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Let $A=\left[\begin{array}{ll}{1} & {2} \\ {5} & {12}\end{array}\right], \mathbf{b}_{1}=\left[\begin{array}{r}{-1} \\ {3}\end{array}\right], \mathbf{b}_{2}=\left[\begin{array}{r}{1} \\ {-5}\end{array}\right], \mathbf{b}_{3}=\left[\begin{array}{l}{2} \\ {6}\end{array}\right]$ and $\mathbf{b}_{4}=\left[\begin{array}{l}{3} \\ {5}\end{array}\right]$ a. Find $A^{-1},$ and use it to solve the four equations $A \mathbf{x}=\mathbf{b}_{1}$ $A \mathbf{x}=\mathbf{b}_{2}, A \mathbf{x}=\mathbf{b}_{3}, A \mathbf{x}=\mathbf{b}_{4}$ b. The four equations in part (a) can be solved by the same set of row operations, since the coefficient matrix is the same in each case. Solve the four equations in part (a) by row reducing the augmented matrix $\left[A \text { b, } \mathbf{b}_{2} \mathbf{b}_{3} \mathbf{b}_{4}\right] .$

See work.

Algebra

Chapter 2

Matrix Algebra

Section 2

The Inverse of a Matrix

Introduction to Matrices

McMaster University

Baylor University

University of Michigan - Ann Arbor

Lectures

01:32

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in this example, we have been provided with a matrix and four different vectors. The first thing that we'd like to do is calculate the inverse of this matrix A. So, to that end, let's start by calculating the determined of A. We know that that the determinant of a is found by taking the product of the main diagonal first. So this will be one times 12 and then subtracting the product of the off diagonal, which is five and two. So we'll take one times 12 minus five times two. And this results in a determinant of 12 minus 10 which is equal to two. So now that we have the the determinant of this matrix, we can say that the inverse of a since its two by two is found by taking 1/2 times the family make tricks were we first go to the main diagonal with 1 12 and interchange the order. So we have 12 and one, then go to the off diagonal which is 52 and multiply the entries by negative one. This gives us a negative five and a negative to which goes here and ordinarily, what I'd like to do is say what this is equal to by scaling all entries by this factor of 1/2. But it turns out this time it's more convenient to leave the A inverse and this form with the scaler on the outside. Let's explain why. So our next step in this problem is to look at the matrix equation eight times X equals B one. We can solve this by augmenting into a two by three matrix and then roll reduction. But now that we have an inverse, we can also say that the solution is of the form X equals a inverse times. Be one now to solve for X, we'll put in a inverse, which is that half we mentioned multiplying 12 negative five negative 21 times be one which is negative one and three. So the reason we left the 1/2 on the outside is that it's easier to deal with this multiplication here when we don't produce fractions from here and here. So for the next step, we can say X is 1/2 of the result of this matrix times. This vector we have all together. The first entry is going to be equal to negative 18 and the second entry is positive. Eight. Would we multiply the Matrix and the vector? Now scale that result by 1/2 and we find that X is going to be equal to negative nine and four. So that solves the first equation that were interested in Let's go to the next interesting equation. Eight times X equals be to copying our method from before we know that X is going to be first a inverse, which is 1/2 times the matrix. 12 Negative five negative 21 then multiply by B two and B two is given here, so we multiply this by one negative five. So altogether X is going to be equal to 1/2 multiplying that matrix times a vector, which results in 22 negative 10. Then scale this vector by 1/2 and X is equal to 11 and negative five. Now let's go to the next matrix equation that we are interested in. It's a X equals. If you haven't guessed B three, then we'll also solve a X equals before next. First, when a X equals B three, we find that X is equal to 1/2 copping our strategy from before. Put in the Matrix 12 negative five, negative 21 And B three is equal to 26 So in this case week obtained 1/2 times the vector, which is going to be a 12 and negative four. So we have that X is equal to when we're looking at a X equals B three a six and negative to once we scale that last vector by 1/2. Let me move some of these over so we can consider this last equation with a little more space. So now that we have more space once again, we can solve this matrix equation by writing X equals a inverse which is 1/2 times the matrix 12 negative five negative 21 and multiply by the vector before, which is 35 In this case, we'll have X is equal to 1/2 times the vector, which will be 13 and negative five. When we multiply this matrix here with the vector, then scale by 1/2 and we see that X must be equal to I made a mistake here should be a 13 negative five here. And so in this vector let me delete some of these after multiplying the Matrix with the victor, it was 26 negative 10. Then Skilling by ah half gives us 13 and negative five. So these are our two. Or excuse me for matrix equations that we have solved here, here, here and here. The respective solutions are negative. 94 We'll have a negative 56 negative two and 13 negative five. And if you've subspace did that, we've worked awfully hard to come up with these solutions. You were mostly correct. If we want to solve all four of these equations simultaneously, we can in fact do that. So it's described the method. First, I'm going to go to a new slide that contains all the original information, and we'll see if we can produce these four different solutions simultaneously. Okay, so now that we're on a clicking pit clear page in order to solve a X equals B one or B I where B I is, either for I equal toe 123 or four. We can first take a and augment with be one be to B three and before let's say what this matrix is going to turn into. So It's a two by Let's see. 1234 to bike. Six matrix were first. I'll put in a so a is 15 2 12 Now this is the portion where we're augmenting in. We're augmenting with vector B one b two B three before which are indicated above. So we have a negative 13 Ah, one negative, five a to six and then a 35 So let's begin row reduction. This looks like it's a horrendous matrix to roll reduce, but really, it's not so bad when you have only two rows. So let's start off as follows. We have a pivot here, and we're going to eliminates the entry immediately below the five. So that means we need to copy Row one. It is 12 negative 11 23 and now multiply row one by negative five. At the results to Row two will obtain a zero to eight negative. 10 negative four and negative 10. Now the next thing we need is a one in the second pivot position in column two. So let's do the row operation were we divide the second row by two. I will say the last matrix Israel equivalent to first a copy row one one to negative 11 23 And now we're dividing row two by two. So it's 01 four negative. Five negative, too Negative. Five. Now we're in good position. This our first pivot with Israel below. And now our second pivot does not have a zero above. So we just need to wipe out to this entry, which is a two. So first, let's copy wrote to row two it 014 negative five negative two and negative five, then Row One, Our operation is going to be to take negative to 10 0 to added to row one, making that entry a zero. So we'll have a one a zero negative nine and 11 Ah, six and then a 13. So our matrix A, which was here, has now been rolled reduced into the two by two identity matrix. But recall we also augmented B one through before, and this is the result of those augmented vectors using these row operations. So now let's state what this means. So far, we have that eight times x one equals B one would be X equals negative 94 hope there did not need to be X one just x there. Then, if we go to this column, we know that eight Times X equals B two is now has a solution. X equals 11. Negative five. Now, if we go to this column well, it can say that eight times X equals B three. Has solution X equal six Negative, too. And finally, the last column tells us that the solution to eight Times X equals before is X equals 13 negative five. So using real reductions, we have solved all together for different systems of equations. And the below are the solutions to each one compared with the results that we had prior when we saw these individually, let me change to this light. We got the same results but involved a lot of matrix multiplication. Honestly, it's kind of a toss up to which the solution is easier. But if you have just two rows, the row operations made this method more fast that, in my opinion,

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