# Thread: [SOLVED] Matrix problem, a bit clueless

1. ## [SOLVED] Matrix problem, a bit clueless

Find the reduced row equivalent to $A$, $R$ where $A=\left( \begin{array}{ccc} 3 & -1 & 2 \\ 2 & 1 & 2 \\ 1 & 3 & 2 \end{array} \right)$. I won't show my calculus but I found $R$ to be $\left( \begin{array}{ccc} 1 & 0 & \frac{4}{5} \\ 0 & 1 & \frac{2}{5} \\ 0 & 0 & 0 \end{array} \right)$.
Now the problem states : Find all the $X$ $\in \mathbb{R}(3\times 1)$ such that $RX=0$. So I had the following system $\big[ \begin{array}{c} x_1+\frac{4}{5}x_3=0 \\ x_2+\frac{2}{5}x_3=0 \end{array}$ I think this implies infinity solutions (but I'm not sure) so I thought I made an error. So I restarted the calculus of $R$ without following what I've done for my first try and I found out that $R=\left( \begin{array}{ccc} 1 & -2 & 0 \\ 0 & 0 & 0 \\ 0 & \frac{5}{2} & 1 \end{array} \right)$. Is that possible?
The problem says "the reduced row equivalent to $A$" so I made at least an error... And also if this calculus was right, I would have the same problem solving for $x_1$, $x_2$ and $x_3$. Please help me!

2. Originally Posted by arbolis
Find the reduced row equivalent to $A$, $R$ where $A=\left( \begin{array}{ccc} 3 & -1 & 2 \\ 2 & 1 & 2 \\ 1 & 3 & 2 \end{array} \right)$. I won't show my calculus but I found $R$ to be $\left( \begin{array}{ccc} 1 & 0 & \frac{4}{5} \\ 0 & 1 & \frac{2}{5} \\ 0 & 0 & 0 \end{array} \right)$.
Now the problem states : Find all the $X$ $\in \mathbb{R}(3\times 1)$ such that $RX=0$. So I had the following system $\big[ \begin{array}{c} x_1+\frac{4}{5}x_3=0 \\ x_2+\frac{2}{5}x_3=0 \end{array}$ I think this implies infinity solutions (but I'm not sure) so I thought I made an error. So I restarted the calculus of $R$ without following what I've done for my first try and I found out that $R=\left( \begin{array}{ccc} 1 & -2 & 0 \\ 0 & 0 & 0 \\ 0 & \frac{5}{2} & 1 \end{array} \right)$. Is that possible?
The problem says "the reduced row equivalent to $A$" so I made at least an error... And also if this calculus was right, I would have the same problem solving for $x_1$, $x_2$ and $x_3$. Please help me!
both are the same. the first answer you got is in the appropriate form

there is no leading 1 in the third column, so the variable for that column, namely $x_3$, is your parameter. you will have an infinite number of solutions based on its value. (i assume you reduced the matrix properly)

so you have:

$\boxed{x_3 = t}$

$x_1 + \frac 45t = 0 \implies \boxed{x_1 = - \frac 45t}$

$x_2 + \frac 25t = 0 \implies \boxed{x_2 = - \frac 25t}$

so that $\bold{X} = \left( \begin{array}{c} - \frac 45t \\ - \frac 25t \\ t \end{array} \right) = t \left( \begin{array}{c} - \frac 45 \\ - \frac 25 \\ 1 \end{array} \right)$

and that's it. you get a new solution for each t, so that's all solutions

3. I cannot understand your difficulty.
You are correct in the first try: $R = rref(A) = \left( {\begin{array}{ccr} 1 & 0 & {0.8} \\ 0 & 1 & {0.4} \\ 0 & 0 & 0 \\ \end{array} } \right)$.

The vectors you want are: $\left( {\begin{array}{r} { - 0.8\alpha } \\ 0 \\ \alpha \\ \end{array} } \right)\;\& \;\left( {\begin{array}{r} 0 \\ { - 0.4\beta } \\ \beta \\ \end{array} } \right)$
where alpha & beta are scalars.

4. the first answer you got is in the appropriate form
And why not my 2nd answer? Isn't it row reduced (the problem didn't say "echelon")?
And for
there is no leading 1 in the third column, so the variable for that column, namely , is your parameter.
so it seems I had to row reduce it echelonly because my third row could have been the second row, for example. Unless it wouldn't be under an appropriate form.

The problem states that there is only one row reduced matrix equivalent to $A$ and I found out at least 2. But Jhevon told me that one is not under the appropriate form, so it seems there's only one of them, now I understand it.
I was also doubting about an infinity of vectors that satisfy $AX=0$ because the next question ask "check out that for all the X found, $AX=0$." I thought there was a finite number of them, but it seems that it's not necessary.

5. Originally Posted by arbolis
And why not my 2nd answer? Isn't it row reduced (the problem didn't say "echelon")?
"echelon" was implied.

or at least, it is the form from which you can "see" the solutions easily.

And for so it seems I had to row reduce it echelonly because my third row could have been the second row, for example. Unless it wouldn't be under an appropriate form.
yes, but like i said. it's good to have those leading 1's.

by the way, both row-echelon (also called row-reduced form) and reduced row-echelon forms have leading 1's

your first answer happens to be in reduced-row echelon form. the second answer was in neither form. but was still "equivalent"

6. Originally Posted by arbolis
The problem states that there is only one row reduced matrix equivalent to [tex]A[/math]
I agree with that statement. I maintain that the first one you found is correct and is the only one.