MHB -311.1.5.8 Ax=b in parametric vector form,

[karush]

Describe all solutions of $Ax=b$ in parametric vector form, where $A$ is row equivalent to the given matrix.
$A=\left[\begin{array}{rrrrr}
1&-3&-8&5\\
0&1&2&-4
\end{array}\right]$

RREF
$\begin{bmatrix}1&0&-2&-7\\ 0&1&2&-4\end{bmatrix}$
general equation
$\begin{array}{rrrrr}
x_1& &-2x_3&-7x_4 & =0\\
&x_2 &2x_3 &-4x_4&=0
\end{array}$
therefore
$x_1=2x_3+7x_4$
$x_2=-2x_3+4x_4$
assume next is $x=x_1[]+x_2[]+x_3[]+x_4[]$
but got ? looking at examples
anyway, so far

 

[HOI]

You say the problem is to solve Ax= b. So where is "b"?

 

[karush]

lets see if x is correct first...
$x=\begin{bmatrix}x_1\\x_2\\x_3\\x_4 \end{bmatrix}
=\begin{bmatrix}2x_3+7x_4\\-2x_3+4x_4\\x_3\\x_4 \end{bmatrix}
=\begin{bmatrix}0\\x_2\\0\\0 \end{bmatrix}
+\begin{bmatrix}2x_3\\-2x_3\\x_3\\0\end{bmatrix}
+\begin{bmatrix}7x_4\\4x_4\\0\\x_4\end{bmatrix}
=x_2\begin{bmatrix}0\\1\\0\\0 \end{bmatrix}
+x_3\begin{bmatrix}2\\-2\\1\\0 \end{bmatrix}
+x_4\begin{bmatrix}7\\4\\0\\1 \end{bmatrix}$

 

[HOI]

Why would x_1 be equal to 2x_3+ 7x_4????

 

[karush]

You say the problem is to solve Ax= b. So where is "b"?

ok very sorry but it looks this was supposed to be Ax=0

 

[karush]

Why would x_1 be equal to 2x_3+ 7x_4??

$\begin{array}{rrrrr} x_1& &-2x_3&-7x_4 & =0\\ &x_2 &2x_3 &-4x_4&=0 \end{array}$

??

 

[HOI]

You have, correctly, You have, correctly,
$x_1= 2x_3+ 7x_4$ and $x_2= -2x_3+ 4x_4$

So $\begin{bmatrix}x_1 \\ x_1 \\ x_3 \\ x_4 \end{bmatrix}= \begin{bmatrix}2x_3+ 7X_4 \\ -2x_3+ 4x_4 \\ x_3 \\ x_4 \end{bmatrix}= x_3\begin{bmatrix}2 \\ 0 \\ 1 \\ 0 \end{bmatrix}+ x_4\begin{bmatrix}7 \\ 4 \\ 0 \\ 1 \end{bmatrix}$.

Since you are given $x_1$ and $x_2$ in terms of $x_3$ and $x_4$ you should have only two vectors times $x_3$ and $x_4$.

 

[karush]

ok i need to practice this more its still to foggy:unsure: