in-the-rest-of-this-exercise-set-and-in-those-to-follow-you-should-assume-that-each-matrix-expressio

Question

In the rest of this exercise set and in those to follow, you should assume that each matrix expression is defined. That is, the sizes of the matrices (and vectors) involved “match” appropriately.
Let $A=\left[\begin{array}{ll}{4} & {-1} \ {5} & {-2}\end{array}ight] .$ Compute $3 I_{2}-A$ and $\left(3 I_{2}ight) A$

Lucas Finney · Accepted Answer

for this problem were given a matrix A which is four or five negative one negative too. We want calculate three times the identity minus a. And we also want to calculate three times the identity kinds of a so three times our identity matrix. It specified as I too, because we want our two by two identity matrix. Oh, I said to And so I thought Thio wrote the wrong number. There we go. We have 3003 minus. So I&#x27;m going to write this, actually, as plus negative a So plus negative four and negative five. Positive one Positive too, which is going to equal. So we have three minus four. Gives us negative 10 minus five. Yeah, zero minus five is going to give us negative. Five zero plus one is one and three plus two is positive. Five. Next doing three times our identity. Identity Identity matrix times A. You have 3003 times. Uh, not negative. A regular A. So we have four or five. Negative one Negative too. It&#x27;s not negative. 12 Negative too. So we&#x27;ll have three times four plus zero times five, then down below. We will have zero times four plus three times five. We&#x27;ll have three times negative one plus zero times, too, and we&#x27;ll end off with zero. A time is negative one zero times negative, one plus three times negative, too. So that&#x27;s going to you? No, these are all pretty easy to evaluate. Here. We&#x27;ll end up with 12 15 a few times. Negative one is negative three and then negative six. And I&#x27;ll point out that that is actually the exact same thing as three times a.

Mike Verwer · Answer

Okay. And for this question, we are asked to compute Uh, thes two piece of a riff. Is it here? Uh, and using this major is a So let&#x27;s begin. So a minus five times, the three by three identity will be the 1st 1 So that will be nine minus 13 minus 87 minus six minus 418 On this, five times three by three Identity, which is 111 000000 And we&#x27;ll start by multiplying in the scaler. Okay? And I&#x27;ll continue over here, right? And so this will result in nine minus five, which is for the other entries won&#x27;t change. Only the diagonal entries will change. Here is the only A things in this matrix are diagonal entries. So this will still be minus eight. This will still be minus 57 Minus five is, too that this was simply minus six. This will still be 18 miles. Five is three. And so we&#x27;re done. And for the next one, we will do, uh, five times three by three. Identity multiplied by a so five times the, uh, three by three identities. This matrix multiplied by a again minus eight seven. My six, Nice for one and eight, and the result will be five times nine plus zero zero. So that will be five times minus 45 and then zero times, or five times minus one plus zero times seven plus zero times one. So that will be minus five and again five times three and the rest of the zeroes. So that will be five times 3 15 and then zero times nine plus five times minus eight, which is minus 40 and zeros for the others, dear. It&#x27;s as minus 15 times 7 35 And so the pattern here is that we only have to multiply correctly, corresponding element by five. And this will be five times minus six. Leaved Maya&#x27;s 30. This will be five times minus four, minus 20. This will be five times one. This will be five times eight. Okay, we&#x27;re done.

Thomas Emment · Answer

Okay, give this matrix A. We want to prove that the identity times by a Z A on eight times a day. So what is the identity? Well, in a three breath in terms of three by three matrix, it looks at this 1000000 What? On it etc. Basically So it has the same properties as what? Do you know how tough games it won times? Anything is just be anything so six times one of six equally like, Yes, When we have anything times, intensity, it should just be except But we&#x27;re gonna prove that it now. Why would you make modification? And you rose off the first matrices matrix against the columns off the 2nd 1 So for the first row First column, you&#x27;ll have first row the first matrix, which is I 100 against the first column of a 3 to 0. About 30 to 200 So we&#x27;re just going to be left with a three now, see, if you have to move on to the second rope, you could be taking the second road. The first caucuses, The rose off the first matrix against the columns of seven Monceau Second Road against the first column 0731 of the tea zero lots of zero. Just get to have an equally if we move on February 001 Lots of 3 to 10 to 300 to 20 And I just want a lot About zero. What could be a zero? Okay, so then equally if you two would be moving down each column 43 156 against each rope. You can see that from the from From this you got 3 to 0 so many times, which is just gonna get our exact column down. But I&#x27;ll show you. This is so say we&#x27;re moving on to our first real here. Certain Colin first row is 100 against 423140 It&#x27;s the other two equally, the third Colin first, where you could have one lakh one zero lots of the five and six now we&#x27;re gonna be taking are ignoring the first element, taking second and ignoring 1/3. Say you&#x27;ve got this one would be just too is ignoring the four of the three. And if this would be the five and then final. This one we&#x27;re nor the 1st 2 elements and just taking the last one. And I hope you can see that this is indeed equal to a So now, for our second proof, we just need to instead take the rows of a against the columns of I for each respective position. So here for the first request. But I got 341 against 1001 of the three. None of the four. Someone&#x27;s second row, 2 to 5 against the first column, you&#x27;re gonna have one of that 20 Lots of tuna five to on the equally hit were first rose, 036 gets first calling one lot. 000 cities so and it will be 40 Okay, I&#x27;m now going on to our second column of ice will be ignoring the first on the third number that we get only really worried about the second. So let&#x27;s say its first road first very 341 Ignoring the end ones and just going through there with four again for the second rose to 25 So is there a lot of this deal? What&#x27;s this? Just left with 20360006 Just one long about three on. Then finally, is beings at the same, but we taking as you can see it. So lost road here on I will just be taking ignore the 1st 2 elements on just taking one. A lot of the last element if each wrote. So that&#x27;s effectively last broke, last element of this rare with one last time. This rose five last time. This was 6156 which again you can see is equal to a

Thomas Emment · Answer

Okay. Dokey. Um, give this matrix A. We want to show that the identity times by a visit to a on eight times the identities e Okay, so I&#x27;m doing major modification. You do the rows of the first column First Matrix by the columns of the second Matrix. So in this case, the first element off the first row First column it&#x27;s going first revived 100 against the first column of a 357 So zeros that 507 just one little about three equally food to move on to the second take. It was the first road go the second column. So what about that 20 to 6 and a on again first right there, calling one of the 10 to 2 or three. Just one moving on to the, uh, second Rhona. There&#x27;s Collett 03157 such to five again zeroes. The two and eight just a six on zero. What&#x27;s the 13 interested, too? On finally, third row. First column. 0 to $3051.7 zero lots. That 2061 of the A plus the +10 to 1. A lot about three. Okay? And I think you can see here. This is indeed equal to R A Matrix. Now let&#x27;s deal with Azov. First Matrix and eyes are second Matrix. Nobody in the Rose of A against the Columns of Ice. I will do first row. First column. So 3 to 1 against 100 So just one out of 30 to 1 of the ones. Which three? Uh, let&#x27;s, um, let&#x27;s let&#x27;s do our first column. So we got the second elder there. Second row they against Still, the first quarter of ice took off one of the 5066 You lost the ticket on Finally, third column forgets. First. We won a lot of 70 ch Tools three. Then, if we&#x27;re moving on to our second column, was zero. What&#x27;s the first element zero so thoroughly just could be having our second element. So I run to our first race of second element in our first words to second. Women are second row six on second element in our servers, eight equally. Well, you know, if this was a two, we did have to lock for Middle Road. If this is too this is a woman of two lots forming a rope, one love authored. I think we&#x27;re having elements are off the roof. It is equally waiting for third calm Here we got zero, lost the 1st 2 elements and then just were not the third. So 321 None of three to just one of the ones number five or six years. 2 to 7 ages of three. And again, this is also equal to our A matrix. We&#x27;ve proven that guy on guy equal to R A matrix.

Thomas Emment · Answer

Okay, we&#x27;ve been given this matrix a on. We want to prove that the identity touched by a single toe and eight times identity is equal to a So what&#x27;s the identity, Clark? The honesty for a two by two matrix is 1001 Where you can think of this matrix is almost the wonder if, like the two dimensional matrix you know, two equally Teoh. Yeah, You configure the one off the to mention round where anything times, right, This is always going to be itself. It&#x27;s got same properties as one is what we&#x27;re trying to set so that we can prove this hit. So when we do major modification, you take the rows of the first Matrix against the columns off the 2nd 1 So he will have 10 against 35130 lots of five. We just We left the three equally guessing with 1st 2nd column one a lot of that 20 lots. That nine. Now we move on to our second road and I&#x27;ll go the second column first. Remember Eyes Our first major accepting second row second column zero loss that to one on that night on Finally, we got were onto art. Second row, First column say second or the first matrix here. One first column of that second matrix 350 Is that 31? More than five. Where does Aziz? I think you&#x27;d see guys equal to a Okay. Now are a czar. First Matrix and our eyes. Our second tricks. So we taking the rose off the A matrix. So you&#x27;ve got three to hear first against 10 So one of the 30 lots. That too. Well, after this three equally stick with the first report when we come to our second column now is hear lots about 31 law. That, too. And now in our second round. First, call him one more. That 50 loss about nine. And then second row, Second column zero. It&#x27;s that 51 law that nine on. I think you can see that both of these are equal to a

Thomas Emment · Answer

okay. We want to show you could get a matrix. A. Now we want to show that the identity terms by A is equal to a on eight times the identity is able to a So what&#x27;s your identity is very simply, 1001 Okay, so this is only so when we do major modification, you take the corresponding rows of the first Matrix Amal&#x27;s pie them by the columns of the 2nd 1 So I say once find, get off First break First column You do First row of the first Matrix on first quarter, the second Matrix. So here we&#x27;ve got 10 It&#x27;s 14 So one times one is 104 is force will get a one next up. Second stick with the first rope will go into a second column, so we&#x27;ve got 10 gets 36 So that&#x27;s just 33. Moving on to a second row First column and, say, second row. First column. Zero lots of that one. But one long at four on, then 01 off the second column. Susilo&#x27;s at 31 on that six, and I think you can clearly see this is going to be equal to a Now we Dickinson&#x27;s Are a Matrix is First will be Taking the Rose are a matrix on the columns of our identity. So let&#x27;s take me there are first row first column. One law that 10 It&#x27;s about three. I&#x27;m gonna be after the one. Then we&#x27;ve got our first race. Second column one. Not 03 lots of that one could be a three second row now. First column. Force that 16 loss that zero just left with a four. And finally, second row. Second column zero is that 41 lot at six. And I think you can see that this is also equal to R A Matrix.

Problem

In the rest of this exercise set and in those to …

Problem 3 Easy Difficulty

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

Linear Algebra and Its Applications

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Absolute Value - Example 1

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

Grace H.

Heather Z.

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

Linear Algebra and Its Applications