D.y = -3/5x - 2
mggfffffffffffffffffffffffffffffffffffffffff
Consider the contrapositive of the statement you want to prove.
The contrapositive of the logical statement
<em>p</em> ⇒ <em>q</em>
is
¬<em>q</em> ⇒ ¬<em>p</em>
In this case, the contrapositive claims that
"If there are no scalars <em>α</em> and <em>β</em> such that <em>c</em> = <em>α</em><em>a</em> + <em>β</em><em>b</em>, then <em>a₁b₂</em> - <em>a₂b₁</em> = 0."
The first equation is captured by a system of linear equations,
![\begin{cases}c_1 = \alpha a_1 + \beta b_1\\ c_2 = \alpha a_2 + \beta b_2\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Dc_1%20%3D%20%5Calpha%20a_1%20%2B%20%5Cbeta%20b_1%5C%5C%20c_2%20%3D%20%5Calpha%20a_2%20%2B%20%5Cbeta%20b_2%5Cend%7Bcases%7D)
or in matrix form,
![\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}](https://tex.z-dn.net/?f=%5Cbegin%7Bpmatrix%7Dc_1%5C%5Cc_2%5Cend%7Bpmatrix%7D%20%3D%20%5Cbegin%7Bpmatrix%7Da_1%26b_1%5C%5Ca_2%26b_2%5Cend%7Bpmatrix%7D%5Cbegin%7Bpmatrix%7D%5Calpha%5C%5C%5Cbeta%5Cend%7Bpmatrix%7D)
If this system has no solution, then the coefficient matrix on the right side must be singular and its determinant would be
![\begin{vmatrix}a_1&b_1\\a_2&b_2\end{vmatrix} = a_1b_2-a_2b_1 = 0](https://tex.z-dn.net/?f=%5Cbegin%7Bvmatrix%7Da_1%26b_1%5C%5Ca_2%26b_2%5Cend%7Bvmatrix%7D%20%3D%20a_1b_2-a_2b_1%20%3D%200)
and this is what we wanted to prove. QED
The answer is probably c if I’m correct