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Find bases for the null spaces of the matrices given in Exercises 9 and $10 .$ Refer to the remarks that follow Example 3 in Section $4.2 .$$$\left[\begin{array}{rrrr}{1} & {0} & {-3} & {2} \\ {0} & {1} & {-5} & {4} \\ {3} & {-2} & {1} & {-2}\end{array}\right]$$

$\left\{\left[\begin{array}{l}{3} \\ {5} \\ {1} \\ {0}\end{array}\right],\left[\begin{array}{c}{-2} \\ {-4} \\ {0} \\ {1}\end{array}\right]\right\}$

Calculus 3

Chapter 4

Vector Spaces

Section 3

Linearly Independent Sets; Bases

Vectors

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in this example, a matrix is provided, and our goal here is to determine the basis for this matrix. The first steps that we have to take in determining such a basis is to put this in row reduced echelon form. And the reason for this is in order to find a basis for the null space we have to solve the matrix equation. A X equals zero. So here's my matrix A. Provided it has not been augmented with the zero vector. So if you prefer, you can say that a matrix a augmented with a zero vector is what will road reduce? This is probably the best practice. That way, when we go to solve the system here will be sure what the actual sizes and not getting mixed up with the very evils. So now that we have augmented with the Zero Matrix keep in mind, our matrix is now of size three by five altogether, where the fifth column consists of entirely zeros that we augmented with next with the original Matrix A. We have a pivot here. The zero is all set below it, but in row three will be eliminating that entry three so Let me copy Row one. It's one zero negative. Three and two. Then multiply. Row one by negative three. Add the result to row three. Row three then becomes zero Negative too. Looks like a value of 10. And then negative eight. And let me copy in row two. So 01 negative. Five and four. Okay, so so far, we have a pivot here when zeros below of a pivot here was rose above, which is good, but we need to eliminate this too. So our next role operation First, I'll copy 10 negative. 32 and zero. The second row 01 negative, 54 and zero. And then our operation will be to multiply this road by two. Added to the third row. Our result is going to be 00000 since Roe to is a multiple of row three. So we're ready to now describe the solution set to a X equals zero. Keeping in mind, we augmented with a zero vector as well as the fact that the variables here are x one x two x three and x for the reason for this is because a zoo of size three by four. So the Vector X must be for by one containing four different vectors now in each row will solve for the corresponding variable that belongs to a pivot. So that would be x one and X two in this case, and we see the X one is three x three minus two x four. Then in Row two, we find x two is equal to five x three, minus four x four, and that takes us next to the free variables, which are X three and X four. Since they do not correspond to pivot to pivots, it's all rights that X three equals x three and X four equals x four. This just helps me to set up what their coefficients are going to be once we go to the North Space, and if you prefer, it's not necessary. But you could put in a zero times x three plus x four and then for X three equals x three. We could say that we have a plus zero times x four. Now we're ready to begin stating the solutions so we can say the solution to a X equals zero is a Vector X, which is equal to first Take the free variables, which are X three. It'll multiply a vector plus x four, multiplying another vector and now pull coefficients for our X three. From here they are 35 There's a coefficient of one here and zero, then coefficients of X four are likewise negative, too negative for zero that we inserted, and there's a coefficient of one here. And the only stipulation for this solution X is that X three and X four are in the set of real numbers. Now, if you look at this set that we've described here very, very carefully, you'll notice that this is just a sip similar way to describe the span of a set of vectors and that set of vectors is the null space. So we're ready to describe this solution we have that the null space of A is equal to the span of the first vector 3510 and the second vector negative to negative 40 and one. Our vector has are vectors in the span, have four entries, and that's because A is of size three by four and we have two vectors, and that is because we found to free variables here. So this is our solutions for determining the knoll of a notice. Though what we really want to do was not just find the null space of a We're looking for a basis. But what's nice about this process is a basis for the null space must span. And that's what we have here. And another interesting issue is this process that we just generated coming up with the two vectors always produces linearly independent vectors. So we can now say that the set 3510 and negative to negative 401 forms a basis for the null space of A. And this is the solution that we're going for and this problem, everything above is just preliminary to get to that conclusion.

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