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?
 A: Start with properties of matrices:
$$\text{det}\begin{bmatrix}a_1+d_1 & b_1 & c_1 \\ a_2 + d_2 & b_2 & c_2 \\ a_3+d_3 & b_3 & c_3\end{bmatrix} = \text{det}\begin{bmatrix}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{bmatrix} + \text{det}\begin{bmatrix}d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3\end{bmatrix}$$
This is true for any column. It is easy to show that:
$$\text{det}\begin{bmatrix}a & 8 & d \\ b & 8 & e \\ c & 8 & f\end{bmatrix} = 8\cdot \text{det}\begin{bmatrix}a & 1 & d \\ b & 1 & e \\ c & 1 & f\end{bmatrix}$$
$$\text{det}\begin{bmatrix}a & 4 & d \\ b & 5 & e \\ c & 6 & f\end{bmatrix} = \text{det}\begin{bmatrix}a & 1 & d \\ b & 2 & e \\ c & 3 & f\end{bmatrix}+3\cdot \text{det}\begin{bmatrix}a & 1 & d \\ b & 1 & e \\ c & 1 & f\end{bmatrix}$$
A: 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}
A: The smart way is knowing that determinate is linear in each column of the input matrix.
Let $$\text{D(C)} := \text{det}\begin{bmatrix}a & C_0 & b \\ c & C_1 & d \\ e & C_2 & f \end{bmatrix}$$.
Then D is a linear function on C.  We have D($\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}$) = 4 and D($\begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}$) = -2 and want D($\begin{bmatrix}4 \\ 5 \\ 6\end{bmatrix}$) = D(3*$\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}$ + $\begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}$) = 3*D($\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}$) + D($\begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}$) = 10.

The dumb way is to just write out the equations.
Given:
$$(1): -1bf + 1ec +2af -2dc -3ae + 3db = -2$$
and
$$(2): -bf + ec + af - dc - ae + db = 4$$
Solve for:
$$(3): -4bf+4ec+5af-5dc-6ae+6db = X$$
Well, start zeroing factors
$$(3)-4*(2): af-dc-2ae+2db = X-4*4$$
now we want to get rid of the af term.  Well, (1)-(2) has af as the leading term:
$$(1)-(2): af-cd-2ae+2db = -2-4$$
As $(3)-4*2 = (1)-(2)$ we get
$$ -6 = X-16 $$
$$ X=10 $$
of course, this is just a noisy, error-prone, first principles version of the "determinant are linear in a column" solution.
