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7 U 1 1 contiuot sv + { 1 11...

Question

7 U 1 1 contiuot sv + { 1 11

7 U 1 1 contiuot sv + { 1 1 1



Answers

$$ (4-7 i)+(2+3 i) $$

To find these two matrices together. Here. First thing we need to do is check the dimensions will work. Okay? First matrix we've got is two rows three columns. That's a two by three. The second matrix is a three row one column matrix. These inner two dimensions have to match which they do. That means is a resulting matrix will be dimension the outer dimensions. Okay, So we're gonna have to buy one. Matrix is our answer just two spots here. Okay, we'll fill in the first spot. This is gonna be row one. Column one. Okay, so everything in row one thinks everything in column one negative one times six plus zero times negative four plus seven. Thanks one. Okay, this should give us negative six plus zero plus seven. That should give us a one in that first little spot right there. Okay, The second guy this is gonna be row two column one. Okay, so we're gonna take everything in row two times that same column right there. So three times six less negative five that was negative for plus two times one spent 18 plus 20 plus two should give us 40. Okay? So one in the first row, 40 in the second round, and that is the result of this modification.

We need to simplify. Seven plus two wired and dove one minus Odo. Nobody using four motored Begin right. Seven minus seven. Iron toe placed, wired up, minus wider square. Now, after a simplification, we will get seven minus seven Ight up. Blessed wire days minus Phi Phi Iota minus. I address square is minus one. So this dome will become too and do minus one. So after for the simplification, we will get seven minus five iota plus tau seven plus two is nine. So the result would be nine minus five. Iron finally began. Say that the on So is nine minus five fired. Oh!

Okay for this one. We have a is equal to 12 negative. One 011 and zero. Negative 11 So the characteristic equation for this problem is given by negative Lambda Cubed plus three Lambda squared minus four. Lambda plus two is equal to zero. And when we solve this equation, it's a cubic equation. So you will get the three argon values as 11 plus I and one minus. I noticed that these talking visor complex congregants. So now if we have the Lambda equals one, then upon solving the system, eh? Minus I times u equals zero. We will get that use equal t times 100 and for Lambda equals one. Plus I, using a similar process, we end up getting that you sequel to a T times I'm minus two negative I and one no, for Lambda Contra Kit equals one minus I, which is given here. Then that implies that you is equal to it. Turns out that the Egan vectors are also conflicts congregates of each other. So you was simply gonna be equal to a T times negative. I'm minus two guy and one

We're given this magic A We're universe first. Me from the convertible meeting room. Without it, the determinant they You could have zero in a but not in veritable works. Check it. A convertible. What do you say? Actually, I read it down here. Let's find a determining a check of the convertible or not. 111 First, I'm gonna high road to buy world one by negative one and had it wrote to So I guess one might one minus one zero. Making one plus 21 No one here. Next I'm gonna multiply growth three by negative one times wrote to I get 111 negative. Q one is negative one. You know, the determining the mortification old the numbers in the pivot in the diagonal interment is clearly not so. Therefore, we can find a neighbor nullifying chambers through the over inside the right chambers on this side. Very eight in the side. You're a here you have the identity matrix for three by three One here is you here alone? Now we're gonna really do until this side here. It looks like this. Once we do that, we will get a members on this side Look for reduced. Well, we already thought before we're finding the determinant. Do it again. First they can about this here. 11 now weaken Can't hold this position here. Negative one. They won negative times. Negative. 101 Negative. 101 Here one. Now weaken. We can scale the throw here. We can divide the group by negative one. We get negative here. Also here. Positive here. Now I can scale road three by minus a few. Added row to cancel this The negative too. Times road Here. That positive you minus one. That's one. Make a few plus one minus one. Two zeros too. You get one Next. We just need get rid of this. We can. He gave the period. Rowing added to the first would be a bit of this one. Here. You never get here from zero and in one one to negative one plus 01 and 101 Now you get a second road out of the first road in negative. One plus two. Just one. You have one plus minus +10 You have minus to plus one. You hear? This is the identity matrix implies on this side. He had a members say in verse, should be one bureau minus one one, minus 12 and minus 11 minus 11


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