In Exercises 39–42, find A^(-1) Check that AA^-1 = I and A^(-1)A ...

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In Exercises 39–42, find A^(-1) Check that AA^-1 = I and A^(-1)A = I <Image>

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Welcome back. I'm so glad you're here. We are given a matrix A equals first row 12, 2nd row three negative one. And we are to find the inverse of this matrix and then we're going to verify our answer. So while we've got a two by two matrix, two rows to columns. And yeah, this is exciting because we recall from previous lessons that there is a formula for finding the inverse of a two by two matrix. So first let's identify what what some of these numbers correspond to in our given matrix. So in the upper left hand corner, that's our A value. And then going across the road. That's our B value. The second row is C. And D. And this will make sense in a moment because our formula for the inverse of a two by two matrix is one divided by a times D minus B times C. And it's really important to note here that A times D minus B times C cannot equal zero because that would be undefined. And we're going to take that scalar value that we determine there and multiply it by a new matrix which will have D. In the first position and then a negative be starting in the second row, negative C. And scooch the A down here. So that's the formula for our inverse are a inverse. And now let's just start filling things in. Alright, A times B. So A. Is one times D. Excuse me is negative one minus A B. Which is two times C which is three one times negative one is negative, one negative two times three is negative six. Okay this is negative seven down here in the denominator it's not zero which means we have the green light. We can go ahead and calculate our a inverse. So this will be one. Well we'll bring that negative up so negative 1/7 instead of one over negative seven. So we've got negative 1/7 times. Let's fill in our D. Will be negative one negative B will be negative too negative C. Will be negative three and A. Will be one. Now we need to distribute that scalar we can distribute that negative 1/7 so negative 1/7 times negative one Is positive 1 7th negative 1/ times negative two equals positive 2/ negative 1/7 times negative three equals positive 3/7 and negative 1/7 times one equals negative 1/7 and that should be our a inverse. But we are going to check it. We're going to take our original A. And multiply that by R. A inverse and see if it equals I. Well once I i is the multiplication identity matrix. That's the one where we've got diagonal of one starting in the upper left hand corner and zeros filling in around them. So if we get that answer we have found the correct a inverse. So now let's take our original A. Which is 123 negative one and multiply that times are a inverse which is 1/7 2 7th 3/ and negative 1/7 to do this multiplication. We are going to take the first row of our A label these a an a inverse. So we'll take the first row over A and multiply that by the first column of our a inverse show you what that looks like. So one from the first row of a times 1/7 from the first column of a inverse plus two from the first row of a times 3/ from the first column of a inverse. Then we leave a little bit of space and we're going to repeat row one from a. Now we're going to go with the second column of a inverse. So this will be one from the first row of a times 2/7 from the second column of a inverse plus two from the first row of a times negative 1/7 From that 2nd column of a inverse. And now we'll proceed with the second row of a and multiply it again by the first column of a inverse. And then the second column of a inverse. So three from the second row of a times 1/7 from the first column of a inverse plus negative one from the second row of a times 3/7 from the first column of a inverse. Leave a little space three again from the second row of a, times 2/7 from the second column of a inverse plus negative one from that. Second row of a times negative. 1/7 from the second column of a inverse. And if you've done this properly you should see things lining up in these parentheses. They should be the same. So now let's go ahead and do that multiplication. So one times 1/7 is 1/7 plus two times three. Seventh is 6/7 1 times 2/7 is 2/7 plus two times negative. 1/7 is negative. 2/7. The second row three times 1/7 is 3/7 plus negative. One times 3/7 is negative 3/7 three times 2/7 is 6/ plus negative one times negative 1/7 is 1/7. Let's do that addition. We've got 1/7 plus 6/7. Will that 7/7 or one 2/7 plus negative two sevens. That's zero 3/7 plus negative 3/7 is zero and 6/7 plus 1/7 is 7/ or one. Hey look at that. That is our multiplication identity matrix. And that means that we have correctly found are a inverse that matches with answer choice C. Well done. We'll catch you on the next one.

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