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$x\cdot\begin{bmatrix}0\\-3\\1\end{bmatrix}$ + $y\cdot\begin{bmatrix}-5\\2\\-4\end{bmatrix}$ + $z\cdot\begin{bmatrix}-20\\-1\\-13\end{bmatrix}$ =$\begin{bmatrix}0\\0\\0\end{bmatrix}$

x = ？ y = ? z = ?

RREF = $\begin{bmatrix}1&0&3\\0&1&4\\0&0&0\end{bmatrix}$

I started with looking for the Reduced Row Echelon Form, but don't know what to do next.

YZY
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  • Do you miss the plus signs? And what methods have you known? –  Oct 05 '19 at 22:17
  • Hi Jack, I tried to make in into RREF, but am I right to start with that? – YZY Oct 05 '19 at 22:19
  • This gives you a system of three linear equations in three unknowns, which can be solved in the standard manner. Have you attempted this, or are you confused in how to set up the problem? – Arturo Magidin Oct 05 '19 at 22:19
  • Since i know these 3 set of vectors are linear dependent, so i'm trying to figure out the scalars. – YZY Oct 05 '19 at 22:20
  • Do you have difficulties in finding the RREF? Or you have found the RREF but don't know how to go on? –  Oct 05 '19 at 22:21
  • I've found the RREF, but don't know what to do next – YZY Oct 05 '19 at 22:22
  • Good. Then you should add that into your post. I assume that you may have seen an example showing how to read solutions (if any) for the RREF. –  Oct 05 '19 at 22:25
  • Welcome to Mathematics Stack Exchange. How about $x=y=z=0?$ – J. W. Tanner Oct 05 '19 at 22:40
  • Hi, since 3 of them are linear dependent, so 0 is not the only solution of x,y,z. I am trying to find solution that is not 0 – YZY Oct 05 '19 at 22:41
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    I don't find the same last row in the RREF. – Bernard Oct 05 '19 at 22:44
  • You got the RREF wrong. One would get the $3\times 3$ identity matrix if one does the calculation correctly and thus $x=y=z=0$ is the unique solution.

    You can use this online tool to see all the intermediate steps.

     –  Oct 05 '19 at 22:58
  • Hi guys.. My bad, the second number in the second vector should be 2 instead of -2.. Then the RREF should be correct. – YZY Oct 06 '19 at 02:48

3 Answers3

1

$x\cdot\begin{bmatrix}0\\-3\\1\end{bmatrix}$ + $y\cdot\begin{bmatrix}-5\\2\\-4\end{bmatrix}$ + $z\cdot\begin{bmatrix}-20\\-1\\-13\end{bmatrix}$ =$\begin{bmatrix}0\\0\\0\end{bmatrix}$

$\implies \begin{bmatrix}-5y-20z\\-3x+2y-z\\x-4y-13z\end{bmatrix}$=$\begin{bmatrix}0\\0\\0\end{bmatrix}$

$\implies y=-4z, -3x-9z=0, x+3z=0$

$\implies y=-4z, x=-3z$

$\implies\begin{bmatrix}x\\y\\z\end{bmatrix}=c\begin{bmatrix}-3\\-4\\1\end{bmatrix}$

J. W. Tanner
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0

Supposin your RREF is exact (which it is not), there remains to interpret it as an equivalent linear system of equations: $$\begin{cases} x+3z=0\\y+4z=0 \end{cases}\implies \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-3z\\-4z\\\quad z\end{bmatrix}=z\begin{bmatrix}-3\\-4\\\quad1\end{bmatrix}.$$

Bernard
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Your matrix is full rank which can be verified with numerical maths software (such as MATLAB).

That means your vectors are in fact linearly independent.

Therefore, only the trivial solution exists: $\begin{bmatrix}x\\y\\z\end{bmatrix} = \bf{0}$

wgb22
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