# Linear algebra system of equations

Given the following linear system:

$x - y + 2z + 3w = 12$

$x - 2y + z + w = 8$

$4x + y + 2z - w = 3$

1) find all solutions of the system. Write the solution in Vector Form
2) find all solutions of the corresponding homogeneous system

I have turned this into a matrix and worked it out two ways resulting in different outcomes, neither of which is solvable...

I'll try to show my work, but I am new to this site and the formatting for matrices is fairly hard to figure out...

1 -1 2 3 12
1 -2 1 1 8
4 1 2 -1 3 (-4 row1)

1 -1 2 3 12
1 -2 1 1 8 (-row1)
0 5 -6 -13 -45

1 -1 2 3 12
0 -1 -1 -2 -4
0 5 -6 -13 -45(+5row2)

1 -1 2 3 12
0 -1 -1 -2 -4
0 0 -11 -23 -65

from here I don't think there's anything else I can do, I'm fairly sure there are infinitely many solutions.

• Let's see the work done so we can help. – Vincent Sep 8 '14 at 20:18
• Wow thanks, but of course. – Daniel Charry Sep 8 '14 at 20:19
• By this computation, it appears that there should be a non-empty solution set. If you show us what you did, we could try to figure out where you went wrong. – Omnomnomnom Sep 8 '14 at 20:24
• I edited the question with my work, any thoughts? I'm thinking that It just needs to be put in vector form? – donsavage Sep 8 '14 at 20:43

Oh of course there is more that can be done!

So starting from where you left, I would first multiple $R_2$ with the scalar $(-1)$ and then add the new $R_2$ to $R_1$ and replace it. Then I get,

$$\left({\begin{matrix} 1 & 0 & 3 & 5 & 16 \\ 0 & 1 & 1 & 2 & 4 \\ 0 & 0 & -11 & -23 & -65 \\ \end{matrix}}\right)$$

$$\underrightarrow{\left({\frac{-1}{11}}\right) R_3 \ \text{and} -1 R_3 + R_2 \ \text{and} -3R_3 + R_1 }$$

$$\left({\begin{matrix} 1 & 0 & 0 & \frac{-14}{11} & \frac{-19}{11}\\ 0 & 1 & 0 & \frac{-1}{11} & \frac{-21}{11} \\ 0 & 0 & 1 & \frac{23}{11} & \frac{65}{11} \\ \end{matrix}}\right)$$

Now translating back to equations you have,

$$x = \frac{14}{11}w + \frac{19}{11}$$ $$y = \frac{1}{11}w + \frac{21}{11}$$ $$z = \frac{-23}{11}w + \frac{-65}{11}$$

Now any value you take for $w$ will solve the system. The method underlined here is made formal in Linear Algebra by Hoffman, Kunze which you should definitely read.

Now the solution set in vector form would be,

$$\left({\begin{matrix} x \\ y \\ z \\ w \\ \end{matrix}}\right) = w\left({\begin{matrix} \frac{14}{11} \\ \frac{1}{11} \\ \frac{-23}{11} \\ 1 \\ \end{matrix}}\right) + \left({\begin{matrix} \frac{19}{11} \\ \frac{21}{11} \\ \frac{-65}{11} \\ 0 \\ \end{matrix}}\right)$$

where $w$ is any scalar in $\Bbb R$.

• Could you explain why when converting it back into the equations the sign of the last two rows change? Shouldn't the last column keep it's sign? for example: z = -23/11w + 65/11 – donsavage Sep 10 '14 at 0:20
• Actually your partly right. The equation should be $z = \frac{-23}{11}w - \frac{65}{11}$ since the last row in the matrix $0 \; 0 \; 1 \; \frac{23}{11} \; \frac{-65}{11}$ represents the equation $z + \frac{23}{11}w = \frac{-65}{11}$. So lslight alteration needed. – Ishfaaq Sep 10 '14 at 1:35