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. 2x + 6y + 6z = 8 2x + 7y + 6z = 10 2x + 7y + 7z = 9 The inverse of is

Linear system equations and inverse matrix for Exercise 37 in college algebra.

Accepted Answer

express them following system of equations as A X equals B. Where a is the coefficient matrix B is the constant matrix installed. The system used to give an inverse of the coefficient agents. Now let's write a. X equals B. First A is our matrix from the coefficients of our linear equations. So if you look here our first one is three. Second one is to third is to you notice it's actually the same matrix as this one. Up here 2 1, 1 -5 -2, 2 multiplying now by matrix X. This is just our variable matrix or X. Y and Z. Finally we're equaling they should speed, which is just our Constance 5 2 and one now define X. We want to take a m first time speed. It actually gives us a inverse up here as this matrix. So we're gonna use that and right are X equals so X equals the a inverse matrix negative 1-0 9, -4, 1 4th negative And -1 4th 1, 1 4th that multiplies matrix P which is R to 1. Now we're gonna multiply these two matrices together matrices rather we take the first index or first row multiplied by the first index of our column and actually speak Other words. Our 1st 1 is -1 Times five. Our next one is plus two times to our last one is plus zero times one. The middle will be 9 4th first now Multiplied by 5 -4 times two -1, 4th times one And our final is negative 1 4th times five Plus one times Plus 1 4th times one. And we can simplify this now on the top, negative five plus four And then plus zero middle 45/4 minus eight minus 1/4. Bottom negative 5/4 plus two plus 1/4. Simplify again. We get -1 on top or four in the middle and one on the butt, Which we get -131. As our matrix. Now earlier we said matrix X. Is X. Y. C. So Our fresh label X is our first number in our matrix negative one. Why would be the three And Z is the one. This is the solution to our problem which makes the answer to our problem, answers seek. Okay. I hope so. If you solve the problem, thank you for watching. Goodbye.

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. 2x + 6y + 6z = 8 2x + 7y + 6z = 10 2x + 7y + 7z = 9 The inverse of is

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