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 October 15th, 2018, 04:09 AM #1 Newbie   Joined: Jan 2014 Posts: 19 Thanks: 0 Linear system Question: Using the fact that $\displaystyle \textbf{Ax=b}$ is consistent if and only if $\displaystyle \textbf{b}$ is a linear combination of the columns of $\displaystyle \textbf{A}$ to find a solution to $\displaystyle \left( \begin{array}{cccc} 1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1 \\ 3 & 4 & 1 & 2 \end{array} \right)$$\displaystyle \left( \begin{array}{c} x\\ y\\ z\\ w \end{array} \right)$$\displaystyle =\left( \begin{array}{c} 20\\ 20\\ 20\end{array} \right).$ My work: I rewrite the given system into $\displaystyle x\left( \begin{array}{c} 1\\ 2\\ 3\end{array} \right)$$\displaystyle +y\left( \begin{array}{c} 2\\ 3\\ 4\end{array} \right)$$\displaystyle +z\left( \begin{array}{c} 3\\ 4\\ 1\end{array} \right)$$\displaystyle +w\left( \begin{array}{c} 4\\ 1\\ 2\end{array} \right)$$\displaystyle =\left( \begin{array}{c} 20\\ 20\\ 20\end{array} \right).$ What should I do next? Please give some hints. Thank you.
 October 15th, 2018, 07:51 AM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics Notice that all 3 of the rows sums to 10.
 October 15th, 2018, 08:15 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,831 Thanks: 2161 As, x = y = z = w = 2 is an obvious solution, the system is consistent. You can use $\text{uAx = ub}$, where $\text{u}$ is the vector (1, -2, 1) if you want to find out more.
October 16th, 2018, 07:35 AM   #4
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Quote:
 Originally Posted by woo My work: I rewrite the given system into $\displaystyle x\left( \begin{array}{c} 1\\ 2\\ 3\end{array} \right)$$\displaystyle +y\left( \begin{array}{c} 2\\ 3\\ 4\end{array} \right)$$\displaystyle +z\left( \begin{array}{c} 3\\ 4\\ 1\end{array} \right)$$\displaystyle +w\left( \begin{array}{c} 4\\ 1\\ 2\end{array} \right)$$\displaystyle =\left( \begin{array}{c} 20\\ 20\\ 20\end{array} \right).$ What should I do next? Please give some hints. Thank you.
x+2y+3z+4w=20
x+3y+4z+ w=20
3x+4y+z+2w=20

Put in augmented matrix form and sove by row reduction to get:

1 2 3 | 20 -4w
2 3 4 | 20 -w
3 4 1 | 20 -2w

1 2 3 | 20 -4w
0 1 2 | 20 -7w
0 0 -2| 0 -2w

z=w
y=20-11w
x=20-9w

$\displaystyle \begin{bmatrix} x\\ y\\ z \end{bmatrix}=\begin{bmatrix} 20-9w\\ 20-11w\\ w \end{bmatrix}=\begin{bmatrix} 20\\ 20\\ 0 \end{bmatrix}+w\begin{bmatrix} -9\\ -11\\ 1 \end{bmatrix}$

You can ck the algebra.

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