Using the determinant to solve a system of equations

Question

Here is the question: Factorise the determinant

$$\text{det}\begin{pmatrix} z & 1 & 2 \\ 1 & z & 3 \\ 1 & 1 & z+1 \end{pmatrix}$$

and hence solve the system of equations

$$\left\{\begin{matrix} zx+y=2\\ x+zy=3\\ x+y=z+1 \end{matrix}\right.$$

I've factored the determinant to $(z-1)(z^2+2z-4)$, but I fail to see how to use the determinant to solve this system.

How can the determinant actually be used for this? After doing some research into potential methods, I came across something called Cramer's Rule (although I do not believe that this would be the intended method). Not to mention, it seems to only be useable for linear/planar equations.

Answer 1

The system of linear equations can be written as $$\begin{bmatrix} z&1&2\\ 1&z&3\\ 1&1&z+1 \end{bmatrix} \begin{bmatrix} x\\ y\\ -1 \end{bmatrix}= \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}\tag1 $$ If the matrix on the left is invertible there is no solution, so the only possibility is that $z$ is such that the determinant vanishes. Once you know the possible values of $z$, you can substitute them in $(1)$ to get ordinary systems of linear equations with constant coefficients.

Answer 2

If you don't mind, I would like to offer a solution without using determinants. If we add the first two equations, $$x+y+zx+xy=5 \implies (x+y)(z+1)=5$$ From the third equation, $x+y=z+1$ $$\therefore (z+1)^2=5 \implies z=-1\pm\sqrt5 \implies x+y=-1\pm\sqrt5$$ You can solve for $x$ and $y$ using the above. Hope this helps.