# Solution of system with matrix

1. Nov 9, 2016

### Kernul

1. The problem statement, all variables and given/known data
Find out for which values of $\alpha$ the system $AX + \alpha X = (1, 1, 1, 1)^t$ has solutions.
$$A = \begin{pmatrix} 6 & 0 & -1 & 2 \\ 3 & 5 & -3 & 6 \\ -2 & 0 & 7 & -4 \\ 2 & 0 & 1 & 0 \end{pmatrix} X = \begin{pmatrix} x \\ y \\ z \\ t \end{pmatrix}$$

2. Relevant equations

3. The attempt at a solution
At first I wanted to proceed with a system, since it's what the problem asks. So I wrote $AX + \alpha X = (1, 1, 1, 1)^1$ like this:
$\begin{cases} 6x - z + 2t + \alpha x = 1 \\ 3x + 5y - 3z + 6t + \alpha y= 1 \\ -2x + 7z - 4t + \alpha z= 1\\ 2x + z + \alpha t = 1 \end{cases}$
And then I should proceed, right?

But the professor showed me that you can do this with matrices this way too:
$AX + \alpha X = (A + \alpha I_4)X$
And that if the determinant of $(A + \alpha I_4) \neq 0 \implies \exists! \alpha$, while if it's equal to 0 the solution depends from $rank(A + \alpha I_4)$ and $rank(A + \alpha I_4 | B)$, which I don't understand what the professor meant with $B$. Would it be the $(1, 1, 1, 1)^t$? In this case, if the two ranks are equal, does it mean it still has more than just one solution? And if they are not equal, does it mean no solution at all?
Anyway, I start by finding the determinant of $(A + \alpha I_4)$ but I find myself with a $4^{th}$ grade equation like this: $\alpha^4 + 18 \alpha^3 + 105 \alpha^2 + 208 \alpha + 40$
I'd like to know if someone finds out a different determinant, because it could be that I messed up while calculating it.

2. Nov 9, 2016

### Ray Vickson

I used the computer package Maple to do all the computations, and did get a different polynomial from yours. Mine is $\alpha^4 + 18 \alpha^3 + 105 \alpha^2 + c_1 \alpha + c_0$ for some $c_1, c_0$ different from yours.

Anyway, what is happening is that for a small number of numerical values of $\alpha$ your equations have no solution; for all other values of $\alpha$ your equations will have a unique solution; see below.

You can take the 4 x 4 system of equations you have written above, and use standard elementary elimination methods to solve for x, y, z, t in terms of $\alpha$. For example, you can use the first equation to solve for z in terms of x and t; then you can put that expression into the other three equations, leaving you with a 3 x 3 system of linear equations in the variables y, z and t. Now you can (for example) use the first equation of this new 3 x 3 system to solve for y in terms of z and t; then you can substitute that expression for y into the other two equations, to get a new 2 x 2 linear system in terms of the variables z and t alone. Then you can solve for z in terms of t, and finally get an equation for t alone. Everything will contain the parameter $\alpha$, and the solution will be valid if you do not divide by zero at the end. That will give you a condition of $\alpha$; it will be a 4th degree polynomial which, when equal to zero, makes your solution invalid. In other words, as long as $\alpha$ is not a root of a certain 4th degree polynomial, you will not be dividing by zero and so will get a valid (unique) solution for x, y z and t (all expressed in terms of $\alpha$).

3. Nov 9, 2016

### Kernul

But isn't finding the root of the 4th degree polynomial what the exercise asks me? I need to know for what values of $\alpha$ the system has solutions.
From what you're saying, it means that $\alpha$ can be anything as long as it is not a root of the 4th degree polynomial in order for the system to have solutions, right?

4. Nov 9, 2016

### Ray Vickson

Yes, that is what I said.

5. Nov 9, 2016

### PeroK

You ought to rearrange that matrix to make the determinant easier. And, avoid generating a 4th degree polynomial. Hint: the second column has three $0$'s and the other entry would be $5+\alpha$, so that gives you one of the factors for a start.

6. Nov 11, 2016

### Kernul

I divide that row by $5 + \alpha$ then? So I have $1$ and end up with a 3rd degree polynomial:
$\alpha^3 + 13\alpha^2 + 40\alpha$
which becomes:
$\alpha(\alpha^2 + 13\alpha + 40)$
and it is equal to $0$ if $\alpha = 0$ or (the solutions of the 2nd degree equation) $\alpha = -5, \alpha = -8$.
Knowing now for what $\alpha$ the determinant becomes $0$, do I have to see the three cases (with the three $\alpha$ found) in which the ranks of $A + \alpha I_4$ and $A + \alpha I_4 | B$ tell me the solutions? This is what my professor told me in case the determinant would have been equal to $0$. If I remember right, if the two ranks are the same, then there are solutions, if they are different, then there is no solution. Am I right?

7. Nov 11, 2016

### PeroK

Dividing by $5+ \alpha$ was not right. What about $\alpha = -5$? You should have kept that as a ready-made factor in your polynomial.

You were lucky that $\alpha + 5$ was also a factor of what was left!

Determinant not equal to 0 gives you a unique solution to your equation. For the values of $\alpha$ that make the determinant 0 you have more work to do to decide whether there are solutions.

For solutions, yes, the two matrices must have the same rank.

8. Nov 11, 2016

### Kernul

But I couldn't simply give $\alpha$ a value of my choice since, after all, I am trying to find them through the determinant itself.

What is the work that I should do? I know I should then delete a row from the system, but then I would have three equations with five unknown?

9. Nov 11, 2016

### PeroK

Anytime you divide by something like $5+ \alpha$ you must consider the case when $5 + \alpha = 0$. In this case, you could have noticed that $\alpha = -5$ made the determinant $0$ and put that to one side, or you could have continued and got $5 + \alpha$ as a factor of your polynomial.

It does look like there is still a bit of work to do to check solutions for $\alpha = 0, -5, -8$.

Last edited: Nov 11, 2016
10. Nov 11, 2016

### Ray Vickson

For most values of $\alpha$ you do not need to do any more work because you know that the system $(A + \alpha I)X = e$ has a unique solution. The question did not ask you to find that solution, but one way of doing the problem would have been to actually find it. However, finding roots of the determinant is faster and easier.

That leaves three values $\alpha = 0, -5, -8$. For each of those values a solution is not guaranteed to exist, but it might! That is, there could be a solution even though the determinant equals 0. The easiest way to proceed would be to try to solve the linear system in each case.

11. Nov 11, 2016

### Kernul

Oh, that's true, because $5 + \alpha$ was the only one in the column, and putting it equal to $0$ means that the column would be all zeros and automatically get a determinant equal to $0$.

Yeah, that's what I thought too about the three values of $\alpha$. What I'm asking is if I should take the entire system or delete one of the equations since the determinant in those three $\alpha$ is $0$.

12. Nov 11, 2016

### PeroK

If I could see a quick way to finish off this problem, I would give you a hint. But, I must admit, I don't see any shortcuts.