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6. (6 points) Find a basis for the space spanned by...

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6. (6 points) Find a basis for the space spanned by

6. (6 points) Find a basis for the space spanned by



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State the part of the Area Postulate that is the basis for your answer to. Exercise 6

Okay, So for this problem, we need to first find the auxiliary equation. That same VP of R is able to r squared plus six r plus nine. And then this can further be factored. Um, we need to find the factors of nine that add up to six. That's gonna be three and three. So that's just gonna actually be r plus three squared. So we're going to get r is equal to negative three twice. So negative three and negative three. So since we have, um, a repeated route are the basis of our solutions. We're gonna first include e to the native three X and then we're going to put in x times E to the negative three x as our second route. Where is the second part of our basis?

Okay. To find the solution Space of this differential equation. Here, we first need to find the auxiliary equation. This is going to u p of our is equal to. Then we have our That should be why double prime here then R squared minus six are plus 25. It's going to be our auxiliary equation. This can, uh, not be factored since the only factors of 25 are going to be one in 25 5 and five, neither of which add up to negative six. So we're going to need to use the quadratic equation, so we set this equal to zero. So we're gonna use quadratic formula. Are is going to be equal to negative. Be so negative. Negative. Six plus or minus square root of them Will have negative six squared minus four times a C here. Okay, All over. Two times a just two times. One simplifying this we get. Ah, this is just positive. Six divided by two. We get three plus or minus, and then this is going to be 36 minus, um 36 minus 100. So 36 minus 100 is going to be negative 64. So we're taking the square root of negative 64. The square root of negative 64 is going to be negative. Course. Ari is going to be eight I and then we have eight. I divided by two. So that's gonna give us plus or minus for. I hear. So our roots are going to be three plus or minus four. I that means we're going to use, um we're going to use the co sign and sign forms. So this is a here. This is B. So our roots are going to look like either the three x so cause I or our solutions are going to be designed for X and then comma heat to the three X and then sign of four X. This is gonna be the basis of our solution space here.

In this video, we're going to find the basis for the road space in the basis for the calm space of the following matrix six. Negative one and the first row 12 negative two in the second row. So let's first start out with finding the basis for the roast space of this matrix, which all call P. Now what is the row space off a matrix? Well, it's the span of the row vectors. It's the span of the row vectors, so we need to find a basis for the span off. Six Common negative one. Me to find a basis for the span of the vectors. Six common negative one and 12 comma native to. And it turns out that the way to do this is to first row reduce the Matrix Rove reduce P two row echelon form. And then after we do that, we take all we take all off. The non zero rose in the new matrix and those non zero rose are are basis. So we the road reduce this guy. We need to reduce p and then take all of the non zero rose and then we'll have the basis for the roast base of P. So in order to ro reduce this matrix, we need to first take the top most pivot position of P and make everything below that per visit. Pivot position zero. So the top most pivot position of P in this case is six. And then we use elementary row operations to make everything below six. So 12 0 So pause the video and rove reduce this matrix to a row echelon form. So I'm assuming you've had a go at it. So this matrix is row equivalent to the following matrix six negative one and 00 And the way we did this was we added to row to negative two times Row one. And now all the fluff to do is to make our pivot position a one instead of a six. So we just need to divide row won by six, giving us the matrix one negative 1/6 and 00 So check this out. Our second step is to take all of the non zero rose in the row. Echelon form off P. So all the non zero roses, just the one row one negative 1/6. And this row vector the set containing this row vector rather is our basis for the road space off P. And since this vector is a vector that lives in our to, then we can think of it as a vector that lives in the plane. So in standard position, it looks like this now moving on to finding the basis for the calm space of P. We need to first remind ourselves off what the calm space is. Well, just like the roast base is the span of the row vectors of the Matrix, the Colin Space. The column space is the span off the column vectors of the Matrix P. So in this case, it's the span off the vectors. Six comma 12 and negative one comma. Negative, too. Now the way to find the basis for a column Space of a matrix is the first row. Reduce the Matrix P in our case to row echelon form. Second, locate the calm in the row. Reduced matrix. I'll call that road is matrix for P p prime. So locate columns of P prime with leading ones and third take the corresponding columns in P. Take the corresponding Collins in P, and what do. I mean by that holiday I had a five by five matrix A. And I wrote, Reduce that to a row echelon form of a, say a prime, and I found that the columns with leading ones in a prime were the columns one and three. So call him one and calm. Three have leading ones in the Matrix a prime. This means that columns one in three of the Matrix A are exactly the vectors in the basis off the calm space of a So columns one in three of the Matrix a form a basis for the calm space of a. So to find a basis for the calm space of our Matrix P. Or actually almost done because we've already wrote reduced P to a row Ashkelon form, namely the Matrix from above one negative 1/6 00 Now all we have to do is to locate the columns in this matrix with a leading one. And there's only one column in this matrix with a leading one, and it's the first column. This column is the Onley column in this matrix with a leading one. So that means that the first column of P is the column vector that forms the basis for the calm space of P and that first column of P is the column. Vector six, comma 12 and the set containing this column vector is a basis for the column space of P And like before, since this vector lives in our two, we can think of it as a vector that lives in the plane and that vector and standard position looks like this.

In this video, we're going to find the basis for the road space in the basis for the calm space of the following matrix six. Negative one and the first row 12 negative two in the second row. So let's first start out with finding the basis for the roast space of this matrix, which all call P. Now what is the row space off a matrix? Well, it's the span of the row vectors. It's the span of the row vectors, so we need to find a basis for the span off. Six Common negative one. Me to find a basis for the span of the vectors. Six common negative one and 12 comma native to. And it turns out that the way to do this is to first row reduce the Matrix Rove reduce P two row echelon form. And then after we do that, we take all we take all off. The non zero rose in the new matrix and those non zero rose are are basis. So we the road reduce this guy. We need to reduce p and then take all of the non zero rose and then we'll have the basis for the roast base of P. So in order to ro reduce this matrix, we need to first take the top most pivot position of P and make everything below that per visit. Pivot position zero. So the top most pivot position of P in this case is six. And then we use elementary row operations to make everything below six. So 12 0 So pause the video and rove reduce this matrix to a row echelon form. So I'm assuming you've had a go at it. So this matrix is row equivalent to the following matrix six negative one and 00 And the way we did this was we added to row to negative two times Row one. And now all the fluff to do is to make our pivot position a one instead of a six. So we just need to divide row won by six, giving us the matrix one negative 1/6 and 00 So check this out. Our second step is to take all of the non zero rose in the row. Echelon form off P. So all the non zero roses, just the one row one negative 1/6. And this row vector the set containing this row vector rather is our basis for the road space off P. And since this vector is a vector that lives in our to, then we can think of it as a vector that lives in the plane. So in standard position, it looks like this now moving on to finding the basis for the calm space of P. We need to first remind ourselves off what the calm space is. Well, just like the roast base is the span of the row vectors of the Matrix, the Colin Space. The column space is the span off the column vectors of the Matrix P. So in this case, it's the span off the vectors. Six comma 12 and negative one comma. Negative, too. Now the way to find the basis for a column Space of a matrix is the first row. Reduce the Matrix P in our case to row echelon form. Second, locate the calm in the row. Reduced matrix. I'll call that road is matrix for P p prime. So locate columns of P prime with leading ones and third take the corresponding columns in P. Take the corresponding Collins in P, and what do. I mean by that holiday I had a five by five matrix A. And I wrote, Reduce that to a row echelon form of a, say a prime, and I found that the columns with leading ones in a prime were the columns one and three. So call him one and calm. Three have leading ones in the Matrix a prime. This means that columns one in three of the Matrix A are exactly the vectors in the basis off the calm space of a So columns one in three of the Matrix a form a basis for the calm space of a. So to find a basis for the calm space of our Matrix P. Or actually almost done because we've already wrote reduced P to a row Ashkelon form, namely the Matrix from above one negative 1/6 00 Now all we have to do is to locate the columns in this matrix with a leading one. And there's only one column in this matrix with a leading one, and it's the first column. This column is the Onley column in this matrix with a leading one. So that means that the first column of P is the column vector that forms the basis for the calm space of P and that first column of P is the column. Vector six, comma 12 and the set containing this column vector is a basis for the column space of P And like before, since this vector lives in our two, we can think of it as a vector that lives in the plane and that vector and standard position looks like this.


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