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Good day, I'm currently studying for an exam and need to learn about matrices too. Well, since I'm not good at English I'll just write what I've done so far. Below is a photo showing the full sheet of paper with the steps I did so far.

The thing I'm wondering about is, that I'd have four unknown variables $(w, x, y, z)$ after writing down the equations, but my matrix has got 3 rows only. Therefor I'm wondering how I'm supposed to find a generic solution for A*x = b

$$A=\begin{bmatrix}0&2&-2&3\\ 1&3&-1&2\\ 2&3&1&0 \end{bmatrix} , b=\begin{bmatrix}-1\\ -2\\ 0 \end{bmatrix}$$

$$A=\begin{bmatrix}1&3&-1&2&-2\\ 0&1&-1&3/2&-1/2\\ 0&0&0&1/2&-3/2 \end{bmatrix} $$

$$z=-3 $$ $$\operatorname{rank}(A)=3 $$

So I actually solved for $z$ but I can't seem to solve for the other unknown variables (or can I?). The thing that confuses me is the fact this matrix is $3 \times 4$ now instead of $3 \times 3$. Any tips would be very much appreciated!

enter image description here

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learner
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beta
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3 Answers3

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We can write this as an augmented matrix:

$$\begin{bmatrix}0&2&-2&3&-1\\ 1&3&-1&2&-2\\ 2&3&1&0&0 \end{bmatrix}$$

Using Gaussian Elimination (Row-Reduced-Echelon-Form), we arrive at:

$$A=\begin{bmatrix}1&0&2&0&12 \\ 0&1&-1&0&-8\\ 0&0&0&1&5 \end{bmatrix}$$

This means our solution can be written as:

  • $z = 5$
  • $x = -8 + y$
  • $w = 12 - 2y$

$y$ is a free variable and this system has an infinite number of solutions.

You can easily test this solution by picking random values for $y$, substituting the values back into the system and validating equality.

Amzoti
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You need to introduce a free parameter (since four variables and rank is three). Looking at your work, there is no leading entry corresponding to the variable $y$. Set $y=s$, where $s$ is any element of the set of real numbers (if you are looking for real solutions). Then use the reduced form of the matrix to write down (equations) expressions for $w$,$x$ and $z$.

AnyAD
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  • I see, many thanks for your answer. Will redo the exercise and ask back if further questions turn up. :-) – beta Nov 10 '13 at 04:11
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Amzoti's answer gives a correct way to solve the matrix equation. I'm adding this answer to elaborate on the situation some more, and hopefully solve your confusion. The matrix corresponds to a linear map $A: \mathbb{R}^4 \rightarrow \mathbb{R}^3$ (or whatever other field you're working with). You have correctly noted that the rank of the matrix equals $3$, this means that the image of $A$ has dimension $3$. This means that the kernel of $A$ has dimension $4-3=1$. If a vector $c$ satisfies the equation $Ac = b$, and $d$ is any element in the kernel of $A$, we get $A(c+d) = Ac + Ad = Ac = b$, so $c+d$ is also a solution to the original equation. Conversely, if $c+e$ satisfies $A(c+e) = b$, then $Ae = A(c+e) - Ac = b - b = 0$, and $e$ lies in the kernel of $A$. So the set of solutions to $Ax = b$ looks like $\{c+e: e \in \textrm{ker}\: A\}$, where $c$ is a fixed solution to the equation. In particular, there will be one free variable, as the kernel of $A$ has dimension $1$. So at the point where you are stuck, you can pick one variable to be free, and proceed to express the other variables in this free variable.

Arthur
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  • Thanks for clarifying the situation. Didn't know I could pick a free variable. And yes you're right, this is in a R3 field, forgot to mention that! – beta Nov 10 '13 at 04:07