Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Identity matrix I = a square matrix C where the main diagonal is 1 and everything else is 0. The identity matrix says that a matrix times I is going to equal the original matrix or the identity matrix times A matrix is going to equal the original. So it doesn't matter which side the I gets multiplied on. Remember, in matrices we don't automatically have the commutative property of multiplication. The square matrix A is called invertible if there exists matrix B such that AB equal BA, which equals IB is the inverse matrix of A. If A is invertible, then there exists precisely one matrix B such that AB equal BA equal I. Hence it's a unique matrix, there's only one that exists. The formula to find the inverse matrix for a two by two square matrix C is that you're going to take and subtract the diagonals, the main diagonal going upper left to lower right. So AD minus BC one over that, we're going to interchange the A and the D locations, and we're going to take the negatives of the B and the C. Now this is only if it exists. So it's important for existence that the AD minus BC doesn't equal 0. And it's also important that none of the rows when we get done equals 0 all the way across some of our properties for matrices. A matrix to the 0 power is just the identity matrix. A to the first power is going to be a, A matrix to the n + 1 power we could think of as the matrix to the N times the matrix. This is true offense greater than or equal to 1A to the negative N is going to be A to the -1 all to the NA to the RA to the SA to the R + S and AA to the R to the SARS. So A to the R * A to the S is A to the R + S. If A&B are the same size and invertible, then a inverse is invertible and a inverse inverse equals the original matrix A. Also, if N is a non negative integer, then A to the north is invertible and AN to the -1 equal A to the -1 N The product AB is invertible and AB to the -1 equal B to the -1, A to the -1. If we think about this one for just a moment and we look at the fact that we're going to have AB being the original, if we had AB times B inverse, A inverse, we can regroup this because we do have the associative property. And BB inverse is really just going to give us the identity. And then the identity with the A or the identity with the A inverse is just going to give the actual functions back, the matrices back. And then a A inverse is just I. If we did this on the other side where we started with our B inverse, a inverse, and then our head, our AB using the same concepts, we're going to also get out the identity matrix. So what this does is this really tells us that the the product of AB has is invertible and that it exists and it is B inverse A inverse. So if I have AB, the invertible is writing it in backwards notation, IE the B 1st and then the A and then to the negative ones. So if the N by N matrix A is invertible, then for any N vector B the system AX equal B has the unique solution X equal a inverse B. If we look at a matrix A * a row column no times a column vector B, we would get. We'd be able to find the inverse. So a inverse is going to equal 1 / 4 * 6 - 3 * 7. We're going to interchange the six and the four, and we're going to make the other two locations turn negative. So when we look at this, we're going to end up with 24 -, 21, so 1/3 six -3 negative 74. If I multiply that 1/3 through, I'm going to get 1/3 of 6/2 negative 7 thirds -1 and 4 thirds. So that's my invertible matrix, and it's going to be a unique matrix. Now we're going to use that concept we just had that says a * X equal B. So I should be able to find X equaling a inverse of B. So if we thought about taking two -7 thirds -1 four third, and we multiply that by 4736, and that's all multiplied by the matrix X. And then if we come over on the right side and we have two -7 thirds -1 Four Thirds, and we multiply that by 10 five, what happens is that the left side now turns into the identity matrix times X and the right side 2 * 1020 plus -7 thirds times 5. Just doing multiplication of our matrices -1 * 10 + 4 thirds times 5. I'm going across the row and down the column for multiplication. If we do this and add them, we can see that our matrix X is really just 20 and -35 thirds, so 25 thirds and -10 and +20 thirds so -10 thirds. So our solution to the original matrix, when we wanted this matrix times our solution matrix of X equaling B would be -25 thirds and -10 thirds. Frequently written as an ordered pair, our X1, our X2. Now a 2 by two matrix is fairly easy. 3 by threes get harder. One method, and there's quite a few, but one method to find the identity matrix first, not the identity to find the A inverse. The inverse matrix is to start by putting the matrix that you're given and then put a square identity matrix at the end. And then what we're going to do is we're going to manipulate the rows until we get the identity matrix up here in the front. And then whatever is left in the back end is going to be the inverse matrix. So when we look at this problem, I'm going to 101 in the first location, first row, first column, but I need zeros to be in these other locations. So I'm going to take row 1 and multiply it by -2 and add it to row 2. And then I'm going to take row 1 and multiply it by -3 and add it to row 3. So I'm going to leave that first row alone for a minute. I take row 2 and row 1 and multiply by -2 and add. We're going to get zero -6 and 8, two -4 and three is -1 negative 210. If I do the same thing to row 1 * -3 I'm going to get 01 -3 zero negative 301. Now I need a one in this middle location because my first column is good. Now my second column I'm going to need a one in the middle and then zeros on the other end. So one method is I'm just going to exchange row 2 and row 3's location. If I exchange row 2 and row 3, then I do have a one where I need it. Then what we're going to do is we're going to take row 2 * -3 and add it to row one and row 2 * -2 and add it to row 3. So we're going to leave the middle row alone and we're going to multiply and add it to the others. So if I multiply and add, I'm going to get 100 times -3 is 0 + 2 nine plus one is 10 zero -3, and then we're going to get 00 negative 141 negative 2. Now this is almost complete. We have a -1 in my original or in my diagonal that I need. So we're going to take row 3 and multiply it by -1 all the way through. So we get 102100, negative 3010, negative 301001, negative 4, negative 1/2. So finally I need to get this two in the first row, third column to go to 0. So we're going to take row 3, multiply it by -2 and add it to row one. So 100182 negative 7010, negative 301, 001 negative 4, negative 1/2. So the inverse matrix of our original is this piece that's down here at the end. So our A inverse is 18 two -7 negative 301 -4 negative 1/2. We could check that by multiplying the original that 132, 283, 310 six times our inverse. If we've done it all right, we should be getting out the identity matrix. I will leave that for you to check now, because we do not have the commutative property for matrices. We'd also check with the inverse coming 1st and multiplying it through by the original matrices. Once again, I'm going to leave that for you to check. So our last problem is giving us an original matrix and giving us a matrix we want it to equal and we want to find the solution matrix. So we have a times our X our solution matrix equaling the given matrix. I'm going to find the inverse of AI. Chose it two by two just to show it quickly. But you can find the inverse of any other matrices with the procedure I just showed. So our diagonal 7 * 7 - 8 * 6 one over, we're going to interchange the sevens and we're going to take the negative of the 8 and the negative and the six. So seven -6 negative 87. If I multiply that on the left side, I'm going to also multiply it on the right side. Now order is important and I'm putting it in the left of each of them. In part. I have to do that because if this B matrices is a two by three and my original square matrix is a two by two, I've got to multiply the inverse in front so that the columns of the first one match the rows of the second. I know my solution matrix is going to be a 2 by 3, so when I multiply on the left side, I'm going to get the identity matrix. If I multiply over here, we're going to get 14 negative 3046, negative 1635 and -53 Remember we multiply across the row while we're going down a column. So 2 * 7 + -6 * 0, then 7 * 0 + -6 * 5. That's how I'm getting these numbers here. So this X is my solution matrix. Thank you and have a wonderful day.