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find the determinant(s) to verify the equation. $$\left|\begin{array}{lll} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{array}\right|=(y-x)(z-x)(z-y)$$

Hello there. Here we need Thio like show the This identity here is true. Basically, we need to compute these determinants to show that indeed they are. Okay, So let's compute the left hand side off this equality if the determinant is equal to the multiplication off the elements on the diagonal w z minus the multiplication off the elements on the other diagonal. That means X y here. And this is equal to so the right hand side of this equation. We need to make the same procedure first. Thesis X Y um Sorry. Hearing forgot in this minus here, minus X y mine is the multiplication off these other two elements. So that see doll view on what happened here is that this is equal to W Z time Z minus X times. Why on this is equals to W the minus X? Why so clearly? These two are equal

Hello there. So we need Thio. Compute the determinant off these two matrices thio show that this equality is, uh, true. So basically here we got, like, one of the properties off the determinant on is that if you some on multiple off of some row or column, that will not change the determined. So, for example, uh, what we got here is this is the determinant for some matrix on Dhere. What we're doing is something the first role, a multiple off the first role, okay? And that will not change the value off the determinant. So we need to show this for a two by two matrix in the general case on Let's see that, indeed, this is correct. So the determinant on the left hand side is equal to W Z minus X y. This is equal to so the right hand side, it's going to be w Z plus z y w minus x y minus. See why w so you can observe here that these two, um, elements are going to cancel out on we end Wheat w Z minus X y is equal to w Z minus X y, which indeed is true.

Mm. Sure we have um within a function uh earlier operator. Uh F. And it is The determinant will be equal to any matrix material representative. And here we here we have a metric that representing F. Yeah, that is equal to a single 13 minus four. 0 2. was heroin one plus five minus three. It's uh mm entries. So we have fundamentally A. Which represents F. Sorry, F. Okay. And it is uh for finding the trend of F. Or solving this function linear function F. We had to find little men. F. Is eagle? Do which uh gentlemen uh stuff. Yeah. It takes a little presents F. So from this we have to find which men we have. We are finding a regiment every 0 -6 plus 21 plus zero plus eight. My s 35 is five and 0 and the answer is getting cold of -8. Thank you.

We already know the determinant of this. Matrix is four. Okay, so let's label this is a column of one. Is column to under this column three. Okay, so here what we want to do is we want to used Column Column three and add Column one times 91 and assigning new value to column three. Okay, so here I I made a mistake. Here is uh w minus you is what we want. Okay, so according to the Syrian um 16 on page 904. Okay. If the entry of any role or any column of a determinative are multiplied by a non zero number K. And the result is added to the corresponding entries of another role or column. The value of determined it remains unchanged. So here. Okay, so according to this cereal, I got X. Y Z minus X. The U U V W. Sorry U V w minus you and 1-2. Okay, so this the determinate of this is unchanged. Okay, so this will still become for okay, this is what we want here. So it is equals to four.