5

O6111 1DOLL...

Question

O6111 1DOLL

O6 1 1 1 1 DOLL



Answers

$\left[ \begin{array}{lll}{1} & {1} & {1} \\ {1} & {2} & {3} \\ {0} & {1} & {1}\end{array}\right]$

Okay, so I'm gonna be multiplying negative one times itself six times so negative one times negative. One time saying good one time single, one times negative, one times negative. One who feel like a broken record. So I'm going to pair each of these up and multiplying. What's nice is all All three of these pears are the exact same thing. So negative one times negative one is a positive one. Same with this. Same with this. Now, when I multiplied those together well, one times one is one times one again is one and they're all positive. So my answer stays positive if you also look, I'm multiplied. Well, 1/1 times. One times one. Right. It just stays one. But notice I had 123456 Because I had an even number of negative ones. The negatives cancel each other out. Right. This two pairs cancelled each other out. Thes two pairs. Cancel each other out. These two pairs cancel each other out. So if you have an even number of negative numbers, if they turn positive because each have a pair of another negative to cancel out

Came this question. I'm multiplying negative one times itself over and over and over. Seven times now, one times one times, one times, one times, one times one times one is just going to be one. Now, these two negatives multiply together, make a positive one. These two also make a positive one. These two also make a positive one. But because of this odd man out here, our answer is going to be negative. If we were to just have even numbers of negative one, we would have had a positive answer. But because of this odd right, we had on number of negative ones that we were multiplying together. My aunts, my final answer will be make divorce.

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

In discussion, we need to calculate the uh universe of forgiven metrics which is maddox. A. Is that two by two matrix having the first row elements as two and one. Second elements as one and one. So first of all we will right Metrics to find this universe as these metrics here. 2, 1, 1, 1 and Here the identity medics as one, Now we will use the row reduction method to convert these metrics to the identity and this one to the metrics whatever it becomes, applying the same row operation or what we will apply to this. So first we will apply the operation are one stores Are 1 -22. So we will get here as Our 2 -1 which will be one only and 1 -1. Here will be zero and second row will be unchanged. So this will be 11. And here in this identity matrix, 1 0 will be one. 0 -1 will become -1 And here zero and one will be we will remain same. And by applying another operation which is our two stores Our 2 -1. We will get Here as faster as 1, 0 as unchanged. And here this will become 1 -1 is zero And 1 0 will be one only. And here in this identity Matrix, this first trial will be unchanged, 1 -1 And here 0 -1 will become -1, 1- of -1. So this will become one Plus 1. That is to So we got these metrics identity metrics and this one, another metrics. It must be the universe of the metrics A. Has inverse must we 1 -1 -1 and two. So now we will check whether it is the universe correct inverse of that. Matics or not. So this must be the universe. Now we will check by multiplying a matrix A to its universe. So now eight times a universe. We'll give cheers 2 1, 1, 1 times 1 -1 -1. And to hear when we will multiply this, we will first multiply the robe onto the column Robin of dramatics here to the column one of metrics universe. And we will get here as Are two times 1 Plus one times -1 which will be -1. This will give one only here. So this will become one here. And when we will melt it by First through to the second column of this matrix, we will get second element of Roman. So here this will become two times minus one And plus one times 2 here as a two times my husband and plus one times two. So this will be one times too. And this will become minus two plus two year zero. So the second element will be zero. And in the third element we will melt it by this road to the first column of this metric. So this will be one times one and ah Plus one times -1. This will be 1 -1 which will be against zero. So this is zero. And when? Now we will multiply row to the second column of Uh this universe. So we will get one times -1. Find out Plus one times 2 plus one times two. So this will be minus one plus two, which will be one. So here this will be one. Now we can see that when we multiplied A. To the universe of it becomes identity matrix of order to. So we verified that this is the correct universe of the matrix. A. I hope all of you got discussion. Thank you.


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