Let a, b, c and d be real numbers that are not all zero. Let ax + by = p cx + dy = q be a pair of equations in the variables x and y with p, q ∈ R.
Show this system of equations has a unique solution if and only if ab − cd != 0.
From Determinant of coefficient matrix, I know (ad -bc) =0 => no unique solution. Have tried substitution of one equation into another and replacement.
=> ab = cd....show that solution is unique
<= solution is unique ....show that ab - cd != 0
Pointers?
On
Assume $ad-bc\ne 0.$ Then
$$\begin{array}{l}ax+by=p\\cx+dy=q\end{array}\implies \begin{array}{l}adx+bdy=dp\\bcx+bdy=bq\end{array}\implies x=\dfrac{dp-bq}{ad-bc}.$$ In a similar way you can get $y:$
$$\begin{array}{l}ax+by=p\\cx+dy=q\end{array}\implies \begin{array}{l}acx+bcy=cp\\acx+ady=aq\end{array}\implies y=\dfrac{aq-cp}{ad-bc}.$$
Conversely, suppose that the system has a solution $(x,y).$ If $ab-cd=0$ then the solution is not unique because $(x+d,y-c)$ and $(x+b,y-a)$ are also solutions. So, if the solution is unique it must be $ab-cd\ne 0.$
Hint: let $b\ne 0$ then we get from the first equation
$y=\frac{p}{b}-\frac{a}{b}x$ plugging this in the second equation we get
$$cx+d\left(\frac{p}{b}-\frac{a}{b}x\right)=q$$ and this is
$$x\left(c-\frac{ad}{b}\right)=q-\frac{pd}{b}$$
multiplying by $b$ we get $$x(bc-ad)=qb-dp$$
Can you proceed?