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If det $ =\begin{bmatrix} a & 1 & d\\b & 1 & e\\c & 1 & f\\\end{bmatrix} = -2$ and If det $ =\begin{bmatrix} a & 1 & d\\b & 2 & e\\c & 3 & f\\\end{bmatrix} = -1$

what is the determinant of $\begin{bmatrix} a & -5 & d\\b & -7 & e\\c & -9 & f\\\end{bmatrix}$

I'm pretty lost here, I've tried multiplying matrices and subtracting them from each other but not getting the correct answer.

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The only thing that varies is the middle column. How does the determinant depend on the middle column? –  Daniel Fischer Feb 12 at 21:21
    
@DanielFischer Not sure, do I have to factor something out? –  Johnathon Pomenrantz Feb 12 at 21:25
    
What are the defining characteristics of the determinant? –  Daniel Fischer Feb 12 at 21:26

3 Answers 3

up vote 5 down vote accepted

We know that the determinant is multilinear which means linear for every column (or row) so

$$\det\begin{bmatrix} a & -5 & d\\b & -7 & e\\c & -9 & f\\\end{bmatrix} =\det\begin{bmatrix} a & -3 & d\\b & -3 & e\\c & -3 & f\\\end{bmatrix}+\det \begin{bmatrix} a & 1\times (-2) & d\\b & 2\times (-2) & e\\c & 3\times (-2) & f\\\end{bmatrix}=6+2=8$$

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(This is essentially a restatement of the earlier answers, but expanding on how the multilinearity can be viewed.)

If we compute these determinants using an expansion along the second column, we find $$\begin{vmatrix} a & 1 & d\\b & 1 & e\\c & 1 & f\\\end{vmatrix} = -1 \begin{vmatrix} b & e\\c & f\\\end{vmatrix}+1\begin{vmatrix} a & d\\c & f\\\end{vmatrix}-1\begin{vmatrix} a & d\\b & e\\\end{vmatrix} =-2 \tag{eq 1}$$ and $$\begin{vmatrix} a & 1 & d\\b & 2 & e\\c & 3 & f\\\end{vmatrix} = -1 \begin{vmatrix} b & e\\c & f\\\end{vmatrix}+2\begin{vmatrix} a & d\\c & f\\\end{vmatrix}-3\begin{vmatrix} a & d\\b & e\\\end{vmatrix} =-1 \tag{eq 2}.$$

The determinant we want is $$\begin{vmatrix} a & -5 & d\\b & -7 & e\\c & -9 & f\\\end{vmatrix} = -(-5) \begin{vmatrix} b & e\\c & f\\\end{vmatrix}+(-7)\begin{vmatrix} a & d\\c & f\\\end{vmatrix}-(-9)\begin{vmatrix} a & d\\b & e\\\end{vmatrix}$$ which is also given by $-3 \times (\text{eq } 1)-2 \times (\text{eq } 2).$ Hence the determinant is $8$.

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Note that the determinant is a multilinear function (or if you prefer in this case a $3$-tensor).

We have that $(-5, -7, -9) = -3(1, 1, 1) - 2(1, 2, 3)$ and so our determinant is $-3\cdot (-2) -2\cdot (-1) = 8$.

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