In Exercises 43–44, (a) Write each linear system as a matrix equa...

Question

In Exercises 43–44, (a) Write each linear system as a matrix equation in the form AX = B (b) Solve the system using the inverse that is given for the coefficient matrix.

Accepted Answer

Welcome back. I'm so glad you're here. We are given the inverse of our coefficient matrix. This is really exciting because it's a lot of work to find the inverse. So that part's done for us. Great. And it's also exciting because we recall from previous lessons when we when we have matrices A times X. So matrix A times matrix X. And it equals B. We know that X equals the inverse of matrix A times B and just zooming out for a second. Okay, if this were algebra, yeah, we could get X by itself by dividing both sides by A. But you can't divide matrices, you can take matrix A and multiply it by its inverse. They would cancel cancel it on the left. And that's why we have an inverse on the right hand side being multiplied by our matrix B. Because what you do to one side, you've got to do the other and that's going to solve for x. Okay, now, what am I talking about with A and X. And be Well, if you look at the system of equations here, thinking about the left hand side, this is a matrix A times a matrix X. What do I mean by this? Okay, if matrix A is the coefficient matrix here, then matrix X would be are variables X, Y and Z. And if you were to do the work and multiply those two matrices, it would look like the left hand side of our system of equations over here. So that is matrices A times X. And they equal baby matrix here. Matrix B. The answer is 12 and three. That's one column. And were given again the inverse of matrix A. Right here with this matrix starting with negative three halves in the corner. And now we're just going to do some old fashioned multiplication of matrices, multiplying the inverse of A, times B. And that will help solve for X. So let's copy these things down the inverse of matrix A. Starting with the first row, we've got negative three halves one half and negative one half. Second row, five halves negative one half and three halves. Third row negative one half, one half and negative one half. Great. So A the inverse of A is being multiplied by B. Well that Is 1, 2, 3, going down that column. So we're multiplying the inverse. And now we'll just do the multiplication, I will write out some of the steps. So we work across the first row and go down the column and add those results together. So the first row, we start with negative three halves times the first column, one plus moving across the first row, one half times going down the column two plus Going across the first row, negative one half, going down the column three. And because we only have one column that we're dealing with, that's as much as it's going in the first row. Now we look at the second row, we've got five halves times starting at the top of the column again, one Plus. working across the for the second row, negative one half times going down the column two plus and moving across the second row three halves times going down the column three, that's the end of that second row. And now going across the third row, we've got negative one half times, starting at the top of the column again one plus, moving across the third row, we've got one half times going down that column two plus and then moving to the end of the third round negative one half times going to the end of the column three. Great now let's do that, multiplication negative three half times one is negative three halves plus two times one half. Well I'm going to write two halves because then that will give us common denominators as we're adding these. And then we've got negative one half times three is negative three halves. We've got five halves, times one is five halves negative one half times two is negative two halves and three halves, times three is nine halves. Now the third row negative one half times one is negative one half, one half times two is two halves and negative one half times three is negative three halves. Now let's do our addition negative three. I'm just going across the numerator is negative three plus two is negative one minus three, brings us back down to negative four halves, five minus two is three plus nine, gets us up to 12 halves and negative one plus two. That's one minus three, gives us negative two halves, simplifying into whole numbers negative four houses, negative 2, 12 houses. Six negative two halves is negative one. We look at our answer choices and that matches with answer choice C. Well done, we'll catch you on the next one.

In Exercises 43–44, (a) Write each linear system as a matrix equation in the form AX = B (b) Solve the system using the inverse that is given for the coefficient matrix.

Key Concepts

Matrix Representation of Linear Systems

Matrix Inverse

Solving Linear Systems

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