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Determine the Null Space of the (3x3) matrix A for which(2j +i @ij (Tj +iif i #j if i =j...

Question

Determine the Null Space of the (3x3) matrix A for which(2j +i @ij (Tj +iif i #j if i =j

Determine the Null Space of the (3x3) matrix A for which (2j +i @ij (Tj +i if i #j if i =j



Answers

Determine the null space of the given matrix $A$. $$A=\left[\begin{array}{rrr} 1 & i & -2 \\ 3 & 4 i & -5 \\ -1 & -3 i & i \end{array}\right]$$

Were asked. Burn the null space of a major. It's pay definition of the null space is shown at the top. We have that. The null space oven MBA in Matrix A is equal to the center of Vectors X, which are in or in such that the Matrix, a times one of those vectors in the set X is equal to the zero factor. And the problem gives the Matrix A is a one by three matrix, which is equal to one minus three and two so you can see that are in value is three. So we're in or three, and that means that our are expected is going to be a three by one matrix. So the vector that's a member of the null space will look something like this. It will be a three by one with X one. Thanks to in the next three, it's No, this is just solving a linear equation, so we'll go ahead and do this matrix multiplication so we'll rewrite some things down here so we have or matrix. A times are vectors. X is equal to zero vector, so our matrix A is one minus three to for X factor next one, thanks to the next three, and that is all equal this year. Vector Multiplying this out. We get an equation. X one that's one times x one So one times x one minus three times x two and plus two times x three is equal to zero. So we have one equation and three influence. Um, and so weaken. Go ahead and sold for X one. Then we have these two free variables x two and x three. So have excellent is equal to three x two minus two x three. It's now we can go ahead and set this x two and x three equal to just some scaler. That is a real number so we can set what's the next to next to rewrite that real quick. So we have x two that that equal to our you will have x three he co EDS and those are just some real Scalea. So now we can write are expected or three by one extractor. So we have from this equation we have our what our X one is equal to use. Our X one is equal to three times x two, which we said as equal to our then minus two times x three, which we set equal to s. Then we have our X two, which is our in our X three, which is that's now we can just break this up into two separate vectors by factoring out the R and s. So then, well, factor out the are you will have that are Victor said So factoring lcr a little bit of leg so factoring out the are okay The expect here is equal to the or times So here in the first through we have three times or so that will be three in the next room We have one or one in the next room Then we just have s so that will be zero and then we'll go ahead and that pulling out our s that'll be multiplied by negative too zero in one So vectors of this form or what make up the null space. So now we can go ahead and write out the answer. So the answer is but the no space still just rate the capital in for the north face No space. A very is eagle Teoh A set of actors rich, thick of the form that we just sold. So rewrite that down. Here you are. Times 310 plus s Waas. What s minus 201 and such? That he's scaler Zoran s that'll clear Scaler Czar and s are real numbers. This is the answer.

Were asked to find the null space of a definition of the no space is given at the top here, with the no space of an MBA and matrix is given by the set of vectors eggs and are in such that the matrix, a times the Vector X is equal to the zero vector. The problem gives the Matrix and he has a three by three matrix shown here. We can see from there that value of N is three, which means we're in or three so we can write our expect er as a three by one matrix with X one next to the next three. So now we can start the soul of this linear equation, but using reduction to get a in a row echelon form. So every Retin A and augmented form down here we can start our row reduction by seeing that we can get, we can get a zero here If we multiply and minus four times the first row and add that to the second room, we're gonna add what is four. I'm drew one and added to that gives our first rule stay the same. 11 is 21 and uh, zero. So I have a minus four times one, and it's a four, which gives zero. We have a minus four times a minus to plus and minus seven, which gives warm. We have minus four times one plus of minus two, which is minus six. Their third road stays the scene minus 134 Then we have a minus four times zero, which is zero in the zero of the third room. So now we can We can get a zero here if we add one. Times were one to row three. So add one times were one the road three. So that will give our first rebels stay the same. Their second rule stay the same on our third row will be one plus and minus one, which is zero. Then we have ah minus two minus to plus three, which is one. And then we have one A C one plus four, which is five. Then a 000 Now we can get let's see weaken, multiply the second row by minus one and that the three and that will give us a zero here, several multiple it or we will we will end minus one one's road to and and that the three So our first row stays the same or a second room will stay the same. 01 minus 60 Then our third rule have a minus one times zero, which is zero minus one minus one times one Edits the 10 the minus one times six plus five which is 11 the most one time 00 which is zero. Now we can dio one more step. Uh, to get a one year, we can multiply that third row but one ever 11 for most playing the third row by one every living We'll show that down here So our first true will stay the same one minus 210 2nd row stays the same and zero minus one minus six zero in our third room will be zero zero and 1 11 times a left. 11 is one and zero. So that's our Rochel Inform Matrix A And now we can multiply that by or expect Cherie because x one x two x three that is equal to zero factory which is a three by one matrix was here s so that we can multiply this out that will have first row x one minus two times x two and a one times X three is equal to zero. Next, we have a zero times x one minus one times x two the minus six times x three 00 and then we have zero times x 10 times X two and one times x three is equal to zero so we can see that x three equals zero. If we played that in here than X two has two equal to Z equals zero. Flex two is equal to zero. This was an ex too. So if x two and x 3/2 the equals zero, the next one has to equal zero that are expected has one solution which is, well, zeros. So we can write the solution to our problem that the North space of a because one solution which is zero factory. That's the answer

Were asked to find a null space of a and the definition of the null space is given at the top here with the null space oven. N by N Matrix A is equal to the set of vectors X in or in such that the Matrix, a times the Vector X is equal to the zero vector. The problem is matrix A as a three by two matrix, with the first column being to one minus three in the second column being minus forward to my Spi we can see from are a matrix that are in value is equal to two, which means we're in or two, Which means our expect er is gonna be a two by one matrix with X one and X two. Now we can go ahead and solve this mystery equation. Um, then we'll do that by putting a in Rochel informed by using road reduction. So we'll start with the reduction and I have the Matrix a re written down here. So this is the matrix. Every rate rewritten and augmented form of zeros on the right hand side. And to reduce this into a relational, inform the first thing we can do is per mutate Row two with row one So and I can abbreviate that with a P for perm et row one into and that will give the new Matrix 12 through minus forward and minus three minus five zero is still on the right. No, the next thing we can do, we can We can get a zero in the bottom left hand corner. If we add three times row one to that third red making abbreviate that within a for ad, we're gonna add three times row one to row three. So that will give us. Everyone will still be seen for a two will still be the same. Now we're adding three times were ones that were three times one plus minus three, which is zero. Let me have three times to, which is six plus minus five, which gives us one and our zero's air. Still in the right hand side, we have three times zero plus zero is zero Now the next thing we can do is we can get zero on the first row are on the first column of our second row. If we add minus two times, red ones are zero to. So you know, abbreviate that is adding minus two times were one. So to that'll give are saving through one. Let me have minus two times one plus two, which is zero. Then we have minus two times two, which is minus four plus minus four, which is minus eight. And our third row stays the same and or zero stay the same. Now this matrix is reduced enough to where we can sold it. I'll rewrite it down Here, begin. We're solving linear equation. The Matrix. Eight times. The X is equal to 00 vector. And so we have our reduced me reduced matrix A as 120 minus AIDS and 01 And we have our expect. Er with excellent next to that's equal to zero. No, that looks like 10. Or rewrite that. Yeah, that's equal to director. So multiplying this l we get x one plus two x two equals zero. We have a zero times x will in which zero minus and eight times x two is equal to zero. We have a zero times x one, which is zero and a one times x Two years of X two equals zero we have this x two equals zero. And if we pulling that into our second equation, that still just zero equals zero. It looked like that appears X two equals zero. We get that X one is also you go to zero so x one equals zero. The next two x two equals zero and that is one solution. So are no space contains only contains the zero factory for eggs is just zero bacteria. It's a two by one matrix. Double zeros. Now we can go ahead and write our solution out for the middle space for a solution will be that the null space abbreviated by capital in in the no space of a is equal to zero vector. This is the answer.

Determine a spanning set for the null space of the given matrix. So our matrix is 123 456 Nope. 123345 I got carried away here. 567 So I'm going to write an augmented matrix, use um, row operations, multiply the first road times negative three and add it to the second row case. Ah, negative. Three times two is negative. Six plus four is negative. Two negative three times three is negative. Nine plus five is negative. Four. Now I'm going Teoh. Multiply by negative five and add to the third row. Two times Negative. Five is negative. 10 plus six is negative. Four. Three times negative. Five. It's negative. 15 plus seven is negative. Eight. Zero king. Now it's going to write my first row. 12 three, zero. Um, zero. Go ahead and divide this one by negative too. 120 And now I'm going to take road to and multiply by Negative too. And added to row three zero zero zero zero. All right, who keep going? Looks like I have some room over here. I'm going. Teoh Multiply. Maybe not. I'll keep going down the page. Easier to go down the page, I think multiply this times negative too. And then I'm gonna add it to the first row. 012 000 102 times negative, too. Two times negative too. It's negative for plus three is negative One. So two times negative two is negative for plus three is negative one. All right, so that gives me a Z X Y z. So axe minus Z is zero. Why? Plus two Z is zero and the last road. It says zero equals zero. Um, so x, Izzy, and why is negative? Two z So are null space would include to negative too t t such that t is any real number. So now we can make a spanning set from this our spanning said, I'm just going to set t equal to one one. Negative, too. One. There are, of course, other options.


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