Answer
See below
Work Step by Step
Given $P(-8)=0\\
L:y=3x-4$
Obtain:
$\\
=1.(-1)+1.1+(-1).2+2.1\\
=0$
$\\
=1.(-1)+1.(-3)+(-1).0+2.2\\
=0$
$\\
=(-1).(-1)+1.(-3)+0.2+2.1\\
=0$
$=0\\
1.a+1.b+(-1).c+2.d=0\\
a+b-c+2d=0$
$=0\\
(-1).a+1.b+2.c+1.d=0\\
-a+b+2c+d=0$
$=0\\
(-1).a+(-3).b+0.c+2.d=0\\
-a-3b+2d=0$
Obtain the system of linear equations:
$\begin{bmatrix}
1 & 1 & -1 & 2|0\\
-1 & 1 & 2 & 1|0\\
-1 & -3 & 0 & 2|0
\end{bmatrix} \approx \begin{bmatrix}
1 & 1 & -1 & 2|0\\
0 & 2 & 1 & 3|0\\
0 & -2 & -1 & -4|0
\end{bmatrix} \approx \begin{bmatrix}
1 & 1 & -1 & 2|0\\
0 & 2 & 1 & 3|0\\
0 & 0 & 0 & 7|0
\end{bmatrix}\\
\rightarrow 7d=0\\
\rightarrow d=0$
From the second row:
$2b+c+3d=0\\
\rightarrow c=-2b$
From the first row: $a+b-c+2d=0\\
\rightarrow a=-3b$
Thus, $A_4=\begin{bmatrix}
-3b & b\\
-2b & 0
\end{bmatrix}=b\begin{bmatrix}
-3 & 1\\
-2 & 0
\end{bmatrix} \forall b \in R$
