<s>[INST] Solving for the determinant only given one column of values. Given
$$ \det
    \begin{bmatrix}
    a & 1 & d \\
    b & 1 & e \\
    c & 1 & f \\
    \end{bmatrix}
= 4
$$
and
$$ \det
    \begin{bmatrix}
    a & 1 & d \\
    b & 2 & e \\
    c & 3 & f \\
    \end{bmatrix}
= -2
$$ 
I am asked to find
$$ \det
    \begin{bmatrix}
    a & 8 & d \\
    b & 8 & e \\
    c & 8 & f \\
    \end{bmatrix}
$$
along with,
$$ \det
    \begin{bmatrix}
    a & 4 & d \\
    b & 5 & e \\
    c & 6 & f \\
    \end{bmatrix}
$$
How would I go about doing this, I understand that the first one would just be 32 since when any row (or column) is multiplied by a scalar the determinant is multiplied by the same value. How do I find the determinant of the second matrix?
 [/INST] You're right about the first question. For the second one, note that\begin{align}\det\begin{bmatrix}    a & 4 & d \\    b & 5 & e \\    c & 6 & f \\    \end{bmatrix}&=\det\begin{bmatrix}    a & 3+1 & d \\    b & 3+2 & e \\    c & 3+3 & f \\    \end{bmatrix}\\&=\det\begin{bmatrix}    a & 3 & d \\    b & 3 & e \\    c & 3 & f \\    \end{bmatrix}+\det\begin{bmatrix}    a & 1 & d \\    b & 2 & e \\    c & 3 & f \\    \end{bmatrix}.\end{align}
</s>