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Determine the dimensions of Nul $A$ and $\mathrm{Col} A$ for the matrices shown in Exercises $13-18$ .$$A=\left[\begin{array}{rrrrrr}{1} & {3} & {-4} & {2} & {-1} & {6} \\ {0} & {0} & {1} & {-3} & {7} & {0} \\ {0} & {0} & {0} & {1} & {4} & {-3} \\ {0} & {0} & {0} & {0} & {0} & {0}\end{array}\right]$$

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Calculus 3

Chapter 4

Vector Spaces

Section 5

The Dimension of a Vector Space

Vectors

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So in this problem, we first need to find find Ah, number three variables forgiven me tricks. And then we need to find the number of people columns up there. Keep meeting. So, um, seems to give them a trick. Please is already, uh, it's already ain't the real reduced form. So we can Well, I will persist right down the matrix. Just just for information. This will be, um 13 inactive for two, 21 81 sticks and 00 one 93 70 And there was Hero one. Sir, we have three zeros for this. What? They call him up with 1/3 row. That is 000 14 Letter three and the last rule his whole zeros. Okay, so for this giving matrix, we can find that the people call him ISS. The first column They're calling and forth color. So that would be Call him a his three. Okay, so what about the number three variables here? So check the three variables we have. First, there are six variables here. We first have X five plus four us for X six plus. Sorry, Linus. Three x so well, Yeah, We start with the export. Sorry should be explore plus or x five minus three x six, which is cereal. Okay, so from the second rule, we have x three minus three. Explore us seven. It's five is zero and from the first rule, we have x one plus three extra checks, too. Minus four x three us two x four minus minus one. Thanks, fi plus six x six. Okay, we'll do this step a step. So from this first question we have explore will be determined by x five and experience. So So we know, we know first X five and x x six are free variables. So what's more, we have x three be determined by explore next fight, but export can be determined by X files. Maggot never have six. So at three can be determined by the pre variable X five with six. So So we're done for the second question and for the first equation, we have x three X for X 56 are all it will be done and next to cannot be done by x five and excess exit. So we have to set It's too as a three variables. So this is our set off three variables. So what? That means no off, eh? ISS three

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