Solving for the determinant only given one column of values.

Question

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?

Answer 1

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}

Answer 2

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:

$$\text{det}\begin{bmatrix}a & 8 & d \\ b & 8 & e \\ c & 8 & f\end{bmatrix} = 8\cdot \text{det}\begin{bmatrix}a & 1 & d \\ b & 1 & e \\ c & 1 & f\end{bmatrix}$$

$$\text{det}\begin{bmatrix}a & 4 & d \\ b & 5 & e \\ c & 6 & f\end{bmatrix} = \text{det}\begin{bmatrix}a & 1 & d \\ b & 2 & e \\ c & 3 & f\end{bmatrix}+3\cdot \text{det}\begin{bmatrix}a & 1 & d \\ b & 1 & e \\ c & 1 & f\end{bmatrix}$$

Answer 3

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.

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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.