Help with a matrix (work shown): can't get nice book answer

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DarkSun

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As part of a bigger task, I have to bring the following matrix to a row-reduced achelon form, then find when it has no solution.

Basically I know how it's done, but I got into trouble with this specific matrix...
I have tried in several ways, but still get something different (and more complex) than the book-solution.

I'll be glad if anyone could tell me where is my mistake (once reduced, I know how to proceed checking the solutions)...
 

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Re: Help with a matrix

Finding the reduced row echelon form, I get:

\(\displaystyle \begin{bmatrix}1&0&0&\frac{2a^{2}+a+2}{4(a+1)}\\0&1&0&\frac{2-a}{4(a+1)}\\0&0&1&\frac{2-a}{4(a+1)}\end{bmatrix}\)

The book sure has a long connected solution.

One obvious 'a' value for which there is no solution is a = -1. See why?.
 
Re: Help with a matrix

Yes, from you solution it's obviously -1,
But...how did you get to this nice form?
 
obviously...
and more specific,
how do you avoid getting the expressions too complex?
 
Initially, I got:

\(\displaystyle \begin{bmatrix}1&0&0&\frac{3}{4(a+1)}+\frac{a}{2}-\frac{1}{4}\\0&1&0&\frac{3}{4(a+1)}-\frac{a-2}{3}+\frac{a}{3}-\frac{11}{12}\\0&0&1&\frac{2-a}{4(a+1)}\end{bmatrix}\)

Then, found a common denominator for the upper two of 4(a+1).

Take the first one:

\(\displaystyle \frac{3}{4(a+1)}+\frac{2a(a+1)}{4(a+1)}-\frac{a+1}{4(a+1)}=\frac{2a^{2}+a+2}{4(a+1)}\)

Do the same for the second one.

Another thing I did was cheated by using my TI-92. These things are ridiculous to do by hand.

If you don't have a calculator capable of this, I would suggest getting one. TI Voyage 200 is the big guy to have. I love it.
 
I'd first do R2-R3, showing that x[sub:15uc8gl6]2[/sub:15uc8gl6]=x[sub:15uc8gl6]3[/sub:15uc8gl6] or a=1/2. Then, do these two cases separately.

If a=1/2, there's no more messy algebra.

If x[sub:15uc8gl6]2[/sub:15uc8gl6]=x[sub:15uc8gl6]3[/sub:15uc8gl6] you only have two variables.

Or, using maxima (a computer algebra package) I get galactacus' answer, which is different from both yours and the book's
 
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