Question
Give an example of a $3 imes 3$ matrix $A$ with all nonzero entries such that det $A=13$
Give an example of a $3 \times 3$ matrix $A$ with all nonzero entries such that det $A=13$
Answers
Find an example of a nonzero $2 \times 2$ matrix whose square is the zero matrix.
Okay, so for problem, tell me too, when you to construct three by three matrix A with none. Zero entries and a veteran be in our three, such that he's not set Spend by columns of a So he's sexually just to construct a matrix A and Victor bees such that the system is inconsistent. So there are a lot of matrices and vectors like this. I'll just give give my own matrix and you can try your own parcel So many trees high gave it so that will be one neck of three and two and or one fight and 271 And the vector B, I consider is next. If I 11 elective seven Okay, just to just to show why this, uh, why's this system is inconsistent is you just need to consider the argumentative Matrix. So his case, it will be one left of three and two and five. Sorry, negative. Five and four or one high. And 11 271 active seven. So if we if you do the you do the couching elimination and to find out a Rory to station form, then I'll just write it down here. So the real redo station form will be 10 17 3rd and zero second. Rory is 01 like 23 13 and zero. So that's rule 000 And the one so clearly this term on the third row should not be one. Because we have all zero back already. There is on the left inside. And if we are given any vector, the nef s I will always be zero. But the right hand side is accounts in one, so the system is not consistent in this case.
In this video, we have a three by N Matrix. So that's three rows and columns, and we're going to assume that the columns of a are going to span all of the space are three. Next, let's addict something extra to this picture. Let's say assume de is a three or N by three matrix, for which eight times D equals I three. So what we want to determine here is what does this matrix de equal in particular? Well, let's start with I three. We know that I three is the three by three identity matrix, so it is formed by columns E one E two e three, where e one is 100 E two is 010 anti three is 001 so altogether were then looking for a matrix D, such that when we multiply a times d, we obtain this three by three identity matrix we see displayed here. First, let's also note that since the columns of a span or three each time we write a matrix equation eight times X equals B, where be is in are three. This matrix equation is always consistent. This is because knowing that the columns of a spin or three is logically equivalent to the statements that a X equals B is always a consistent system of equations. Let's use that fact next to construct our matrix de here, we can say that the systems eight times X equals e one eight times X equals e to eight times X equals e three were e one e two e three are the columns of our identity matrix here. Each one of these systems are consistent, since we could have placed anything in these positions on the right hand sides and they're still consistent. Let's suppose then that let's call this d one de to d three our solutions. So in other words, let's say what we mean exactly were saying that since eight times X equals e one e two or three here is always consistent. It follows that a Times D one must be equal to the one. Since this is our first solution. Eight times D, too, is E to and likewise, eight times D three is E three. Now let's then define the matrix D By columns D one do to d three, where these came from. The solutions that we had constructed. Then when we take a times D, this is what will produce by the definition of a matrix times another matrix. We take a Times first D one, which came from here. Then the second column of the product is a times D to and lastly, we have eight times d three for the third column of this product. So this is what eight times D looks like. But we also know that eight times d produces e one e two, any three so we could make a substitution here, here and here to get the following. We have the one e two and e three, but this is now the three by three identity matrix. So we've just shown how can to construct the particular matrix de such that eight times d equals I three.
So one way we could have a three by three matrix and a vector B Such that B is not in the calm space of a would be as if we constructed, um, a Matrix that in this case has only one pivot column. So an example of this would be 100 and then we'll just have a bunch of zeros. So this is a simple example. Um, but here we see that there is only one pivot column. Um, and we can call this a so that way we know that the column space of a or at least the basis of the column space of a is 100 which means that it could be 200300 But there's always going to be a number here, and then zeros are on the bottom. Um, then let's just pick a B vector to be 010 We know that this is not possible because there's no way that we can make that second value in the second row. There's no way we can make that value one. So this would be a very simple example, but an effective one
In this example we're dealing with the three by 30 matrix A that's provided here for such a matrix. We can ask interesting questions such as if we were to solve a matrix equation eight times X equals, say, zero vector. What would be the solution set to such a system? Well, this is kind of a strange deal, since our Matrix A has nothing but zero entries. So if we go to the augmented matrix for this system, then it would have this form. First. I'm inserting a as green. So there's a three by 30 matrix and I'll use blue for the augmented portion. So there it is. This is the augmented matrix for eight times X equals zero. Interestingly, the variables are x one x two x three, and we have that x one x two x three are all free variables, since if a is zero matrix, there is just no pivot columns, and that makes all variables free in this situation. Well, in this case, if we were to describe the solution sets to eight times X equals zero matrix, the solution would be X, which is equal to x one x two x three and if we write this in Parametric Victor form, it takes on the form of X one times 100 plus x two times 010 plus x three The last three free variable time +001 where x one x two and x three are in our the set of real numbers. So this is the solution set in Parametric Vector form. But possibly this is just a descriptive f x one x two x three could be any real number whatsoever. And this is the solution, then the solution set to our matrix equation is Theo. Entire set are three. So that's the solution. Sets to a X equals zero when we're dealing with the zero matrix.