use-the-algorithm-from-this-section-to-find-the-inverses-of-leftbeginarraylll1-0-0-1-1-0-1-1-1endarr

Question

Use the algorithm from this section to find the inverses of 
$$
\left[\begin{array}{lll}{1} & {0} & {0} \ {1} & {1} & {0} \ {1} & {1} & {1}\end{array}ight] 	ext { and }\left[\begin{array}{llll}{1} & {0} & {0} & {0} \ {1} & {1} & {0} & {0} \ {1} & {1} & {1} & {0} \ {1} & {1} & {1} & {1}\end{array}ight]
$$

Michael Jacobsen · Accepted Answer

In this example, we have obey tricks A, which is given to be lower triangular and the entries in the lower triangular portion are all exactly one. This matrix will be in vertebral and so let&#x27;s calculate its inverse. So to do that are algorithm is to make a matrix formed by a and augmented by the three by three identity matrix. So this is going to be equal to first let me place a in blue. It&#x27;s 100 110 and then 111 Then we&#x27;ve augmented with the three by three identity matrix, which is 100 010 and 001 So this is our matrix, and our job is to roll reduce until this portion before being augmented becomes the identity matrix. So now to that end, we have a pivot here, and our operations will be to eliminate these two entries. So first we copy Row one, it&#x27;s 100 100 Then take negative one times Row one and add it to the second row will attain 010 negative one, one and zero and for the next operation will copy multiply the first row by negative one and added to road to we&#x27;ll obtain this time 01 one. You obtain a negative one zero and one altogether. Now, for this new matrix, we have a pivot. Here we have a zero above, which is good, but we need to eliminate this entry, making it a zero. So will say that this is roll equivalent to first copy rose one into So we have 100100 and row one row two is 010 negative. 110 And now our operation will be once again Multiply wrote to buy a negative one and added to row three will obtain 001 Then we have a zero here than a negative one and a positive one. So now let&#x27;s bring back the fact that we&#x27;ve augmented. Since this is I three, we&#x27;ve now shown that a inverse exists and it&#x27;s equal to 100 negative. 110 and zero negative one and one. Then what you can do to check that we really found the inverse compute a times a inverse and show that you get the three by three identity matrix. Let&#x27;s redo this example, where instead of a three by three upper triangular matrix consisting of ones, let&#x27;s make it be a four by four case. So let&#x27;s see what the inverse will be in this situation. Well, this is a tricky matrix, not because it&#x27;s going to be overly difficult to take a augment with the four by four Identity Matrix and ro reduce. It&#x27;s just a very long process. And when we make many, many different with medical operations, the probability of one tiny mistake is quite high. So let&#x27;s go back to the Matrix we looked at in the three by three case. Recall that we had this interesting form. Row one is 100 Let&#x27;s go back to our case and hope that that still works out. Let&#x27;s say for a inverse our row one will be one, then zeros. It&#x27;s of size four by four. So it&#x27;ll be about this large. Okay, let&#x27;s pop back to a inverse and note that in the second row what we&#x27;ve done is we have negative one, then one and then zeros thereafter. So let&#x27;s make it negative one, then one in our case here and zeros thereafter, hoping that this works. So let&#x27;s go back to the three by three case and notice. Now. In the last row, we had zero than a negative one and a positive one. Let&#x27;s maybe use a different highlighting color. So let&#x27;s try that out for our case. Let&#x27;s go. Zero negative one and one in the four by four. So zero negative one and one. And I hope maybe a zero could go here. Now we have kind of a staircase pattern. It goes negative 11 Then go down a row. Negative 11 And my hope is there&#x27;s gonna be a negative one and one here, Let&#x27;s try it out. Place a negative one and one and zeros elsewhere. So this is our form for the Matrix. And I&#x27;ve never did roll reduction to verify this. We just looked at the pattern that worked out in the three by three case. Now for this matrix A provided here and the a inverse we&#x27;ve just calculated you can check that when we multiply. We, in fact get the four by four identity matrix that tells us our guests is correct

William Scherer · Answer

number 431. They want us to find the inverse of this four by four matrix here. Okay, left the bracket open. Because the way we&#x27;re going to do this, we write the identity matrix on the right. Okay, so four by four identity matrix on the right and we want to perform role operations until we get the identity matrix on the left. Okay, so we&#x27;re gonna work one column at a time. We already have a one in the upper left and all zeros in the first column. Okay, Second column. We want to get a one here and the rest zeros. So since we want a zero right here, I am going to subtract row two from row three and put that in Rose three. Okay, so we&#x27;re doing Rose three. Mine is wrote to All right, so rewrite the first two rose. No. And each element in row three subtracted from row 20 minus zero is 01 minus 01 minus 101 minus zero is one and so on. Yeah, and just rewrite Row four. All right, So, row, uh, column two is finished as we want it. Okay, Column three. We want one right here, and we want the rest of them zero. So I&#x27;m gonna dio row one minus row three and put it in Row one and row four, minus row three and put it in a row for Okay. So I&#x27;m killing two birds with one stone here because I want to get a zero right here and a zero right here. And when we subtract these ones from this one that will give us the zeros we need. So each element and row one minus each element in row three and that will give us 1001 and 11 11 negative 10 re copy wrote to re copy Row three and each element in Row four minus each element and row three one minus negative. One is going to give us a to 01 negative one and one. Okay, so our third column is complete Now for the fourth column. We want this value here to be a one, and we want the rest of these to be zero. Okay, so I am going to multiply row for times one half, and that will give us a one right here. Okay. And then I am gonna add half. I&#x27;m gonna add that to Row three. So half of our four plus row three. And I&#x27;m gonna subtract half of our four from row two. And I&#x27;m going to subtract half of our four from row one, minus half of our four. Okay, so when we do, half of our four will get a one here. And then we only added to this one will get a zero when we do, one minus half of two will get 01 minus half of two will get zero. So that&#x27;s what we want. So pretty easy, operations. But we&#x27;re doing four rows. All right, so in the first row, first row, let me see. We&#x27;re doing our one. Okay? 10 All zeros said for the top left, and we&#x27;re getting one half negative. One half negative. One half, second row. Um, we&#x27;re getting a zero half. Half negative. One half, Third row, zero negative. One half. If half and forth. Row zero, half negative. One half. Half. Okay. So now we have the identity matrix on the left. See, we have all ones in the main diagonal and old zeros for the rest of the elements. So this this matrix on the right is your inverse matrix. So if the original matrix was a this is how we would write the inverse matrix. Now, in their answer, they factored out a one half. Okay, so let&#x27;s just do that. So a half times what will give us one? Well, I have times two will give us one. And this is one negative one negative one. This is zero one one negative one. This is zero negative. 11 one. And this 01 negative. One one. Okay, so we factored out a half. So wherever there&#x27;s a one half, it&#x27;s just one, because a half times one will give us a half where there&#x27;s negative one half its negative one and so on. Okay, so this matrix with the one half factored out is our inverse major

Ernest Castorena · Answer

Okay. This time they&#x27;re giving a the 100 zero negative. 10 101 And they want us to find So it&#x27;s argument right at the gate by three Identity gun reform row reduction. But she&#x27;s the 1st 1 is there. Give it. We&#x27;re gonna subtract it from the bottom round. We&#x27;ll end up with their 01 negative. 101 100 zero one. There, there and all we have to be really is flip the sign of the centre. 10 zero negative one Europe. So we put the center up. I&#x27;m multiplying it by negative one. And now we&#x27;re the identity of the left hand side. And we can read off Are inverse negative one, Okay.

William Scherer · Answer

number 435. They want us to find the inverse of this matrix here on the left. Okay. And the way we&#x27;re going to do that is to write the identity matrix here on the right and do elementary row operations until we get the identity matrix here on the left. Okay, so this is what we were given. This is a six by six identity matrix, okay? And we almost have the identity matrix here. But we have to get rid off these five ones and row six, okay? And we can do that by subtracting each of these rows from row six. Okay, so we don&#x27;t really need to keep rewriting, um, the whole thing, because we&#x27;re only gonna be working with row six. So I&#x27;m gonna do Row six minus row five, and then I&#x27;m gonna do rose six minus row four, rose six minus row three rose six minus road to and rose six minus roll one. Okay, um, so let&#x27;s start with grow six minus row five. Okay? Everything will stay the same. Except for this right here will be one minus one. I&#x27;ll get a zero right here. So let me let me just cross this out for now and rewrite it as a zero. And also right here. I&#x27;m gonna end up doing zero minus one. So I&#x27;m going across this out and write negative one, Okay? Nothing else changes because you&#x27;re just subtracting zeros, except for right here and right here. Okay, then let&#x27;s subtract Rose six from row four. Well, everything will stay the same. Except for one minus one. Right here. Somebody across this out and right zero. And also zero minus one right here will give me a negative one. Someone across this cell and write negative one. Okay. And it keeps going like that. Rose six minus row three. The only one that changes is right here. When I do one minus one, this becomes zero. And when I do zero minus this one, This becomes negative one. Okay. Rose six minus wrote to one minus one. I get a zero right here and zero minus one. I get a negative one right here. And finally Row six, minus row one one minus one. I get a zero right here and zero minus one. I get a negative one right here. So now we have the identity matrix on the left. We have all zeros. Except for the main diagonals, all ones. So this matrix here is our inverse matrix. Okay, so, um, just to clean this up, everything on the bottom, everything on the bottom row is a negative one, except for the bottom. Right. So these are all negative ones on the bottom row, except for the bottom. Right. And this matrix here is, let&#x27;s say the original matrix was a So this is the inverse matrix. A eight of the negative one. Okay. And of course, this matrix on the right. I mean, on the left is the identity matrix. So this one here on the right is the answer.

Chun Yeung · Answer

for the inverse I&#x27;ll be nature&#x27;s to exist to determine the has been non zero. So for that given matrix, we can find the determinant and check if the aim for success and we can finally determined by reducing matrix to did bro. That&#x27;s a long form, and they just do that first. I&#x27;m going to subtract the third row by the first row. Then we get 0100 Then I&#x27;m going to subtract it there, rolled by the sudden role. Then we get I want. And then I&#x27;m a subjected last wrote about the first rule. Get through one so in one. But we&#x27;re not done yet because and they wrote actual on form. The last row has to be all zero and we can reduce and get rid of that ones by subtracting the last rule by the second row. And then we get it&#x27;s gonna be zero zero and we&#x27;re done. So the determinant is going to be one times one time zero time zero And I got that by multiplying that that I Agnone Ellemann off the matrix. What you see equals zero because the determinant is zero. The inverse does not exist

Sriram Soundarrajan · Answer

good going. Bizarre problem. About 80 here they given metrics is 1000 minus one serial one serial one. So we need to find its interest. So first, finally dominant. Dominant. This word needed minus minus minus one. So when you go, Friday was so far equals where they do my It could buy this one fever because minus one, it does their own little because it&#x27;ll see about three. It was like they do zero last Christmas. Si tu one equals minus one. It do. Zero minus zero equals zero. See toe What they took, What might have zero? Because it&#x27;s not State Street regarding minus one does he could for this the Group three. But see, even you could smart a zero. It could c 32 equals zero. It could settle. The Eagles are buying. It&#x27;s a little it&#x27;s It was his. My excellent set up. One little one little zero is it&#x27;ll minus one

Problem

Repeat the strategy of Exercise 33 to guess the i…

Problem 33 Hard Difficulty

Answer

When we multiply any other row from $A$ with column from $B$ result is zero.

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

Linear Algebra and Its Applications

Anna Marie V.

Alayna H.

Maria G.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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When we multiply any other row from $A$ with column from $B$ result is zero.

Linear Algebra and Its Applications

Anna Marie V.

Alayna H.

Maria G.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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

Anna Marie V.

Alayna H.

Maria G.

Michael J.

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

Linear Algebra and Its Applications