#5. Let TA: R3 _ R3 bean isomorphism with [0 1 k A= 1 k 2 where k is real: k 2 0Find all possible values for k...

(a) For what value or values of $k,$ if any, is $M_{m n}$ isomorphic to $R^{k} ?$ (b) For what value or values of $k,$ if any, is $M_{m n}$ isomorphic to $P_{k} ?$

Here's our given matrix equation. So to find the values of K that would satisfy this well, we have a product of three matrices and we can't do e mean three matrices all at once. We have to multiply thes first two matrices and then take that obtain product and then multiply by our third matrix. So if you multiply this first row matrix by this matrix here, what do we get while we multiply? We have a I mean, this matrix is a one row by three columns of a one by three matrix. And then this is a three rose by three columns have a three by three matrix. So therefore, the inner numbers here matchup right. The columns of the first matrix match the rows of the second matrix. So therefore, yes, the multiplication is defined. It can actually go ahead and multiply these two matrices. What we get is the outer numbers. We get a one by three matrix, so we just go. We actually do. It's like doing the dot product multiply. Um, we get four a two plus four because we get two times one plus two times two plus K time zero so That's two plus four. So our first entry is six. Okay. And then we get, well, four, um, plus three K, right, Two times, two plus two times zero plus k times three. So that's four plus three K. So our second entry in The Matrix here is four plus three K. And the third entry well, is, um, to time zero plus two times three plus k times one. So that Z six plus K. So six plus king. Okay, so there's our obtained new role matrix, and then we take that and multiply it by the column matrix here to came. Okay? And that's still equal to zero. Okay, so we go ahead and do the location now where our first entry is? Um yeah, well, we get six times two and then plus four plus three k times two and then plus K. Um, time six plus came. Okay, six plus. Okay. Okay. And that's still equal to zero. Uh, it should be okay. Plus plus six of the book is mistake now, plus plus six times six plus cave. Plus the quantities. People's K is equal to zero. Okay, So thats becomes what? We have 12 and then we get a, um, June or 12 we get plus eight plus six K plus another six K plus K squared is equal to zero eso um, this becomes while we get k squared plus 12 K plus 20. So k squared plus 12 K plus 20 which is that equal to zero. Well, we saw okay here, getting back to this, and we find that we got K is equal to either, um, negative too. And 10 native 10. So therefore, the values of K which satisfy our given are given initial matrix equation are negative. Two and 10 negative. 10. They have to and negative 10. All right.

If you consider this following matrix equation and we're looking to find all values of K that would satisfy the equation. So we have to multiply these three matrices now. Major complication is not communicative. Meaning if we have, you know, two matrices a times B, that is most likely not necessarily the same as B times eight. But it is associative, right? We can make it. We have three matrices. We can multiply the first two first, or we can multiply the second to first, right? We just can't multiply in the opposite border. So in our community, but we are associative. So, um, to simplify this, we're gonna go ahead and multiply the second matrix by the column matrix. So what we get here, so we just we have K 11 and then times well, multiplying the middle matrix here by our column matrix, we get, um, well, one times k. So it's K plus one times one plus zero times one. So that just becomes K plus one. And then our next entry, we've got, um, one times k. That's K plus zero times one So, plus zero plus two times one that becomes K plus to our next entry. Here is K plus two. And then we have zero times K plus two times one plus negative. Three times. One that becomes, um, just to minus three, which is negative one. Okay, so now we have K 11 times this column metrics here. Okay. And if we do the application now, what do we get? Well, first we get k Times K plus one. Okay. And then plus, um, plus one times K plus one. So we get k Times K plus one plus well, plus K plus two plus. Negative one. Right. So we get k Times K plus one plus K plus two minus one. So U k oh, Kate. And keep us one plus K plus two minus one, which becomes okay, Can't keep us. One gives us a k squared, and then we get K squared, plus K plus K. That's K squared. Plus two K and then plus two minus one. Is this plus one? So we get k squared. Plus two K plus one. Okay. And this is, of course, still equal to zero. This is equal to zero. And then well, we saw Okay. Well, weaken back of this case square because two K plus one that's K plus one time K plus one or K plus one squared is equal to zero. Well, solving for K Could that K either equal to negative one or positive one? Therefore, K is equal to negative one. Um And what? So therefore, um, on hope that native one end? Um, sorry. Native one and a negative one. Right? The only value of K here that works is this is yeah. Um que Plus one times k plus one. The only value of K that works is, well, negative one. I mean, you could take native one occurs twice, but that means the only value of K that would satisfy thing given makes this equation would be k being equal. Thio negative one. She's thank you.

We want to find the values. Okay, for which matrix A is convertible, matrix A is a three by three. Matrix throws 1 to 0 K. One K. M 0 to 1. As shown in the upper right. To solve this problem, we're gonna use the relationship between the determinant and matrix convertibility. Remember that to find the determined about three by three matrix with rows A B c, D e, f g h I. We use the formula here on the left. Then we can use the fact that A is inevitable if and only if it's determined does not equal zero. So we want to find where from the determinant formula. Certain values of K. Make are determined zero and go from there. So first applying the formula, the left, the determinant of A is one times one minus two K minus two. James, t minus 00 or two minus four K. So solving for zero, we have two minus four K equals zero or 14 equals two or k equals one half. That means that are determinant is zero and that is not inevitable when K equals one half. So we must have that Matrix A is convertible whenever K does not equal one half.

We want to find the value of K. For which the matrix A. Is inevitable with A is given by Rose k minus three, negative two and negative two K minus two. So remember the determinant? About? Two by two. Matrix with a. B. C. D is simply a. D minus C. B. And that matrix A is in veritable if and only if the determinant is not equal to zero. So we want to solve for the determinant of A and then determine for what values of K. This determinant does not equal zero. Such a Matrix A is convertible. So first determine the way is just k minus three times minus three minus 16. We set this equal to zero so k minus three times minus 16 equals zero or k minus three times came last week was 16 which gives K is equal to 18. K is equal to 19, so K is equal to 18 and 19. When this determinant is equal to zero. We want to find the values of K for which A. Is not convertible. So we have that. All values of K excluding 18 and 19 allow major K. Two convertible or A is in vertebral, as long as kate is not equal 18 or 19.