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Suppose 4 ad are nxn. Show that if AB is invertible; ABW WAB = I. |areand Bj [Hint: There is a matrix W such that...

Question

Suppose 4 ad are nxn. Show that if AB is invertible; ABW WAB = I. |areand Bj [Hint: There is a matrix W such that

Suppose 4 ad are nxn. Show that if AB is invertible; ABW WAB = I. | are and Bj [Hint: There is a matrix W such that



Answers

Suppose that $A$ and $B$ are $4 \times 4$ invertible matrices. If $\operatorname{det}(A)=-2$ and $\operatorname{det}(B)=3,$ compute each determinant below. $$\operatorname{det}(A B)$$.

All right. So we're told that supposed that A and B are four by four in vertebral matrices if the determinant of a is negative two and the determinant of B is three than compute the determinant of b squared times The inverse of hey, so first thing the determined of the inverse of a is one over the determinant of a so determinant of the inverse of a is one over the determinant of a, um that is postulate 10 now the determinant of a times B According to postulate, nine is the determiner of a times the determinant of B So we really have here the determinant of be times b times the inverse of a So that's the determinant of B times the determinant of B times the determinant of the inverse of a According to postulate nine determined B is three time determining Toby again. Times determine it of the inverse of a And so our answer would be negative. Nine has. So we only used postulates nine and 10 in your book

Okay. This is a continuation that started a problem. Number 11. Um so let's ah, let's work on this one. Um, if the determinant of a is negative two than the determinant of negative A considering that it is a four by four matrix, if you take negative one times a then according to postulate four, the determination determinant of K times A is K to the end power times the determinant of a So we're multiplying by negative one the whole matrix by negative one. So negative one to the fourth power because it is a four by four matrix is one. So the whole fact that that you've got a negative here means absolutely nothing. So now we just need to dio According Teoh Um according Teoh order my thinking here. You know what? Let's do this. Mina Right this out. We already figured out that negative means nothing, So I'm just going to write eight of the third power to be and then another be OK, so we're multiplying be times two right here. Okay. So, um, determinant of to be So this is the same thing. Um postulate four. That's going to be two to the fourth power times the determinant of V. Whoops. Okay, so now we really have to To the fourth power times the determinant of a to the third power be to the second power. All right, so that's two to the fourth power times. Um, the determinative A three times. So that would be three times three times three and determine it to be two times. Oh, man. A was negative to my bed back. Um, determined of a It was negative. Two negative, too. Times negative, too. A three times. Okay. And be two times and three. Okay. To to the fourth power times. Negative. Two times negative. Two times negative. Two times, three times three. Give me 1152. But I seem to think that the answer should be negative. The back of the book shows that the answer is positive, and so I'm wondering where I went wrong. So I'm gonna pause and think about where my mistake is and ah, I will get back to you. All right. I cannot find the mistake. Now, the fact that this is a four by four matrix would mean that this up here, negative one to the fourth Power is positive. Um, we need to take negative one to the end power and the end powers for because the four by four matrix, that would be negative one to the fourth power, which is one which means that this right here becomes meaningless. So that's going to give us a to the third power to be to the second power. And eight of the third power is eight times eight times a. That's the only place the negative can come in now. And that's gonna be negative times negative have negative, which is positive. I mean, negative. Negative, negative, negative. So I'm going with the answer in the back of the book is wrong. Um, negative. 1152. Unless I'm missing something here. Thank you.

All right, I'm going to start by finding the determine int. And by the way, this is a continuation from problem number 11. We're trying to figure out the determinant of this thing down here. So I'm going to begin by doing the determinant of be to the negative one power. And so a postulate 10 says the determinant of the inverse of a is one over the determinant of a So the determinant of the inverse and B is one over the reef. Okay, We also need to know the determinant of the inverse of a so likewise that's going to be negative. 1/2. Okay, two times the determine it. I mean the determinant of two times the inverse of B. When you multiply an entire matrix by two impossibly four, that's going to put in a factor of ah que to the end power. So it's gonna be two to the end power. So the determinant of to be to the negative one power is going to be the determinant of via negative one power times two to the end power, which is four and it's four. So two times two is four. June's report into its eight times to a 16. That's going to give us 16 3rd Okay, the determine its, uh, the inverse of a times be So I'm gonna do that now the determine it of the inverse of a times Be well, that's going to be the determinative in inverse of a times the determinative be the determinants of the inverse of a is negative 1/2 the determinant of B is three so negative 1/2 times three is negative. Three halves. Okay, so now we need to transpose that. So it says in postulate five when you transpose it, it doesn't affect the determined. So determining of all of that transposed is still negative. Three house. Okay, so now we just need to take this one times this one, the inverse of a times B transposed. Ah, to be transposed determinant of those two together so that the determinant of the whole thing is negative. Three halves times 16 third's threes. Cancel out 16. Divided by two is eight. Our answer is negative.

So we have be minus C that multiplies the equal to zero. So since the is in vertebral, we can multiply from the right by the inverse of the so right B minus c the and then the inverse. And that's equal to zero times Demers. Now the product the D in verse is just it. Entity metrics. So this becomes B minus c that multiplies the again. Did he is equal to now Any metrics multiplied by 00 So on the right hand side, we just have zero and now must be bribed by the identity matrix does nothing. So on the left hand side we have B minus c and on the right inside we have zero Now we some ah si on both sides. So that's right, B minus c and then we add plus C on the right. Inside we have zero Blasi Well now minus C plus C is just zero. So are we on the left inside? We just have B and on the right inside something zero does nothing. So we're just left with C, which is what we wanted


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