Find the Determinant of a 3x3 Matrix | Class 12

We go over determinants of 3x3 matrices, and specifically - how to find them. The determinant of a 3x3 matrix is found by taking a sum and difference of products of 2x2 matrix determinants. Pay close attention to see the pattern! #Class12

The determinant of the final example is 138.

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[kingbeauregard]

I'm going to put this here, just in case it helps someone who stumbles upon it. Recently I had reason to write a program that calculated determinants, and there were two algorithms I tried out. One was a recursive algorithm that worked like the process illustrated here: step through the top row of the matrix, multiply each of those elements by the determinant of the matrix you get when you remove the row and column in question, and go plus / minus / plus / minus.

The other way to calculate the determinant was to sum the products of every permutation of one element from each row, and this will take just a little explaining. Let me explain my notation here. Let's say one of the permutations is the 2nd element in the first row, the 3rd element in the second row, and the 1st element in the third row ... I'm going to call that (2, 3, 1). Okay, so this permutation technique would involve adding the product of row elements (1, 2, 3), and elements (1, 3, 2), and elements (3, 1, 2), and elements (3, 2, 1), and so on for all the permutations. The trick is managing the signs, and here's the part that it took me a while to find an answer on. Here are the rules:

- Any permutation where there are two elements in the same column, the sign is "0" (i.e. ignore it). So (1, 1, 3) can be ignored.

- For any other row, the plus or minus is calculated by counting the number of transposed pairs, and is it even (which means a sign of 1) or odd (sign of -1). What I mean is, take a sequence like (2, 3, 1). Now look at each element in that sequence, and count how many elements to its right are lower. Comparing 2 its successors, it turns out that the 1 is lower; and comparing 3 to its successors, it turns out the 1 is again lower, for a grand total of two transposed pairs. Two is even, so the sign of the (2, 3, 1) permutation is 1. On the other hand, the number of transposed pairs on (3, 2, 1) is three, so the sign is -1. This works the same even on bigger matrices; so if it were a 5x5 matrix and we were looking at permutation (3, 2, 5, 1, 4), the number of transposes would be five, so its sign would be -1.

Now, at least in my particular implementation, the first technique (recursively masking rows and columns) was three times faster than the second technique (iterating through permutations and counting the transposes). That surprised me, and it feels like it could depend upon particular coding decisions. So, try both ways if you dare, and see which one is faster for you.

Oh, uses for this sort of thing ... ? Cramer's Rule, for one. You can solve massive sets of linear equations via the magic and wonder of determinants.

[immortalspiritualbeing]

I am starting to preview linear algebra these days, and this video really helps a beginner like me, thank you!