Parametric system solved with Cramer's rule

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Given the system with parameter $a$: $\begin{cases} -2y+ax=2\\ x+y=1 \end{cases}$

• (a) solve it with Cramer’s rule for $a=1$;
• (b) find a value of $a$ for which the system is impossible, again using Cramer’s method.

Solution

Write the system in standard form: $\begin{cases} ax-2y=2\\ x+y=1 \end{cases}$, with coefficient matrix $A=\begin{pmatrix} a & -2\\ 1 & 1\end{pmatrix}$ and $\det A = a+2$.

(a) With $a=1$: $\det A=3$. $x=\frac{\begin{vmatrix} 2 & -2\\ 1 & 1\end{vmatrix}}{3}=\frac{2+2}{3}=\frac43,\qquad y=\frac{\begin{vmatrix} 1 & 2\\ 1 & 1\end{vmatrix}}{3}=\frac{1-2}{3}=-\frac13.$ $\boxed{x=\frac43,\quad y=-\frac13}$

(b) The system is impossible when $\det A=0$ but the numerator determinants are not both zero. $\det A=a+2=0 \implies a=-2$. With $a=-2$ the numerator of $x$ is $\begin{vmatrix} 2 & -2\\ 1 & 1\end{vmatrix}=4\neq 0$: no solution, so the system is impossible. $\boxed{a=-2}$