I had this question on my exam and I thought I could solve it using the row echelon method. Well, I couldn't, and I still don't know how to solve it. We were asked to "determine $a$ such the equation system has a unique solution" and to determine $a$ itself. The system had three equations:
$2x - y + az = 3$
$3x - 4y + 2az = 1$
$x + y - z = 2$.
Could you provide me with a step-by-step of how to solve this?
Thank you
On
You could also demonstrate the value of $a$ that makes the system incompatible, this means that the system will have solution for any value distinct from that value of $a$.
This is proven taking the determinant of the coefficient matrix $A$ and looking for which value(s) of $a$ is zero.
$\det(A) = \begin{bmatrix} 2&-1&a \\ 3&-4&2a \\ 1&1&-1 \end{bmatrix} = 0 \Rightarrow 5+a=0 \Rightarrow a = -5$
When $a=-5$ the determinant is zero thus $A$'s rank is less than $A^*$ (augmented) rank, so the system has no solution as stated in Rouché-Capelli theorem.
For the rest of the problem, use Gauss-Jordan by putting the matrix in row echelon form, or use Cramer's Rule since it would be easy to calculate in this context.
After some row operations, your augmented matrix becomes
$$ \left[ \matrix{ 1 & -1/2 & a/2 \cr 0 & 1 & -a/5 \cr 0 & 0 & -1-a/5 \cr} \right| \left. \matrix{3/2\cr 7/5\cr -8/5}\right]$$
In order for this to have a solution, what you need is $-1-a/5 \ne 0$, i.e. $a \ne -5$.