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. x - y + z = 8 2y - z = - 7 2x + 3y = 1 The inverse of is

Linear system of equations and its coefficient matrix inverse for college algebra exercise.

Accepted Answer

express the following system of equations as a X equals B. Where a is the coefficient matrix and B is the consummation. Then solve the system used to give an inverse the coefficient matrix, let's first write up our X equals B. C. A X equals B. Matrix A. It's just the coefficient matrix of our system of equations. So we're gonna take each coefficient and put them into this matrix. The first one we have X. The first variable is one negative two for the Y. There's no Z. So it's zero. Next row we have three X. Three for a. Y negative two for C. Last 10 for X. There's no X. Six for y negative two for C. And our order variables was X. Y and Z. 91 R. B matrix which is our Constant Matrix. That is just everything on the right side of the evil side to 5 -4. Now we want to solve for matrix X. This is given by the equation, X equals a inverse times B. You'll regret this now, as X equals R, a, inverse is given to us up here right here negative 12 3rd negative two thirds negative 11 3rd negative one thirds and negative negative three halves. This is multiplied By R.B. Matrix to 5 -4. Now we can do our multiplication. So you'll have the matrix. Now we're multiplying. So we're gonna take the first index of our first row in matrix A inverse multiply it by the first index. The first column matrix B. That is -1 times two. Our next one Plus 2/3 multiplied by five And -2/3 multiplied by -4. Now to our 2nd row negative one times 2 Plus 1/3 times five -1/3 times negative four. Our last row negative three times 2 Plus one times 5 minus three halves times negative four. Now we're gonna simplify our matrix. You were gets negative two plus 13th Plus 8/3 on the top. Middle negative two plus five thirds plus four threats On the bottom. Get negative six plus five And then plus six. Now we can simplify it even further. 75. We should get 12/3. The middle You get 3/3. The bottom you'll just get back Simple five. Once more we get 415. Now we said earlier our X. Matrix is the order X, Y. Z. our first number because in the X which is four second number, why is one third number? Z is five. This is the answer to our problem which means the answer to our problem is answer. A. Okay, I'll help 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. x - y + z = 8 2y - z = - 7 2x + 3y = 1 The inverse of is

Watch next