determine-the-dimensions-of-nul-a-and-mathrmcol-a-for-the-matrices-shown-in-exercises-13-18-aleftbeg

Question

Determine the dimensions of Nul $A$ and $\mathrm{Col} A$ for the matrices shown in Exercises $13-18$ .
$$
A=\left[\begin{array}{rrrrr}{1} & {-6} & {9} & {0} & {-2} \ {0} & {1} & {2} & {-4} & {5} \ {0} & {0} & {0} & {5} & {1} \ {0} & {0} & {0} & {0} & {0}\end{array}ight]
$$

Jillian Rae Villa · Accepted Answer

the question asked to determine the dimensions of null A and call A for the Matrix shown below, so as a reminder to find the dimensions of calm. A, um, all you need to do is find the dimension in which thesis abs, space spans or, in other words, find the number of linearly independent columns in the subspace on or, in other words, the number of fact there&#x27;s in basis. So looking back at the Matrix, we just need to look for the linearly independent columns. And in order to do that, we have to reduce this special inform. But lucky for a surgery is the national in form, so we could just count the pivot columns. So in the first column, we find Pivot Column going down. Then, right there&#x27;s another pillow pivot going down, and then right there&#x27;s another pivot, and so far there on Lee three pivot columns. So the dimensions of calm space a equals three. On the other hand, to find the dimensions of null space a being all we have to do is count the number of a free variable columns because, as a reminder, the null space of A is just a set of solutions of the homogeneous equation. Um, a X equals zero. So when we reduce this to reduce ash on farm and we, um, right the Parametric vector form, we get the free variables as part of the its own column. So, for example, writing in Parametric form, the few variables would end up being, Let&#x27;s say, X three, uh, Times column, plus explore times a column and the number of, um, the number of free variable columns determines the dimensions of no space. A going back to our problem. We find that the third column and the fifth column contain are free variables, and there are only two columns of free variable, therefore, the dimensions of no space a equals two.

Runpeng Li · Answer

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&#x27;s already ain&#x27;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&#x27;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&#x27;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&#x27;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&#x27;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&#x27;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&#x27;s too as a three variables. So this is our set off three variables. So what? That means no off, eh? ISS three

Runpeng Li · Answer

problems. We&#x27;re keeping a three by three matrix. So that is one or collective 107 zero. Okay, so, um, we can first observe how many people call him. So we&#x27;re there. Be So there are one and two conway on Colin to so that that is too so about free variables. We first have seven of x two Cyril and x one plus four x two minus x three. It&#x27;s cereal, so that gives X one can be determined by X three. So that is X one equals x three and x two is always do so free. Better balls. We have only one, so no a it&#x27;s one.

Runpeng Li · Answer

a queue of matrix is two by two matrix, which is three four connective, 6 10 Okay, so, in order to determine the dimensions, um, first transformed this matrix thio reduced ashen of form. So this will be three. I&#x27;m sorry. Three or and we use second role. Plus, why sell the first room? So that would be zero. I saw a bit of personal eight. 18. So that gives X two. It&#x27;s cereal and X one could be ex want us or ex too, is you? So that means everything is cereal. Because we have X to zero and plugging into this equation, we have x one will be zero to so no A it is zero. What about a column? Obey. This dimension will turn out to be, uh, too, because we have two columns, have two columns with the people positions. So we have two people columns so that that is too

Thomas Emment · Answer

case. We want to determine the dimensions off this matrix on dimensions. Matrix are It&#x27;s rose on dhe, its columns. So first off was done with the rose. How many we got? 123 So three rows on. Now he&#x27;s finally columns we&#x27;ve got Well, I&#x27;ve got 12344 columns. So this is gonna be a three by four matrix.

Mark Augustyn · Answer

we&#x27;re gonna take The Matrix. Negative. Three negative for eight and 10 and we&#x27;re going to multiply by the identity matrix 1001 Okay, for my matrix multiplication. Remember, we multiply across and then down. So negative three times one plus eight time zero and then on the bottom row, we have negative four times, one plus 10 times zero, and then we&#x27;re going to the same thing. But with the second column, So negative. Three times zero, They&#x27;re negative. Three times zero plus eight times one and then negative. Four times zero plus 10 times one. Okay, They were gonna simplify this in the first spot. Their negative three times one is negative, three plus zero. So this would be negative. Three. And the bottom spot here, negative four times one time or plus 10 times zero would just be the negative four in the top spot. Negative. Three times there are zero plus eight times. One which is just eight, and then negative. Four times 00 plus 10 which would be 10. So this was matrix eight times the identity matrix. Now I need to do the other order. I need to take the identity times the original. Okay, so the identity matrix is 1001 and then a matrix A is negative. Three negative 48 intent. Okay, so following its similar set of steps, we&#x27;re going to multiply across and down, so we&#x27;ll have one times negative. Three plus zero times negative. Four zero times negative, three plus one times negative. Four one times eight plus zero times 10 and then zero times eight plus one times 10. Okay, so then we just need to simplify this. So one times negative three is negative. Three plus zero. So that first spot there&#x27;s negative. Three. Let&#x27;s go down zero times negative. Three plus negative. Four would be negative for one times eight plus zero is eight, and then zero plus 10 is 10. Okay, So what we&#x27;ve shown here is that a times I your identity matrix is equal to your identity Times matrix A in which is just equal to matrix a

Problem

Determine the dimensions of Nul $A$ and $\mathrm{…

Problem 13 Easy Difficulty

Determine the dimensions of Nul $A$ and $\mathrm{Col} A$ for the matrices shown in Exercises $13-18$ .
$$
A=\left[\begin{array}{rrrrr}{1} & {-6} & {9} & {0} & {-2} \\ {0} & {1} & {2} & {-4} & {5} \\ {0} & {0} & {0} & {5} & {1} \\ {0} & {0} & {0} & {0} & {0}\end{array}\right]
$$

Answer

The dimension of the column space is 3
The dimension of the subspace Null A is 2

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

Linear Algebra and Its Applications

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The dimension of the column space is 3The dimension of the subspace Null A is 2

Linear Algebra and Its Applications

Lily A.

Catherine R.

Heather Z.

Kristen K.

Vectors Intro

Vector Basics - Example 1

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

Lily A.

Catherine R.

Heather Z.

Kristen K.

The dimension of the column space is 3
The dimension of the subspace Null A is 2

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

Linear Algebra and Its Applications