find-the-inverses-of-the-matrices-in-exercises-1-4-leftbeginarrayll3-4-7-8endarrayright

Question

Find the inverses of the matrices in Exercises $1-4$ 
$$
\left[\begin{array}{ll}{3} & {-4} \ {7} & {-8}\end{array}ight]
$$

Angelo Rendina · Accepted 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

Michael Jacobsen · Answer

in this video, we&#x27;re going to be calculating the inverse of the matrix A that we see displayed here. The first step in our computation is to find the determinant of this matrix a offer the determinant of a we look to the main diagonal. It&#x27;s entries are eight and four, so we first take eight times for then we subtract the product of the off diagonal. The off diagonal is five and six. So right minus five times six then the result is going to be 32 minus 30. So we find that the determined of this matrix A is to now we&#x27;re ready to state the inverse. We&#x27;re going to use the formula since we have a two by two matrix displayed here that the inverse of this matrix A is equal to one over the determined of a provided the determinant is not zero. We have a value of two here multiplying a particular matrix. The way we find the Matrix is we look to the main Dagnall Britain and yellow and we interchange its values. So four goes in this position and eight goes in this position. Next, look to the off diagonal which contains elements five and six. They&#x27;re going to stay put. But we take negative one times those elements. So negative five goes here and negative six goes here. So this becomes altogether 1/2, which is the determined of A and the Matrix. We multiply that by has entries for negative five and negative 68 So now we see that the inverse is equal to two negative five halves for the first column, then negative three and four for the second column. So this is how to calculate the inverse. But keep in mind that this information here is very important. It only works as a formula if this is a two by two matrix.

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

Siddharth Shenoy · Answer

blue. It&#x27;s ascension Oi! And today will be talking about how to talk like the inverse of a major and the others may trick is the key, sweet, key way to solve systems of linear equations. So let&#x27;s begin. So what exactly is an emperor&#x27;s? So, for example, let&#x27;s take number five. The inverse of five is actually one over. If you multiply, this time&#x27;s the inverse, you get one. Similarly, you do the same train, same thing in major cities. For example, if you have a matrix day and written like this and then you take the inverse of it written this&#x27;ll CE and then you would actually get a one is not actually want to be a couple there in boats, it&#x27;s actually called and identity matrix. And then what that looks like is a stream of one&#x27;s in a diagnosed me and our like. So if it was that if both a cz if but the matrices here are two by two matrix, then you get a string of dangle ones in as the answer. Now you can&#x27;t you can&#x27;t actually divide maitresse, ese. So that&#x27;s why you take the inverse of a matrix. So let&#x27;s say for example, a Matrix A is written as so where you have these elements a be seen and deed t actually take the inverse of this. There&#x27;s a simple formula and that iss one divided by the determinant of a times. And what you&#x27;re going to be doing is you&#x27;re me swamping A and D So that&#x27;s d comes here a is here and then you&#x27;re you&#x27;re going to then take the opposite of being So the opposite of being was here and then the opposite of C. And if you multiply this out, you will get the inverse of this. So, for example, um, Bruce, why didn&#x27;t really write up a Terminator is that&#x27;s actually the first thing that you want to look at is if this determine it is, it&#x27;s a zero, Then there is no empress. Sometimes you get that. And so the very first test you want to figure out is what is the term that of a And so the formula. There is just simply a times d minus. Be time, See? Okay. So let&#x27;s say, for example, the matrix is if a is equal to eight five, minus seven and minus five first, let&#x27;s figure out what the determinant is or this matrix and see whether we can actually get in first. And that is going to be eight times minus five. My minus by times minus seven. So this is going to be minus 40 and this is going to be plus five and then so are determined. It is to be minus five now to get to get the inverse. Well, since this is not zero, we can evacuate the inverse of this matrix on DDE. And to do that, that would be the universe of a won over minus five times use want positions of A and E. So this will be minus five and eight, and then you take the opposite of five, which is minus five. And in the opposite it&#x27;s C, which is a husband. And so when you multiply this out off he multiplied. This fraction were to hear what you&#x27;ll get is one and one, and this will be minus 7 50 and that ISS minus 1.4 and thens the last the year the multiplying minus 1/5 to 8 and get minus one point stinks. So this is our in bursts of this this matrix and to double check your work. Whether you actually get the identity, you could actually do that. ISS. And that&#x27;s simply a 51715 The a times thean verse is what we got 11 minus 1.4 on But it&#x27;s blend boy six and this equals to multiply matrices Oh, you want Thio Initially Figure out what&#x27;s the size of you matrix and it&#x27;s very simple. So this is this is a two by two matrix and this is also two by two matrix. And what happens is that the middle? It really don&#x27;t really middle to look at the outside. So I&#x27;m gonna expect a two by two. Uh, matrix has a solution then, um, to multiply these matrices in springs you. So check the road the first road and you multiply by the first comma and then that becomes the first element of the solution matrix. And so that&#x27;s simply a times one hate. Then you add the signalman five times minus 1.4. Then we go to the next element. The next element is simply first room and in the second call, and that is eight times one, which is eight plus five times minus 1 to 6. Ah, third element here is minus seven times one itches when a seven plus minus, five times minus four. So the minuses skacel so used to have lost value. So the tiny five times on four. And then the last element is this second row times the second column. And so that&#x27;s going down seven points. One is when it&#x27;s seven minus phi times minus 16 which is a pasta five times 1.6. And so you multiply this out, this&#x27;ll becomes eight minus seven. And this is my seventh plus seven. And this is eight minus eight points. Eight. Really? This is my seven plus eight, and in fact, you it&#x27;s a practice and add everything. It becomes 15 months in an zero. So, in fact, we did, um, get the This is our solution, as are in verse for this matrix. Thank you for watching

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

Michael Jacobsen · 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.

Problem

Use the inverse found in Exercise 1 to solve the …

Problem 4 Easy Difficulty

Find the inverses of the matrices in Exercises $1-4$
$$
\left[\begin{array}{ll}{3} & {-4} \\ {7} & {-8}\end{array}\right]
$$

Answer

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

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

Linear Algebra and Its Applications

Grace H.

Heather Z.

Kayleah T.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

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

Linear Algebra and Its Applications

Grace H.

Heather Z.

Kayleah T.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

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

Grace H.

Heather Z.

Kayleah T.

Kristen K.

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

Linear Algebra and Its Applications