = 1 2 0 4 -3 0

which in turn gives me $$ \left( \begin{array}{ccc} 1 & 2 & 0 & 4 & -3 \\ 0 & 0 &1 & 1 &-1 \\ 0 &0& 0&0 &0\\ \end{array} \right)$$

Can you form the matrix of coefficients of the system? Well, now reduce (by row, by columns: as you wish) this matrix. Write down in your question what you get and from now we'll continue... — DonAntonio, May 06 '14 at 16:37
Well, I can't know what you did but that looks incorrect. I did it and the third row becomes all zeros, meaning the solution space has dimension three... — DonAntonio, May 06 '14 at 16:48
Where do you get that from? You really need to show your work...I can't understand those $;0,,,4,,,-3;$ in the first line. — DonAntonio, May 06 '14 at 17:26

score 1 · Answer 1 · answered May 06 '14 at 17:00

$$\begin{pmatrix}1&2&-2&2&-1\\ 1&2&-1&3&-2\\ 2&4&-7&1&\;\;1\end{pmatrix}\stackrel{\begin{cases}R_2-R_1\\R_3-2R_1\end{cases}}\longrightarrow\begin{pmatrix}1&2&-2&\;\;2&-1\\ 0&0&\;\;\,1&\;\;1&-1\\ 0&0&-3&-3&\;\;3\end{pmatrix}$$

Clearly, the third row becomes all zeros in the next step (do it), so...etc.

find a basis and the dimensions of the solution space w

1 Answers1