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1 1 'JJuo y ocSumt Which of the following uthcr patnogens 1 Iskare affccting tnc urogcnitI tact | op Aj4i TRUE about N 1 L 1 about futurc cplsoda 1 Check S Jsi...

Question

1 1 'JJuo y ocSumt Which of the following uthcr patnogens 1 Iskare affccting tnc urogcnitI tact | op Aj4i TRUE about N 1 L 1 about futurc cplsoda 1 Check S JsiloutueliIu u 1 rcsult Oriniectni

1 1 'JJuo y ocSumt Which of the following uthcr patnogens 1 Iskare affccting tnc urogcnitI tact | op Aj4i TRUE about N 1 L 1 about futurc cplsoda 1 Check S JsiloutueliIu u 1 rcsult Oriniectni



Answers

1 lerc, yellow prccipirare (S) is of (a) $\mathrm{Na}_{2} \operatorname{CrO}_{i}$ (b) $\mathrm{BaCr}_{2} \mathrm{O}_{7}$ (c) $\mathrm{Na}_{2} \operatorname{Cr}_{-2} \mathrm{O}_{1}$ (d) $\mathrm{BaCrO}_{4}$

Hi everyone saw in this question, they asked choose the pair in which even the first element is greater than even of the second element, but in case a year already is river urged. So it's B S B S F O ces, magnesium, aluminum, D C N O. So in this case option A and B are correct option A and be odd cutting. Okay, uh huh. Now this P S and N. O. And it's fast chris so far. Mhm And and oh, are correct curry and it's now phosphorus has greater ammunition potential than sell for similarly nitrogenous, greater ammunition potential than oxygen. Mhm. First generation energy.

Okay, we're given a three by three Matrix A is 010 001 012 And we want to determine if it in verse exists before going on. You might think right off the bat that because they're zeros that this in in bearer, this matrix is not in veritable. Well, first thing to check is the rank. Remember that For a matrix to be in veritable and M by N matrix will be in vertebral if and only if the rank of that matrix is equal to end. And in this case, the rank of this matrix is three. And so it is a three by three, and therefore the rank of A is equal to three. So we do have to continue on and see if there is an inverse that we can find. 1 may exist. Um, but we're not sure. So we're gonna go ahead and use our guys Jordan method. So we will take our matrix a and we will adjoin it with the three by three identity matrix. Um, and so I am going to start by switching Wrote to and row three. So 010100 012001 and then 001 010 And the reason I did that was I see that I have a one here and then in this position, I have a one as well, so it might make things a little bit in here. Okay, so let's go ahead and look at the guy. Was Jordan method here? We're gonna need Since we have a one in this position, everything below it now needs to be a zero. So we need to make that a zero. And so I'm gonna take row one and subtract off road to and then replace it. That road two So 010 100 So we get zero minus 01 minus one zero minus 21 minus 00 minus zero and then zero minus one. Okay, so now what I want to do is make that a zero, and hopefully you can already tell that we're not going to be able to get to a identity matrix because we don't have a one in this column. But let's finish this one off just to verify so I would take two times. Row three added to road to and replace road to. So we get zero zero zero one to negative one 010100 Then if I go ahead and switch road to Withrow three I get 010 100 001010 and then 000 one to negative one. However, as we already kind of assumed that what happened, this guy here is not the identity majors. Okay, way need to get to the three by three identity matrix, which looks like this. And of course, we don't have a zero in this column here at all. So therefore, matrix A is not in vertebral and is considered to be a singular matrix.

So we're given a two by two matrix a one one plus I one minus I and one. And we want to determine if an inverse exists for this matrix. We're going to use the KAOS Jordan method. So the girls Jordan method says we take our inverse a and then we're going to augment it with the two by two identity matrix. If we can perform row operations, which turns the left hand side into the two by two identity matrix, then whatever we have left over here will be the inverse. So performing a row operations, the first step is going to be turning this value to zero. So we'll take one minus. I times Row one will subtract off row to and will replace Row to with that new value. So row one isn't changing. So we take one minus. I times one minus one minus I That gives us zero. And now we're gonna take one. Plus I times one minus I, which is one minus y squared. So it's one minus negative one or two and then to minus one is one one minus. I times one plus zero is going to give us one minus I and then one minus I times zero plus one. I'm sorry, minus one. Well, give us negative one. Now, our next up then is going to try to be to turn this guy into a zero. So in this row operation, I'm gonna take the negative of one. Plus, I multiply that by row two, add it to row one and replace my row one so wrote to isn't going to change. And so now we have negative one. Plus I times zero plus one. So that's gonna remain a one negative one. Plus, I times one plus one. Plus I this will turn to zero. Will have negative one. Plus I times one minus I So we're gonna get negative, and from above, we know that that multiplication is too. So we're gonna get negative too, and so negative too. Plus one will give us a negative one, and then we're gonna have negative one, plus i times negative one, and that is going to give us just a positive one. Plus, I so notice we have successfully turned the left hand side into the identity matrix, which means that what is remaining on the right hand side is are inverse matrix. So the inverse matrix is negative one one plus I one minus I and negative one. So let's go ahead and verify that the answer that we got is in fact the inverse to remember to do that, we're gonna take our matrix, multiply it by its inverse. And if we receive the identity matrix as a result that we know we calculated it correctly. So matrix a is 11 plus I one minus I anyone multiplied by its inverse so performing matrix multiplication We get one times negative one plus one plus I times one minus I we get one times one plus I plus one plus I times negative one negative one times one minus I plus one times one minus I. And then finally we get one plus I times one minus I plus one times negative one. And so in our world one com one We know this value here is to So we have a minus one plus two, which is one in row. One column to we have one plus high minus one. Plus I. That gives us zero and wrote two column one. We have negative one minus I plus one minus I, which is zero and in road two column to We know again that this is a value of two and to minus one gives us one. And this is the two by two identity matrix. So, in fact, we have checked our answer.

Oh, question. Question. Why every national on this white back? That's why it supports GP y X and why that attractive deal in school moving towards the solution for together Question complained last formal. You both decided so. And with why did the next best buy that yes body And it's last off GDP itself could be for as the square as part of this aspire zero My next. Right. Check Tito minus twice. Explain. Yes, my next via zero. Let's buy s t. Yes. So this people get bias as the most G? Yes, they did. That's his square minus Always Just lightness s minus three basis for us. That's about what? This is a different station. I listen to you. Let's explain minus the last one car. Yes, Well bar as honest one city. Yes, Gs. That's right. That's right. Minus quantify it. Yes, by s point as well. That respect not taking just like this was why, yes. Anyone school you love blood. Plus Paul. Well, big question. You know, again he said nothing more Glove Yes and yes, G s, which is indignation forms here. So a point dp if we go there ffs for this one. Most definitely will be using this Are fighting with beautiful nation from its get beat. Being like this being even people this week, baby financing I would be less Yeah, question, yeah.


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