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The problem I was given: Calculate the value of the following determinant:

$\left| \begin{array}{ccc} \alpha & 1 & \alpha^2 & -\alpha\\ 1 & \alpha & 1 & 1\\ 1 & \alpha^2 & 2\alpha & 2\alpha\\ 1 & 1 & \alpha^2 & -\alpha \end{array} \right|$

For which values of $\alpha \in \mathbb R$ is the following system of linear equations solvable?

$\begin{array}{lcl} \alpha x_1 & + & x_2 & + & \alpha^2 x_3 & = & -\alpha\\ x_1 & + & \alpha x_2 & + & x_3 & = & 1\\ x_1 & + & \alpha^2 x_2 & + & 2\alpha x_3 & = & 2\alpha\\ x_1 & + & x_2 & + & \alpha^2 x_3 & = & -\alpha\\ \end{array}$

I got as far as finding the determinant, and then I got stuck.

So I solve the determinant like this:

$\left| \begin{array}{ccc} \alpha & 1 & \alpha^2 & -\alpha\\ 1 & \alpha & 1 & 1\\ 1 & \alpha^2 & 2\alpha & 2\alpha\\ 1 & 1 & \alpha^2 & -\alpha \end{array} \right|$ = $\left| \begin{array}{ccc} \alpha - 1 & 0 & 0 & 0\\ 1 & \alpha & 1 & 1\\ 1 & \alpha^2 & 2\alpha & 2\alpha\\ 1 & 1 & \alpha^2 & -\alpha \end{array} \right|$ = $(\alpha - 1)\left| \begin{array}{ccc} \alpha & 1 & 1\\ \alpha^2 & 2\alpha & 2\alpha \\ 1 & \alpha^2 & -\alpha \end{array} \right|$ =

$(\alpha - 1)\left| \begin{array}{ccc} \alpha & 1 & 0\\ \alpha^2 & 2\alpha & 0 \\ 1 & \alpha^2 & -\alpha - \alpha^2 \end{array} \right|$ = $-\alpha^3(\alpha - 1) (1 + \alpha)$

However, now I haven't got a clue on solving the system of linear equations... It's got to do with the fact that the equations look like the determinant I calculated before, but I don't know how to connect those two.

Thanks in advance for any help. (:

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If you want to use your determinant calculation, first note that if the determinant is not zero, there cannot be a solution. (Otherwise, the fourth column would be a linear combination of the first three.) That leaves only three possibilities for $\alpha$, and you can analyze those three systems of equations directly without much difficulty. –  Michael Joyce Nov 4 '12 at 23:52
    
I added the (matrices) tag since this involves determinants, and solutions to systems of equations. –  Cameron Buie Nov 5 '12 at 16:00
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up vote 3 down vote accepted

Let me first illustrate an alternate approach. You're looking at $$\left[\begin{array}{ccc} \alpha & 1 & \alpha^2\\ 1 & \alpha & 1\\ 1 & \alpha^2 & 2\alpha\\ 1 & 1 & \alpha^2 \end{array}\right]\left[\begin{array}{c} x_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c} -\alpha\\ 1\\ 2\alpha\\ -\alpha\end{array}\right].$$ We can use row reduction on the augmented matrix $$\left[\begin{array}{ccc|c} \alpha & 1 & \alpha^2 & -\alpha\\ 1 & \alpha & 1 & 1\\ 1 & \alpha^2 & 2\alpha & 2\alpha\\ 1 & 1 & \alpha^2 & -\alpha \end{array}\right].$$ In particular, for the system to be solvable, it is necessary and sufficient that none of the rows in the reduced matrix is all $0$'s except for in the last column. Subtract the bottom row from the other rows, yielding $$\left[\begin{array}{ccc|c} \alpha-1 & 0 & 0 & 0\\ 0 & \alpha-1 & 1-\alpha^2 & 1+\alpha\\ 0 & \alpha^2-1 & 2\alpha-\alpha^2 & 3\alpha\\ 1 & 1 & \alpha^2 & -\alpha \end{array}\right].$$

It's clear then that if $\alpha=1$, the second row has all $0$s except in the last column, so $\alpha=1$ doesn't give us a solvable system. Suppose that $\alpha\neq 1$, multiply the top row by $\frac1{\alpha-1}$, and subtract the new top row from the bottom row, giving us $$\left[\begin{array}{ccc|c} 1 & 0 & 0 & 0\\ 0 & \alpha-1 & 1-\alpha^2 & 1+\alpha\\ 0 & \alpha^2-1 & 2\alpha-\alpha^2 & 3\alpha\\ 0 & 1 & \alpha^2 & -\alpha \end{array}\right].$$

Swap the second and fourth rows and add the new second row to the last two rows, giving us $$\left[\begin{array}{ccc|c} 1 & 0 & 0 & 0\\ 0 & 1 & \alpha^2 & -\alpha\\ 0 & \alpha^2 & 2\alpha & 2\alpha\\ 0 & \alpha & 1 & 1 \end{array}\right],$$ whence subtracting $\alpha$ times the fourth row from the third row gives us $$\left[\begin{array}{ccc|c} 1 & 0 & 0 & 0\\ 0 & 1 & \alpha^2 & -\alpha\\ 0 & 0 & \alpha & \alpha\\ 0 & \alpha & 1 & 1 \end{array}\right].$$

Note that $\alpha=0$ readily gives us the solution $x_1=x_2=0$, $x_3=1$. Assume that $\alpha\neq 0,$ multiply the third row by $\frac1\alpha$, subtract the new third row from the fourth row, and subtract $\alpha^2$ times the new third row from the second row, yielding $$\left[\begin{array}{ccc|c} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & -\alpha^2-\alpha\\ 0 & 0 & 1 & 1\\ 0 & \alpha & 0 & 0 \end{array}\right],$$ whence subtracting $\alpha$ times the second row from the fourth row yields $$\left[\begin{array}{ccc|c} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & -\alpha^2-\alpha\\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & \alpha^3+\alpha^2 \end{array}\right].$$ The bottom right entry has to be $0$, so since $\alpha\neq 0$ by assumption, we need $\alpha=-1$, giving us $$\left[\begin{array}{ccc|c} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 0 \end{array}\right].$$

Hence, the two values of $\alpha$ that give the system a solution are $\alpha=0$ and $\alpha=-1$, and in both cases, the system has solution $x_1=x_2=0$, $x_3=1$. (I think all my calculations are correct, but I'd recommend double-checking them.)


The major upside of the determinant approach is that it saves you time and effort, since you've already calculated it. If we assume that $\alpha$ is a constant that gives us a solution, then since we're dealing with $4$ equations in only $3$ variables, we have to have at least one of the rows in the reduced echelon form of the augmented matrix be all $0$s--we simply don't have enough degrees of freedom otherwise. The determinant of the reduced matrix will then be $0$, and since we obtain it by invertible row operations on the original matrix, then the determinant of the original matrix must also be $0$.

By your previous work, then, $-\alpha^3(\alpha-1)(1+\alpha)=0$, so the only possible values of $\alpha$ that can give us a solvable system are $\alpha=0$, $\alpha=-1$, and $\alpha=1$. We simply check the system in each case to see if it actually is solvable. If $\alpha=0$, we readily get $x_1=x_2=0$, $x_3=1$ as the unique solution; similarly for $\alpha=-1$. However, if we put $\alpha=1$, then the second equation becomes $$x_1x_2+x_3=1,$$ but the fourth equation becomes $$x_1+x_2+x_3=-1,$$ so $\alpha=1$ does not give us a solvable system.

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