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I got 4 equations, where are $x,y,z,u$ variables , and $a,b$ parameters. So the solution from the book is like this:

$$x+(a+1)y-(a+1)z-au=1$$

$$ax+(a+1)y+az-2u=2$$

$$ax+(a+1)y-2z+au=b$$

$$(a-1)x+3(a+1)z-4u=3-b$$

So when I do all transformations it gets this form:

$$(a+1)y+x-(a+1)z-au=1$$

$$(a-1)x+(2a+1)z+(a-2)u=1$$

$$(-a-2)z+(a+2)u=b-2$$

Now, next steps are these:

1. If $aneq-1 , aneq1 , aneq-2$: System is indefinite




2. If $a=-1$: System is indefinite




3. If $a=1$: System is indefinite




4. If $a=-2$: System can be either indefinite or impossible

My question is how did we got that first part where $a$ can't be $1, -1$ or $-2$. Why it can be $2$ or $0$ or some other number?

Suppose $a_{i,j}, b_k in mathbb{Z}$.

I will talk about what you have. I do not know if your computations are OK (I think they are not).

In your case, you have done changes and you have obtained:

$$begin{bmatrix}1&a+1&a+1&-a\ a-1&0&2a+1&a-2\0&0&-a-2&a+2\0&0&0&0\end{bmatrix}begin{bmatrix}x\ y\z\uend{bmatrix}= begin{bmatrix}1\ 1\b-2\0end{bmatrix}$$

That is telling you that you have a degree of freedom since you have $4$ variables and $3$ equations. Hence, one of your variables will depend of the others. Hence, the system will have infinite solutions. Hence it is 'indefinite'. But you have got that it is indefinite for all $a$. (Since you got this reduction for all $a$ and there is a $1$ that will never be zero. Then you will never have a zero matrix for any $a$.)(Changing values of $a$ you may get more zero rows but it will be still 'indefinite' since you can not make a whole row zero with a fixed value of $a$.).

Moreover,a system like
$$begin{bmatrix}a_{1,1}&a_{1,2}&a_{1,3}&a_{1,4}\ a_{2,1}&a_{2,2}&a_{2,3}&a_{2,4}\a_{3,1}&a_{3,2}&a_{3,3}&a_{3,4}\a_{4,1}&a_{4,2}&a_{4,3}&a_{4,4}\end{bmatrix}begin{bmatrix}x\ y\z\uend{bmatrix}= begin{bmatrix}b_1\ b_2\b_3\b_4end{bmatrix}$$

is an 'impossible' system or has no solution when you find that a whole row of $a_{i,j}$ is zero for a fixed $i$ and the respective $b_i$ is not zero.

In your case, you have done changes and you have obtained:

$$begin{bmatrix}1&a+1&a+1&-a\ a-1&0&2a+1&a-2\0&0&-a-2&a+2\0&0&0&0\end{bmatrix}begin{bmatrix}x\ y\z\uend{bmatrix}= begin{bmatrix}1\ 1\b-2\0end{bmatrix}$$

And that is not 'impossible'.

HINT:

1. Try to put your system this way.


2. Reduce it until you get an upper triangular matrix.

When you get something like:

$$begin{bmatrix}b_{1,1}&b_{1,2}&b_{1,3}&b_{1,4}\ 0&b_{2,2}&b_{2,3}&b_{2,4}\0&0&b_{3,3}&b_{3,4}\0&0&0&b_{4,4}\end{bmatrix}begin{bmatrix}x\ y\z\uend{bmatrix}= begin{bmatrix}c_1\ c_2\c_3\c_4end{bmatrix}$$

you will see that $b_{4,4}$ and $c_4$ will be certain expressions.

If $b_{4,4} = 0$ and $c_4=0$, the system will be 'indefinite'.

If $b_{4,4} neq 0$, you will be able to find $u$ and solve your system. And it will have a unique solution.

If $b_{4,4} = 0$ and $c_4neq 0$, the system will be 'impossible'.