# [SOLVED] Matrix problem, a bit clueless

• September 14th 2008, 03:13 PM
arbolis
[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!
• September 14th 2008, 03:28 PM
Jhevon
Quote:

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
• September 14th 2008, 03:34 PM
Plato
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.
• September 14th 2008, 03:58 PM
arbolis
Quote:

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
Quote:

there is no leading 1 in the third column, so the variable for that column, namely http://www.mathhelpforum.com/math-he...a53b1844-1.gif, 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.

Quote:

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.
• September 14th 2008, 04:03 PM
Jhevon
Quote:

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

"echelon" was implied. :p

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

Quote:

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"
• September 14th 2008, 04:13 PM
Plato
Quote:

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.