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Determinant when multiplying a matrix by a constant


2m read
·Nov 11, 2024

So let's say that I have matrix A and its elements, it's a 2x2: a, b, c, d. We have a lot of practice taking determinants of matrices like this. The determinant of this matrix, same thing as the determinant of a, b, c, d, it's going to be equal to a times d, a d, minus b times c.

Now, what would happen if we multiply one row of this matrix by a constant? What would happen to its determinant? Well, let's try it out. So you have your original, the determinant of your original matrix a, b, c, d. I'm just rewriting what I just did up here: that's a d minus b c.

Now, if I were to multiply, let's say, this first row by a constant k, how would that change the determinant? Well, instead of this being just an a, this is now a k a. Instead of this being just a b, it is now a k b. And so this is equal to k times a d minus b c, which is the same thing as it's equal to k times our original determinant of our matrix A.

So that seems pretty interesting, and I encourage you to see that if you, instead of doing the first row, you did the second row, you would have gotten the same outcome. And then you can also verify that look, if I multiply both of these rows by that constant k, how would that change things? Well then, I'm going to have k a times k d, so you're going to have a k squeezed in there, and then you're going to have k b times k c.

And so this would actually be, you could factor out a k squared, and so this is going to be k squared times the determinant of A. And that can be extended to a generalized property that if I have some n by n matrix A, then the determinant, the determinant of k times that n by n matrix A, the determinant of this, when I multiply a constant times an entire matrix, I'm multiplying that constant times all the rows, you could say all of the elements.

Well, this is going to be equal to... pause this video, see if you can intuit what this general formula is going to be. You might be tempted to say it's k times the determinant of A, but remember that's only if I multiply one row by k. But if I multiply the entire matrix by k, well then this determinant is going to be the constant k to the nth power times the determinant of our n by n matrix A.

And you could see this play out in a three by three case. In fact, I encourage you to try it out with some three by threes, and you could also do a generalized proof for an n by n case. But I won't do that now; this is really just to give you the idea.

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