m-use-as-many-columns-of-a-as-possible-to-construct-a-matrix-b-with-the-property-that-the-equation-2

Question

[M] Use as many columns of $A$ as possible to construct a matrix $B$ with the property that the equation $B \mathbf{x}=\mathbf{0}$ has only the trivial solution. Solve $B \mathbf{x}=\mathbf{0}$ to verify your work.
$A=\left[\begin{array}{rrrrrr}{12} & {10} & {-6} & {-3} & {7} & {10} \ {-7} & {-6} & {4} & {7} & {-9} & {5} \ {9} & {9} & {-9} & {-5} & {5} & {-1} \ {-4} & {-3} & {1} & {6} & {-8} & {9} \ {8} & {7} & {-5} & {-9} & {11} & {-8}\end{array}ight]$

Viswesh Uppalapati · Accepted Answer

and this question, we&#x27;re gonna be selling ball number 42 of, um, the literature textbook from section 1.7, which is based on a linear independence. So the problem here is asking us Thio use as many columns of a provided ah, that I wrote out here So constructive matrix equation be where such that B X equals zero has only the trivial solution, which in this case, um, but it&#x27;s just a zero calling Becker with as many variables as possible from from a so that this means that be as linearly independent. So it&#x27;s asking us to construct a set by using the columns of a thio constructive, linearly independent set of B using as many calls a ce possible which are linearly independent themselves. So to do this, you just have the solve the ah matrix equation X equals zero, which I set up right here. Azan augmented matrix where I just contaminated a with a zero vector and you wrote, is it? I recommend using a R e f calculator or just a regular echelon formed calculators because, um, if it&#x27;s any bigger than like a four by four or four by three as it would take a long time to get this in echelon form. So I haven&#x27;t written out the echelon form of Matrix. But basically, the rose would earn the columns without a pivot position, which is basically it, um, basically a leading entry in every column. So the columns without a leading entry was a pivot position are, um, for your variable. So in the in the final matrix that is reduced Columns three and five are free variables. So we cannot include this these columns and be as road using them would give us. Ah, would end up making them free variables. So making the making the entire set, which would make the entire set linearly dependent. So if you exclude these two columns, these other four columns 124 and six are linearly independent themselves because all of them contain, uh, leading entries that air pivots, uh, making it so that there are no free variables and be making the entire set of be linearly independent. So be could be written as dog went emitters that contains the columns 124 and six. So in this case it would be a 12 negative seven nine negative 48 10 Negative six nine Negative 37 We&#x27;re just calling to and calling three are calling for It is negative. 37 Negative 56 Negative nine and column six is 10 if I make it a born mine negative idiot, which will be the Matrix for B, and it would literally independent.

Michael Jacobsen · Answer

In this exercise, we have a very large matrix A which is of size four by five. The first thing we need to do with they is to roll reduce this into row reduced echelon form. But to do this by hand takes a very substantial amount of work. That&#x27;s why it&#x27;s probably a good idea in this case to use a calculator or computer software to roll reduce in a calculator. I roll reduced and got this particular matrix. So now let&#x27;s consider certain things about this matrix. A. If we take eight times X and set it equal to the zero vector, then we know the following that this affect our matrix equation has nontrivial solutions. Well, we would have been able to determine this without row reducing A because a zoo of size four by five telling us that we have more vectors than entry per vector in this case, and whenever that happens, the columns of a are linearly dependent, which further implies that there will be nontrivial solutions to this equation here. So let&#x27;s suppose we wanted to have only the trivial solution. Well, we would have to change out our Matrix A into a different Matrix B. Let&#x27;s make our Matrix B a similar as possible. So the question now becomes Which columns of a do we take to place inside B so that this equation replaced with B has on Lee the trivial solution. Let&#x27;s determine that next, the key to this kind of situation is to look at the pivot columns of a. So in my role reduced echelon form, there&#x27;s a pivot column and calm 12 and three. So that tells us for our Matrix B that we&#x27;re building. If we want an equation where B X equals zero has on Lee the trivial solution, then we must go to this were reduced form and pick column, one from The Matrix, Say, because there&#x27;s a pivot and calm to here. Pick calm to as well, then the next two columns are non pivot columns, which means not that these entries are specifically bad, but we don&#x27;t want them to satisfy our equation. Below the next Pivot column is column four. So we need to grab those entries here. So now we know our Matrix B should have in column one eight negative 965 Calling to should be negative 34 negative to negative one and call in three should be too negative. 74 and 10 with this matrix be provided. We now know that when we take be Times X equals a zero vector, we get on Lee the trivial solution. And that&#x27;s because when we roll, reduce be we would obtain just thes entries. So all comes of B R now pivot columns, meaning there&#x27;s no free variables, so this exercise is complete. But also note what this tells us. If you start with a matrix, say with linearly dependent columns and you roll reduced to find the pivot columns. Then you can build a new matrix such that the columns are now linearly independent.

Nyah Kshatriya · Answer

and this problem were given a matrix a B C D, which is equal to three identity matrix 1001 And we&#x27;re multiplying that by, um, this matrix right here. Nine negative seven negative five and four. And so because this has two rows and two columns, our average our final matrix is going to have she rose in two columns. And so A is going to be equal to a matrix with four entries, two on the top and two on the bottom. And so I&#x27;m going to begin by writing out R A B c d. Because we know that that is equal to 1001 And so to multiply the matrices, we&#x27;re going to start with this first top left country right here and is going to be equal to nine times one plus negative seven times zero. We got that by taking the first row in the first column of each respective matrix. And so then we&#x27;re gonna do the same thing for the next entry. We&#x27;re gonna take nine times zero this time Organized the second column 9 10 0 plus negative seven times one and then we&#x27;re going to repeat this process for the next two entries, The sun is going to be the bottom row on the first column. So negative five times one plus four times zero. And then lastly, we have the bottom row in the second column Negative five times zero plus four times one. And so we can see that the zeros all cancelled out and they&#x27;re all equal to zero. Which means we&#x27;re only left with the entries. I have, ah, one better multiplied by one. So nine negative seven negative five and four which you can see is the same thing that we multiplied the identity matrix by. So this property shows us that if you have a matrix multiplied by the identity matrix, it&#x27;s going to be equal to that original matrix.

Kushal Mohnot · Answer

All right, So for this question, we&#x27;re asked to figure out what is a times B. So this is matrix A in this matrix B. Um, First, you figure out if we can actually multiply these matrices, and for that we need to figure out they&#x27;re rows and columns. So this one has two columns. Andi, it has two rows. So this is Roland. This is row two. So the dimensions of this matrix are two times to I mean, Chrissy&#x27;s us. This the dimensions of this matrix is 1 to 3 columns and two rows. So let&#x27;s see that 12 rows on and point to three columns. So we write these dimensions as 30 it&#x27;s actually too two times three now to figure out if we can actually multiply them, I would write, like so for trying to a Times B. I would write them right next to each other. So two times two on two times three. And if the clothes numbers are the same, that means it&#x27;s fine, and the resulting matrix will have the dimensions of these outer numbers. So the resulting matrix will have will be a two by three matrix. All right, So how do we find the elements of each of those? Um, sorry. I&#x27;m already done this problem because the first time I did it didn&#x27;t upload for some reason. So I just did it again. Um, and erase the board are until for the first element, you take this row, multiply with this call. So you get negative one times three just negative. Three and five times Negative. Eight. Just negative. 40. That&#x27;s gonna be your first element. And negative three minus 40 is equal to negative. 43. So that&#x27;s your first element for your next element. You&#x27;re still on the first row. Now you multiplied with the second call. So it&#x27;s just negative. One time six and 00 And those you have negative six over here. Similarly, for the last element of that row, you multiply the first road with the last column. So it&#x27;s negative. One times four plus five times negative. 12. So negative. One times four is negative for and five times 12 is 60 When you add that and you get 56 Next you have three times three on two plus two times negative eight. So three Times Street was nine on and plus two time native, it was negative. 16. So nine minus 16 is equal to negative. Seven on, then. Same thing, this whole row with this whole row and you get B two C three times six is 18. You just have 18 here. We&#x27;re right, wolf. Well, I guess it&#x27;s good. Edited twice. All right. On for the last row, right? You do last rule with the last column to get the last element. So that&#x27;s three times for which is 12 on two times 12 which is 24 and you get 36 when you add a boat up. So this is your answer.

Kushal Mohnot · Answer

right. This question will find a product of C A And to do that before something right out. What? See? And they are O. C. Is 4 10 negative to six, five and nine. And a is native 15 three in two. All right. First, we have to figure out their dimensions or keep compatible or not. So this is 123 as three rows. This one has 1 to 2 rows. And this has two columns on. This has 1 to 2 columns. So if these inter numbers are the same, that means they&#x27;re good, so we can multiply. These and the resulting matrix will have three rows and two columns. All right, that gives us only need to know to perform the multiplication. All right, so first step, we take this rope and multiply each column that will be each of our elements. So that row with this column would be four times negative one. All right, so we have a three by two. Seems a little longer to all right. So yeah, well read. Negative. Four negative for 10 times three is 30 um, four times five is 20. We only have two calls. Andi 10 times to was 20. All right, now this role with Mitch call so negative two times negative one is gonna be too, Andi, six times three is gonna b e t negative two tanks. Five is gonna be negative. 10 on negatives. Signed just six times two is gonna be 12 on five times negative. One is gonna be negative. Five and nine times three is gonna be 27. Five times I was 25 and nine times two is 18. All right, so our final matrixes and you just add them up, make them look nice and neat. So 30 plus negative forges 30 miles four, which is 26. 20. It was 2040 to 18 20 10 plus 12 or negative. 10 plus 12 which is to five or 27. Minus five. Rather 22 on 25. Plus 18 to target them ahead. Five plus eight. There&#x27;s 13. 1234 43 Doesn&#x27;t look like the product isn&#x27;t before you three, but yeah, this is

Ankit Gupta · Answer

in discussion. We are given with B equals toe. This metrics on a plus B equals two dramatic six. Well, zero minus RL minus foreign 11. I went to find a so we can give a equals two. The result in metrics 6 12 0 And then you do have been negative for an 11 minus B. Hence, we can give you a equals. Two. Six. Well, Tzeitel, Negative. 10 Negative. Full 11 minus 46 minus five minus 63 No, the two mattresses are of the same order. Here it is off to closely here. It is also too closely and we can subtract and to subtract the mattresses, we have to subtract the corresponding elements. Hands were six minus full, well minus six zero, minus negative. Fife Negative N minus and negative. Six. Negative. Four minus three and 11 minus two. Hence the vo to six. Fight minus 10 plus six is a negative for negative for negative. Three is negative. Seven and 11 minus two is nine. That is

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$[\mathbf{M}]$ With $A$ and $B$ as in Exercise $4…

Problem 42 Hard Difficulty

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Linear Algebra and Its Applications

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David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications