I want to know for which values of constant a the following system of equations has exactly one solution, infinitely many solutions or no solution:
$x + y + z = 0$
$x + 2y + az = 1$
$x + ay + 2z = -1$
My approach to solving this problem was to construct an augmented matrix, then reducing it. I end up with the following:
$\left[
\begin{array}{ccc|c}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & \frac{-1}{(a – 2)}\\
0 & 0 & 1 & \frac{1}{(a – 2)}
\end{array}
\right]$
What I am now struggling with is how to interpret this. My reasoning is as follows:
From the last row, I end up with $1 = \frac{1}{(a – 2)}~$.
Thus, for $a = 2$, there should be no solution or infinitely many solutions as it would result in division by $0$.
If $a \neq 2$, there should be exactly one solution.
Is this correct? Would there be a more appropriate way to determine the value of constant $a$ such that it satisfies these conditions?
Best Answer
As a general rule, you will find it helpful to evaluate the determinant of the matrix of coefficients and identify the value(s) of $a$ for which this determinant is zero.
The determinant works out to be $$2a-a^2$$
So there is a unique solution provided $a\neq2, a\neq0$.
When $a=2$ the equations are evidently inconsistent: the second and third equations are parallel and distinct planes, so there is no solution.
However, when $a=0$ the equations are linearly dependent: the first equation may be obtained by adding the second and third. This means that the planes intersect in a common line, so there are infinitely many solutions. You can solve any two of these equations and obtain the line of intersection.
You can check for yourself, but I got $$\underline{r}=\left(\begin{matrix}-1\\1\\0\end{matrix}\right)+\lambda\left(\begin{matrix}-2\\1\\1\end{matrix}\right)$$