How Does Parameter 'a' Affect the Solution of This Matrix System?

[Hernaner28]

Hi. I want to reduce this matrix so then I can see what happens for each value of a (indeterminate, incompatible, etc) but I don't know how.

\displaystyle\left( {\begin{array}{*{20}{c}}<br /> a&amp;1&amp;1 \\ <br /> 1&amp;a&amp;1 \\ <br /> 1&amp;1&amp;a <br /> \end{array}} \right)

Thanks!

 

[HallsofIvy]

Do you not know anything about "row- reduction"? If you don't show any attempt at all, we won't know where you need help.

 

[Hernaner28]

Yes, look, I've have this system:

\displaystyle\left( {\begin{array}{*{20}{c}}<br /> a&amp;1&amp;1 \\ <br /> 1&amp;a&amp;1 \\ <br /> 1&amp;1&amp;a <br /> \end{array}\left| {\begin{array}{*{20}{c}}<br /> 1 \\ <br /> 1 \\ <br /> 1 <br /> \end{array}} \right.} \right)

And I want to reduce it to see what happens when I change a by a number. But I've tried to reduce it and I couldn't. Where can I start?
Thanks

 

[HallsofIvy]

You say you've tried to reduce it. Great! What have you tried? What have you done?

 

[Hernaner28]

\displaystyle\left( {\begin{array}{*{20}{c}}<br /> {a - 1}&amp;0&amp;1 \\ <br /> 0&amp;{a - 1}&amp;{1 - a} \\ <br /> 1&amp;1&amp;a <br /> \end{array}\left| {\begin{array}{*{20}{c}}<br /> 0 \\ <br /> 0 \\ <br /> 1 <br /> \end{array}} \right.} \right)

 

[HallsofIvy]

You appear to have subtracted the third row from the first and second rows. Okay, why? What was your objective?

 

[Hernaner28]

PLease I need help. It's obvious that my objective is to reduce the matrix to see what happens if for example a=2 ... if it is compatible or not. What else do you want me to tell you? How old I am?

 

[DryRun]

PLease I need help. It's obvious that my objective is to reduce the matrix to see what happens if for example a=2 ... if it is compatible or not. What else do you want me to tell you? How old I am?

HallsofIvy already told you to convert the matrix to row echelon form, in other words, use Gaussian elimination.

What else do you want me to tell you? How old I am?

That's the wrong way to ask anyone for help. Learn some manners and we might respect you enough to help you.

You have to be patient if you want to understand mathematics and learn how to fix your mistakes or misunderstandings. We have a great group of people on this forum who offer to help for free, but we are not here to do your work for you (that would be easier for us, actually). Instead, we do more than that -- we help you to understand, so with our guidance, you can solve your problems on your own, as you would have to do in the test/exam.

 

[Hernaner28]

But I've already tried Gauss elimination! Oh my god. But I just don't realize how to make a diagonal matrix, I am saying this since the first post. It seems you are pulling my leg asking me "what have you tried?" "Oh great! Tell us, what's your objetive?" Sorry but Let's leave it here, thank you anyway.

 

[DryRun]

Your Gaussian elimination is obviously wrong.

 

[Hernaner28]

when I have parameters in the diagonal it's difficult for me to reduce it. Can you show me how it would be?

Thanks

 

[DryRun]

When it's all about numbers yes but in this case when I have parameters in the diagonal it's difficult for me to reduce it. Can you show me how it would be?

Thanks

The variable a might throw you off at first, but you just have to stick to the same principles when converting to row echelon form. Work it out step by step.

\displaystyle\left[ {\begin{array}{*{20}{c}}<br /> a&amp;1&amp;1 \\ <br /> 1&amp;a&amp;1 \\ <br /> 1&amp;1&amp;a <br /> \end{array}\left| {\begin{array}{*{20}{c}} <br /> 1 \\ <br /> 1 \\ <br /> 1 <br /> \end{array}} \right.} \right]
The first step is to get zero under the matrix element a in the first column.

After the first step, the augmented matrix becomes:
\displaystyle\left[ {\begin{array}{*{20}{c}}<br /> a&amp;1&amp;1 \\ <br /> 0&amp;\left(\frac{a^2-1}{a}\right)&amp;\left(\frac{a-1}{a}\right) \\ <br /> 0&amp;\left(\frac{a-1}{a}\right)&amp;\left(\frac{a^2-1}{a}\right) <br /> \end{array}\left| {\begin{array}{*{20}{c}} <br /> 1 \\ <br /> \left(\frac{a-1}{a}\right) \\ <br /> \left(\frac{a-1}{a}\right) <br /> \end{array}} \right.} \right]
Work it out and check if you get the matrix above for the first step. Then, the second step is to get 0 under the matrix element \left(\frac{a^2-1}{a}\right) in the second column.
If you don't get the matrix above, then you will obviously not be able to proceed, so you need to go over your notes and review the method of doing Gaussian elimination again.

 

[Hernaner28]

Aham.. I see, so you had to divide the first row by a? So a cannot be zero. That's what I wanted to avoid, to divide a row by the parameter. But there's no other way, is it?
Thanks!

 

[Ray Vickson]

Aham.. I see, so you had to divide the first row by a? So a cannot be zero. That's what I wanted to avoid, to divide a row by the parameter. But there's no other way, is it?
Thanks!

What you really want to do is to solve the following system for x_1, x_2, x_3:
\begin{array}{rcl}<br /> a x_1 + x_2 + x_3 &amp;=&amp; r_1\\<br /> x_1 + a x_2 + x_3 &amp;=&amp; r_2\\<br /> x_1 + x_2 + a x_3 &amp;=&amp; r_3<br /> \end{array}<br />
where r_1, r_2, r_3 are some arbitrary parameters which we regard as "given".
Does the system have a unique solution? No solution? Infinitely many solutions?

I don't understand what is preventing you from getting started. For instance, you can start by using the first equation to solve for x_3 in terms of x_1 and x_2; then when you substitute that expression into the other two equations you will have two equations in the two unknowns x_1 and x_2. For some values of 'a' you might be stuck at that point and be unable to proceed further. However, for almost any value of 'a' you will be able to do one more step, say solving the second equation for x_2 in terms of x_1. Then, substituting that expression into your expression for x_3 and both of these into the remaining equation, you will have a single equation in x_1. For some values of 'a' and for some right-hand-sides this will incompatible (i.e., of the form 0*x_1 = r, where r ≠ 0); for most values of 'a' you will get a solution.

So, just go ahead and do the work.

RGV