In Exercises 37 - 42,
a. Write each linear system as a matrix equ...

Question

In Exercises 37 - 42, 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. w - x + 2y = - 3 x - y + z = 4 - w + x - y + 2z = 2 - x + y - 2z = - 4 The inverse of is

Accepted Answer

express the following system of equations. As A X equals B. Where A is the coefficient matrix and B is a constant matrix. Then solve the system is the given in verse of the coefficient matrix. So let's go ahead and write up our matrix first. We want A. X equals B. First. Now the A matrix is going to be every coefficient in our system of equations. So the A let's say we have negative two first 12 negative one. Now we have no W on the second row. So zero negative one negative one and two negative three negative four negative 110121. We noticed that matrix the same as this one up here. Now we're gonna find our X. Matrix. Our X matrix is just our list of variables. We're going from W two Z. So we have W X, Y and Z. Let's sell this is equal to r B matrix. R B matrix is just are constants. So everything on the right side of the equal sign -5, -4 negative three. So that's our eggs equals B. Matrix. Let's go ahead and find what the system of equations will be. Or the solution. Rather we want to make use of the n groups. So we want to find X equals, we can make use of inverse matrix A one times B. Where B is our constant matrix A one is our interest. So we actually were actually given the inverse of our matrix, hey that is given by this right here. So we can rewrite this now S X. Which is Are negative 8/ -7/13. 1/13. By over 13 15. Number 26 nine. Number -5/13 -11. Number 26 -9. Number 26. Number 13 three. Number 13 17. Number 3/26 7/13 negative 1/ 3/26. That is our matrix a inverse. This is multiplied by our constant matrix negative 53 negative four negative three. Now we want to multiply these two matrices together and evaluate so I can almost pipe. We'll take the first index of our first row and matrix a inverse and multiply it by the first index of our column and be so. Other words negative 8/13 Times are -5. Next one will be negative 7/13 times three -1/13 I say plus 1 13 times negative four. Alright the plus first And plus five or 13 times negative three. That all goes on the top rope. Now it's the same for the next trips. Our next 1. 15/26 by negative five Plus 9/13 times three -5 or 13 times -11/26 times negative three. Next one look it -9/26 times negative five -8/13 times three Plus 3/13 times negative four plus 17/26 times negative three. Finally we have one more row 3/26. Multiplied by 9-5 Plus 7/13 times three -1/13 times. I needed four plus 3/26 times negative three. Now open to simplify everything in our matrix X equals it's half one. 40/3 minus 21. I'm sorry 40 for 13 I should say that's 40 for three but 40 for -21/13 minus 4. 13 -15/13. And we have negative 75/ plus 27. 13 Plus 20/ Plus 33/26. Third row 45- managed 24/13 -12/ -51/26. Finally Look it negative 15/ plus 21/ Plus four for 13 -9/26. You simplify all of that On the top. We will get zero 2nd 1 we get 52 or 26. Third one we get native 78/26. And the fourth one we get 26/26 which we can simplify even further. 20 to -3 and one Now we send our X. Matrix was W. X, Y. And Z. In that order. That means our first number zero is W second number is X. Two. Third number Y negative three. Fourth number Z is one this is the solution to our Matrix. This makes the answer to our problem. We noticed 02 -3 and one answer. D. Okay. I hope to help you solve the problem. Thank you for watching. Goodbye.

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In Exercises 37 - 42, 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. w - x + 2y = - 3 x - y + z = 4 - w + x - y + 2z = 2 - x + y - 2z = - 4 The inverse of is