Consider the contrapositive of the statement you want to prove.
The contrapositive of the logical statement
p ⇒ q
is
¬q ⇒ ¬p
In this case, the contrapositive claims that
"If there are no scalars α and β such that c = αa + βb, then a₁b₂ - a₂b₁ = 0."
The first equation is captured by a system of linear equations,
[tex]\begin{cases}c_1 = \alpha a_1 + \beta b_1\\ c_2 = \alpha a_2 + \beta b_2\end{cases}[/tex]
or in matrix form,
[tex]\begin{pmatrix}c_1\\c_2\end{pmatrix} = \begin{pmatrix}a_1&b_1\\a_2&b_2\end{pmatrix}\begin{pmatrix}\alpha\\\beta\end{pmatrix}[/tex]
If this system has no solution, then the coefficient matrix on the right side must be singular and its determinant would be
[tex]\begin{vmatrix}a_1&b_1\\a_2&b_2\end{vmatrix} = a_1b_2-a_2b_1 = 0[/tex]
and this is what we wanted to prove. QED
