In Exercises 13 - 18, use the fact that if
a b d - b
A = then A^(...

Question

In Exercises 13 - 18, use the fact that if a b d - b A = then A^(-1) = 1/(ad-bc) to find the inverse of c d - c a each matrix, if possible. Check that AA^(-1) = I_2 and A^(-1)A = I_2. 2 3 A = - 1 2

Accepted Answer

Hey everyone in this problem we are asked to find the multiplication inverse of the matrix given a K. By using the formula for a inverse that were given. We were also asked to verify that a inverse is equal to I two and a inverse A is equal to I two? Where I two is a multiplication identity matrix of order to Okay, so our matrix A. Has first column negative four negative three. And second column 61. We're given this formula for a inverse or A A sorry for a inverse. Where A in verses one over a d minus bc times of matrix with first column d minus C. And second column minus B. A. Okay, so the first thing we want to do is take our matrix A and identify what this little a little b. Little C little D R. Okay. And in our case little A is going to be negative for Little B is going to be six. Little C is gonna be negative three and little D. Is going to be one. Okay, so we work across the first row A. B. And then down to the second row C. D. Okay, so using these this means that a inverse is going to be one over a negative four times D. One minus B. Six times C negative three times the matrix with first column one, three and 2nd column -6 -4. If we work this out we get a inverse is equal to one over we have negative four plus 18. So we get 1/14 times that matrix. First column 13. 2nd column negative six, negative four. Moving that factor of 1/14 inside of our matrix. Okay, when we have a constant multiplied outside of our matrix, it applies to every single term. When we bring it inside. So we get 1/14. 3/14, -3/7. And -2/7. Okay, Alright, so this is the inverse matrix we've found and go ahead and highlight it. So we don't lose track of it. And now what we're asked to do is verify a inverse and a inverse A are equal to the identity matrix. So let's start with a inverse. Okay, a inverse. This is going to be the matrix. A negative four negative three in the first column 61 in the second column times a inverse of matrix we just found above. Okay, now when we're multiplying matrices, this isn't a term by term. Operation like adding and subtracting. Okay, when we multiply them, we take a role of the first matrix multiply it to a column of the second matrix. Okay, and put that in the corresponding row and column of our new matrix. Okay, so what does this look like? So for a inverse let's start with the first row of a the first column of a adverse. Okay, we're gonna take We're going to be in row one column one of our resulting matrix. Because that's the row and column we're choosing to take the first element in the row negative four multiply it by the first element in the column. 1/14. Okay then we add we move across the road and down the call. Next element six times the next element of the column 3/14. So that's what we do for the first real first column. Moving down to the second row, let's do the second row. And we're gonna stick to the first column. The first element in the row negative three times the first element in the column. 1/14. Okay then we add next element in the row, one times next element in the column 3/14. All right, we're gonna go back to the first row. We're gonna move across to the second column. We're gonna be working with the first column of the first matrix. Or sorry, the first row of the first matrix. The second column of the second matrix. This is going to go in row one column two of our resulting matrix. First element of the row is negative four. Let me make that negative more clear, negative four times the first element in the column negative 3/7. And then we add we work across the row. The next element is six and down the column, the next element is negative 2/7. Okey. Alright, we got three terms done. Now, we just need to do the last one. The last column in the last row we get negative three times negative 3/7 plus one times negative. 2/7. Okay, great. Now if we work this out, we're done the hard part. We're done the multiplication. We just need to do some arithmetic. Okay, so here we're going to have negative four over 14 Plus 18/14. This is going to give us 14/14 which is one. Mhm Second row, first column we have negative 3/14 plus 3/14. That's gonna give us zero. Okay, first row, second column we're going to get 12/ -12/7. That's going to be zero. Last row. Last column we're gonna have 9/7 minus 2/7. This is going to be 7/7 which is one. Okay, and so for a inverse we do in fact get the identity matrix multiplication, identity matrix of order to just like we wanted. Okay, so that's a big happy face. We got exactly what we were looking for when we did our check of a inverse. Okay, now let's do the check for a inverse. A We're gonna do the exact same thing. A inverse. A Okay, now remind you, a inverse is 1/14, 3/14 in the first row or sorry? The first column and the second column is negative 3/ Times -2/ times a A has first column negative four negative three and second column 61. Okay, Alright, Now this is going to be the exact same process we did for the other matrix multiplication. Okay, so we take the first row times the first column we get 1/14 times negative four. Okay? And then we add across the road negative three. Sevens. Down the column negative three. Moving on to the next row, same column 3/14 times negative four plus negative 2/7 times negative three. Okay. Moving to the second column back to the first row, 1/14 times six plus negative 3/7 times one. And second column. Second row 3/14 times six Plus -2/7 times one. Alright, so again, the hard part's done, The multiplication is done. Now we just need to do some adding subtracting and normal multiplying. Okay. We are going to get negative 4/14 plus 9/7 which is 18/14. Okay, 14/14. That's gonna give us one. Second row. First column we're going to get negative 12/ plus 6/7 which is 12 over to 14. Okay, this is going to be zero. Top right, we get 6/14 minus 3/7 which is minus 6/14, that's zero. And the bottom right, We get 18/14 minus 2/7 which is 4/18 K. We get our sorry for over 14. We get 14/14 which is going to be one. Okay. And so for this one a inverse A. We also get the identity matrix multiple identity matrix order to which is what we wanted to double check that this is indeed the inverse of a. Okay, so happy face there to that check worked out. And if we go back up the answers for this question, we double check that the matrix we found was indeed the identity. And so the answer to this one is that the identity matrix? Or sorry? The inverse matrix we found. Sorry, not the identity. The inverse matrix we found has first column 1/14. 3/14. 2nd column negative 3/7 2 7th. That's answer. C Thanks everyone for watching. I hope this video helped see you in the next one.

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In Exercises 13 - 18, use the fact that if a b d - b A = then A^(-1) = 1/(ad-bc) to find the inverse of c d - c a each matrix, if possible. Check that AA^(-1) = I_2 and A^(-1)A = I_2. 2 3 A = - 1 2