Solve the system..

Solve the system

$$4x_1 - 5x_2 + 2x_3 + 2x_4 = 1$$

$$-x_1 + x_2 + 2x_3 + 3x_4 = 2$$

$$3x_1 -4x_2 + 4x_3 + 5x_4 = 3$$

$$3x_1 - 3x_2 -6x_3 - 9x_4 = -6$$

$$\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix} = \begin{bmatrix}\\\\\\\\\end{bmatrix} + \begin{bmatrix}\\\\\\\\\end{bmatrix} s + \begin{bmatrix}\\\\\\\\\end{bmatrix} t$$

so... I threw this horrid system of equations into a augmented matrix and solved the RREF form to get

$$\begin{bmatrix}1&0&-12&-17&-11\\0&1&-10&-14&-9\\0&0&0&0&0\\0&0&0&0&0\end{bmatrix}$$

but I'm not sure now how to solve for the three vectors. (also are these vectors I'm filling out or are these $$4 \times 1$$ matrixes??

• Good, you've got it in RREF. Now what do you think you should do?
– user137731
Jun 25 '16 at 20:36
• Try to solve for $x_1, x_2 , x_3, x_4$ for each row? Jun 25 '16 at 20:40
• The columns with out the 1 are the "free variable" columns right? Jun 25 '16 at 20:41
• How about we set some of the $x_i$ equal to a free variable? No -- it's the columns without a pivot. So it'll be your third and fourth columns (the last is of course a little different because this is an augmented matrix).
– user137731
Jun 25 '16 at 20:41
• Nope. Start with $\pmatrix{x_1 \\ x_2 \\ x_3 \\ x_4} = \pmatrix{-12s-17t+11 \\ 10s+14t-9 \\ s \\ t}$. Then just split this into separate columns $$\pmatrix{-12s \\ 10s \\ s \\ 0}+\pmatrix{-17t \\ 14t \\ 0 \\ t} + \pmatrix{11 \\ -9 \\ 0 \\ 0}=\pmatrix{-12 \\ 10 \\ 1 \\ 0}s+\pmatrix{-17 \\ 14 \\ 0 \\ 1}t + \pmatrix{11 \\ -9 \\ 0 \\ 0}$$
– user137731
Jun 25 '16 at 21:06

The row reduced form says \begin{align*} x_1 & =12x_3+17x_4-11,\\ x_2 & =10x_3+14x_4-9. \end{align*} In other words, setting $x_3=s$, $x_4=t$, the solutions can be written as $$\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}=\begin{bmatrix}12\\10\\1\\0\end{bmatrix}s+\begin{bmatrix}17\\14\\0\\1\end{bmatrix}t\,- \begin{bmatrix}11\\ 9\\0\\0\end{bmatrix}.$$

• Ok, i get confused on the parameter part with $s$ and $t$ . should it now be like $\begin{bmatrix}-12\\10\\s\\t\end{bmatrix}$ + $\begin{bmatrix}-17\\14\\s\\t\end{bmatrix}s$ + $\begin{bmatrix}11\\-9\\s\\t\end{bmatrix}t$ ? Jun 25 '16 at 20:56
• Am I incorrect? Jun 25 '16 at 21:02
• The only problem is that the OPs reduced form does not match his original equations. Jun 25 '16 at 21:04
• my RREF is incorrect? @almagest Jun 25 '16 at 21:05
• @almagest good catch, edited it. was a typo when i typed in the RREF form Jun 25 '16 at 21:08

When you have free variables, then you can express the non-free variables in terms of them.

So you will get some equation formed by the rows in RREF for instance:

$x_1 - 2x_3 = 3$

$x_2 + x_3 = 1$

Now solve for x1 and x2...

$x_1 = 3 + 2x_3$

$x_2 = 1 - x_3$

So now the vector $<x_1,x_2,x_3> = <3+2x_3,1-x_3,x_3> = <3,1,0> + <2,-1,1>x_3$

Does the process make sense to you now?

You need to remember that each row of an augmented matrix is an equality relation. So when you place it into row reduced echelon form, you recieve a series of identities. Each row as you read down will give the identity for that index. (Row 1 gives the expression you write in place of x1, row 2 for x2, etc.). Empty rows mean that the variable is free. This is assuming that the matrix is square. If it is not square then variables that are unlisted are free. I am sure you can read it close enough to understand. Just solve for each variable such that the number of variables in the relationship has been limited.

You also asked if this was vectors or matrices. Sorry kid... I hate to break it to ya. Vectors are matrices. An augmented matrix is just a way of writing the matrix equation Ax = b. Nothing explicitly states that x must be a vector. Thats just the convention. Vectors are matrices with one column.

In fact, matrix multiplication is defined as special vector multiplication.

Let b1... bn be vectors:

A[b1, b2, b3, bn] = [Ab1, Ab2, Ab3... Abn]

As you can see, this makes matrix multiplication the same as vector multiplication.

• OK. No worries.
– user137731
Jun 25 '16 at 21:22