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Once again I'm doing a math team practice packet. The problem was: I found the determinant of a 3x3 matrix the way I know how to, which is:

Determine all values of $x$ for which the determinant of $\begin{pmatrix} 1 & x & 2\\\ 3 & 4 & 5\\\ 6 & x & 7 \end{pmatrix}$ equals zero.

I found the determinant of a 3x3 matrix the way I know how to, which is: $$\det\begin{pmatrix} a & b & c\\\ d & e & f\\\ g & h & i \end{pmatrix} = a \times \det\begin{pmatrix}e & f\\\ h & i\end{pmatrix} - b \times \det\begin{pmatrix}a & b\\\ c & d\end{pmatrix} + c \times \det\begin{pmatrix}d & e\\\ g & h\end{pmatrix} $$

I solved the problem using the way I know how, and I got some random fraction when the answer is supposed to be 2. I then looked at the way that they solved the problem and the formula they used was:

$$determinant\ = -b \times \begin{pmatrix}d & f\\\ g & i\end{pmatrix} + e \times \begin{pmatrix}a & c\\\ g & i\end{pmatrix} - h \times \begin{pmatrix}a & c\\\ d & f\end{pmatrix} $$

Using those numbers, it simplified to $0=10x-20$ which means $x=2$

Going back to my original question, is there a different way that you need to find the determinants of matrices based off of what you are given or am I missing something? I'm mostly wondering why they used the formula that they did instead of the standard 3x3 determinant finder.

Bernard
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CoolDoodShooz1
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  • It's the same formula, only expanding the determinant across a different row or column. You can do it across any row or column, so long as you take care with the signs. See e.g. Laplace expansion. There is often, depending on context, some row or column that will end up more convenient than another (e.g. if one row or column has lots of zeros, or in this case if one row or column has the unknowns you're after). – Jakob Streipel Dec 03 '21 at 18:38
  • $det\begin{pmatrix} a & b & c\\ d & e & f\\ g & h & i \end{pmatrix} = a \times det\begin{pmatrix}e & f\\ h & i\end{pmatrix} - b \times det\begin{pmatrix}a & b\\ c & d\end{pmatrix} + c \times det\begin{pmatrix}a & b\\ c & d\end{pmatrix} $ is a wrong formula. Are you sure you didn't want to do this: $det\begin{pmatrix} a & b & c\\ d & e & f\\ g & h & i \end{pmatrix} =a\times det\begin{pmatrix}e&f\h& i\end{pmatrix}-b\times det\begin{pmatrix}d&f\g&i\end{pmatrix}+c\times det\begin{pmatrix}d&e\g&h\end{pmatrix}$ ? Othewise, you can expand any row or column, the result will be the same! –  Dec 03 '21 at 18:39
  • @StinkingBishop yes that is what I meant... thank you – CoolDoodShooz1 Dec 03 '21 at 18:41
  • @CoolDoodShooz1 Now, when you use the right formula, do you get the same result ($x=2$)? (You should!) –  Dec 03 '21 at 18:43
  • You can use row reduction and obtain a matrix with $0$s in the first column, under the first row, then use the expansion along the first column. – Bernard Dec 03 '21 at 18:45
  • You can expand off of any row or column (not just the first one - be careful about signs). Often, we choose one that will make some part of the calculation easier. – Michael Burr Dec 03 '21 at 18:49
  • @StinkingBishop Ok I'm just dumb on multiple levels... I wrote down the right formula originally on my paper, but for some reason I thought $7 \times 4$ was equal to $21$ – CoolDoodShooz1 Dec 03 '21 at 18:51
  • @CoolDoodShooz1 Happens to everyone! No worries, and good luck! –  Dec 03 '21 at 18:51

1 Answers1

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One approach avoids computing the determinant completely.

The determinant is zero if and only if the column vectors are linearly dependent. Since the left column and right column are independent, this means the middle column, $\begin{pmatrix}x\\4\\x\end{pmatrix}$ must be a linear combination of the left and right column.

This means there must be an $a,b$ such that: $$a+2b=x, 3a+5b=4, 6a+7b=x.$$ Then you have three equations in three variables, and it is easily solved, $a=-2,b=2,x=2.$

This works because all the occurrences of $x$ are in one column. The same would work if all occurrences of $x$ were in a row.

Thomas Andrews
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