find-the-inverses-of-the-matrices-in-exercises-29-32-if-they-exist-use-the-algorithm-introduced-in-3

Question

Find the inverses of the matrices in Exercises $29-32,$ if they exist. Use the algorithm introduced in this section. 
$$
\left[\begin{array}{rrr}{1} & {0} & {-2} \ {-3} & {1} & {4} \ {2} & {-3} & {4}\end{array}ight]
$$

Michael Jacobsen · Accepted Answer

for this example, we have a three by three matrix A that&#x27;s provided here for this matrix. We want to be able to determine what it&#x27;s in verses. If it does indeed have one. And one way we can tell if there is an inverse is reduce A until we get to the identity matrix. This will tell us that this is an in vertical matrix. If we never make it to the identity matrix, then A is not in vertical. So here&#x27;s our trouble, though with this approach we want to find the inverse so fits equal to the identity matrix. Then the next step is to find that in verse, which is a lot of trouble. So we&#x27;re going to take an alternate approach. What we&#x27;ll do is take a matrix a then augment with a three by three identity matrix. So if we do Rover operations and we get the identity matrix here, then we have an inverse, and it turns out the Arian verse will be in this position. So let&#x27;s begin row reduction. First, this matrix will be equal to all place A, which is one negative. 32 01 negative, three negative 244 and now we begin augmenting. So the augments break will be here. And for the I three identity matrix, we have 100 010 and 001 So this is the matrix we must reduce. This matrix is of size three bike six. So it feels a little bit daunting, but it&#x27;s not too bad. Really? Unless it had something like four rows to consider. First, we&#x27;ll begin with this pivot position. Eliminate the negative three and the positive two below. So let me re copy Row one of this matrix 10 negative too. 100 and multiply. Row one by three. Added to row two. We&#x27;ll obtain 01 Then we have a four or excuse me. Negative too. A three. A one and a zero for a next operation. Multiply row one by negative too. Added to row three. Then we&#x27;ll obtain zero negative. Three eight Negative too zero and then a one. So for this pivot position were all set because we have zeros below for this divot position. We need to eliminate this entry and then call him too will be sorted out. So will say that this is row equivalent to the following first I&#x27;m going to copy Rose one end to so 10 negative too. 100 Then Row two is 01 negative. 231 and zero. And this is what our operations going to be. Multiply row two by three Add the results to row three. The result is going to be zero that we have a zero A to seven, a three and a one. So now looking at the pivots Columns one and columns to are the 1st 2 columns of the three by three identity matrix. But we still have some trouble here for this pivot position. We need to eliminate the entries above. And now, ordinarily, I would take the step of dividing this to here by two to make it a one. But that would give us, ah, whole bunch of fractions to work with. So let&#x27;s make that our final step in a row operations. So if we copy row three, which is 002731 then our operations are going to be the same. Add row three to row to and will have 01 Then we have obtained zero 10 four and a one. Then add row three to Row one, the same operation and we get one 008 three and one. Now we have one final roll operation. We need to take this row and divide it by two so that we have the first portion of this matrix row equivalent to the identity matrix. So the last step is that this is roadkill in equivalent to 00 one seven halves, three halves and one halves for our third row. Then I&#x27;m just going to copy the rest. So 010 10 41 makes Row two and Ra one is 100 83 and one. So here we have the three by three Identity matrix. And don&#x27;t forget we have arguments. Break here. This matrix here is the main matrix we want. Let&#x27;s pause for a moment, though we&#x27;ve really shown two things in this row operations that a here is raw equivalent to the three by three identity matrix. This proves a inverse exists. And the second thing we&#x27;ve shown with these row operations is that a inverse is this matrix 831 10 41 seven halves, three halves and 1/2. So this is the method for calculating the inverse of a three by three matrix. There are other methods, but doing by hand computations. Typically, this is a fastest route.

Angelo Rendina · Answer

if we want to invert the medics, eh? We can write the metrics, eh? All right. And then produced by Rose to obtain identity and somatic sacks on the right. And if these reduction works So if we get the identity here on the left, then we know that X is actually the inverse of a So on the top left, we have the matter that we want. Invert, which is one minus 214 months, 73 months to six months for So we write the extra metrics with these ended entity on the right, and then we proceed by role reduction. So we see that in the top left already have a one. And so we know that that&#x27;s gonna be our people. And so I have alighted in red the first end of the second row, which is the four. And in green, the first answer of the third row, which is minus two. And so we&#x27;re gonna produce the second and third row by subtracting suitable multiple of the 1st 1 So the new second draw in red down here is going to be the old pro to minus four times. Throw one and you see that that is four. He&#x27;s actually the one that I&#x27;ve highlighted in red in the top left and also for the third row. The new third row in green. In the bottom, right is the old their draw, plus two times the first row. Again, the two comes from the island, it minus doing green. So if we do this reduction, we get one minutes to one and then 100 and then the second row becomes 01 Minus one is Justin both some and then on the right of minus 41 and zero on their third rolling green after the computation is zero to minus two and then to a one. So we&#x27;ve simplify the first column and the first year, the first column. And now we want to also simplify the second column, so we&#x27;ll write the same metrics in the top left. So woman&#x27;s 21011021 is two and now we want to reduce the last throw. So for the last row again, I&#x27;ve alighted the two in blue because we have our people here, these one here. And so we replace the third row with blue said Raw, which is the previous third Raw man is twice the second row. So if we do that, we end up with in the last row 000 and 10 minus 21 Now the issues that these last row here is all zeros and that means that our original medics, eh is not in veritable. If he were, then we would have got something like 001 And then we could have proceeded by transforming the left side of the metrics into the identity matrix. But since we got the zero raw at the end, we know the A is not invulnerable, so I mean

Snigdha Chandra · Answer

Hello and welcome to this video where we are going to find the inverse off a two by two matrix. In order to do so. Let&#x27;s begin vit an example matrix and which is a two by two matrix, with the elements being one 24 on seven before find the inverse off. A matrix is very important to understand that we can only find in verse off a square matrix. On that the square metrics should be non singular, meaning that the determinant after metrics should not be equals 20 So let&#x27;s start the process by first. Fine finding the determinant off metrics eight. Written something like this we find the determinant by taking the product off the elements in the leading diagonal one times seven subtracting it from the product off the elements in the other diagonal. Two terms for if you simplified further seven miners, it gives you a negative one. So in this case, the determinant Bernard Bang zero. We can proceed and find the inverse off. The matrix involves off matrix is written with the same letter a wood exponents off. Negative one on the process is to take the reciprocal off the determinant on, multiply it with another two by two matrix, which we derive from the original metrics by scrapping the position off the elements in the leading diagonal. So he&#x27;ll come seven earned. Here goes one on done by negating or flipping the sign off the elements in the other diagonal. So two becomes negative two or becomes a negative for in the next step, I&#x27;m going to replace the determinant off a Vitter&#x27;s value. It is negative one leaving everything else the scene. No. One over negative one. I can simplify and write it as negative one. So basically, in the next step, I just have to multiply every element in the metrics with negative one. And if I do so negative one times serving his negative seven negative one times negative two is still unlike wise for negative one. So, Darcy involve us. Uh, the given metrics. Thank you for watching

Angelo Rendina · Answer

So we remember that to find the verse automatic Say we ride the block Metrics, eh? Identity And we read use by Rose to obtain identity on the left and a matrix X on the right. And when we do so, it turns out that X is actually the inverse of a So we&#x27;re going to use these algorithm to compute the inverse off the metrics. A five, then four seven. So we started with the medics. 5 10 for seven on the left and we ride the identity matrix on the right. So 1001 So now the first step is to simplify the first people on the top left. So what we do here is we divide by five the first line. So we have one too. Then 1/5 zero, then 4701 Now we replace the second roll with the second roll minus four times the first throw. So we have 12 1/5 0 on the top line and the second row is four minus for zero seven minus eighties, minus one. Then we have minus 4/5 and one where we recall here that the second line are too then you are too. Is the old are too minus four times the green are one and also here are one The new or one was the older one divided by five. Now we merely change sign on the second row. So we have 1 to 1/5 zero. And down here we have 01 Which is why we wanted to see on the left and 4/5 minus one. Now the final step here to replace. So the new first role he&#x27;s going to be the green are one minus two. Red are too, so that the first line is now woman 01 to minus 20 And then we have 1/5 minus 8/5. So that&#x27;s minus 7/5 and then zero plus two is two. And down here we have 01 4/5 minus one and Saul dramatics minus seven. Fifth two for 1/5. Minus one is the inverse off. The medics say

Angelo Rendina · Answer

So we have the metrics, eh? Given by three two, seven, four. Since we need to complete its inverse, we started with computing. It&#x27;s determinant. So remember that the determined today he&#x27;s given by the product of the elements on the men diagonal minus the product off the elements on the secondary diagonal. So this determined is three times for minus two times seven, which of course, is minus two. Now the inverse of a he&#x27;s given by one over the terminal today that multiplies the matrix obtained from a bye, swapping the elements on the main diagonal so for three and negating the elements on the second of diagonal so minus two here and minus seven. And we have completed the determinant already. So the days in verse is given by 47 minus two minus two one seven ofs and minus three l&#x27;s

Angelo Rendina · Answer

we have the metric say given by three minus four seven minus eight. Since we have to complete its inverse, we start with computing is determinant. So the determinant of a he&#x27;s given by the product off the elements on the main diagonal minus the product of the elements on the secondary diagonal. So on the main diner we have three times minus eight. Pay attention to the signs minus and on the second there that are gonna We have seven times minus for and it is minus 24 minus minus 28 which is blast for now. The inverse of a is given by one over its determinant that multiplies the metrics off then from a by swapping the elements on the men diagonal and negating the elements on the secondary diagonal. So the minus four becomes blast for and the seven becomes minus seven. Now we have already computed the determinant of A, which is four. So we can write. The final result as minus eight divided by four is minus two 45 before is one. Then we have minus seven over four and three over four

Problem

Find the inverses of the matrices in Exercises $2…

Problem 31 Hard Difficulty

Find the inverses of the matrices in Exercises $29-32,$ if they exist. Use the algorithm introduced in this section.
$$
\left[\begin{array}{rrr}{1} & {0} & {-2} \\ {-3} & {1} & {4} \\ {2} & {-3} & {4}\end{array}\right]
$$

Answer

$A ^ { - 1 } = \left[ \begin{array} { c c c } { 8 } & { 3 } & { 1 } \\ { 10 } & { 4 } & { 1 } \\ { 7 / 2 } & { 3 / 2 } & { 1 / 2 } \end{array} \right]$

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

Linear Algebra and Its Applications

Grace H.

Catherine R.

Heather Z.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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$A ^ { - 1 } = \left[ \begin{array} { c c c } { 8 } & { 3 } & { 1 } \\ { 10 } & { 4 } & { 1 } \\ { 7 / 2 } & { 3 / 2 } & { 1 / 2 } \end{array} \right]$

Linear Algebra and Its Applications

Grace H.

Catherine R.

Heather Z.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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

Grace H.

Catherine R.

Heather Z.

Michael J.

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

Linear Algebra and Its Applications