Finding unique solutions to simultaneous equations using matrices.

Question

Question: When does the system have a unique solution?

\begin{cases} kx + 3y = -6\\ x + (k+2)y = 2 \end{cases}

I've been showed questions similar to this, I've attempted to construct this into multiple matrices whereby $AX=B$, but get lost due to the $k$ variable.

Any help to get the ball rolling appreciated!

Answer 1

So you have the system in the form of ${\rm A} \boldsymbol{x} = \boldsymbol{b}$

$$\begin{bmatrix} k & 3 \\ 1 & k+2 \end{bmatrix} \pmatrix{x \\ y} = \pmatrix{-6 \\ 2} $$

which is invertible when ${\rm det}({\rm A}) = k^2+2k-3 \neq 0$, or when $k \neq 1$ and $k \neq -3$.

The solution is

$$ \pmatrix{x \\ y} = \begin{bmatrix} k & 3 \\ 1 & k+2 \end{bmatrix}^{-1} \pmatrix{-6 \\ 2} = \tfrac{1}{k^2+2k-3} \begin{bmatrix} k+2 & -3 \\ -1 & k \end{bmatrix} \pmatrix{-6 \\ 2} = \pmatrix{ \tfrac{6}{1-k} \\ \tfrac{2}{k-1} } $$

Answer 2

Determinant of the coefficient matrix ${\rm A}$ is \begin{align*} {\rm det}({\rm A})= \left|\begin{matrix} k & 3\\ 1 & k + 2 \end{matrix}\right|=(k-1)(k+3) \end{align*} It is necessarily nonzero, as the system has unique solution. In this case is Cramer's rule appropriate. For each $\; k\neq 1, k\neq -3,\;$ the solution is the pair $$(x,y)=\left(\frac{{\rm det}({\rm A_x})}{{\rm det}({\rm A})},\frac{{\rm det}({\rm A_y})}{{\rm det}({\rm A})}\right),$$ where \begin{align*} {\rm det}({\rm A_x})= \left|\begin{matrix} -6 & 3\\ 2 & k + 2 \end{matrix}\right| \quad \text{and} \quad {\rm det}({\rm A_y})= \left|\begin{matrix} k & -6\\ 1 & 2 \end{matrix}\right|. \end{align*}