In Exercises 21–38, solve each system of equations using matrices...

Question

In Exercises 21–38, solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.

System of equations for exercise 33 in college algebra, chapter on matrices.

Accepted Answer

Welcome back. I am so glad you're here. We are given the system of equations and we're asked to solve it. We recall from previous lessons that when we have a system of equations, we can set up a corresponding matrix with its coefficients and constant terms and use that to solve it. So for our first row of our matrix, that's going to be that first equations, coefficients and constant terms. So that will be 21 negative 17. And the second equation will be the 2nd row 1, -1, 1. And the third equation, the third row one negative two negative 2, 16 antes Gaussian or Gauss Jordan elimination. I'm going to demonstrate Gauss Jordan elimination. We're going to convert this matrix into a matrix that has a diagonal row, well diagonal ones for the first three columns. So the first column will read 100. The second column will read 010 and the third column will read 001. And then our fourth column will be where we will have our x, y and Z solutions. So in order to get the matrix on the left looking like the matrix on the right. We're going to start off with getting this one in position in the first column. So we want to convert this to to a one. We're going to be replacing row one and to do that here. I'm going to take row one minus row two and have that replace row one because we're replacing row one, we can copy down Rose two and three. So that'll be 11 negative and one negative two negative 2, 16. Now taking row one minus row 22 minus one is the one that we want. One minus one is zero, negative one minus negative one is also zero and seven minus one is six. All right. Now, we need zeros in the rest of the positions in column one. So let's work on Roe two's one. We're going to replace road to in order to do that. We use the road that has the one in the desired position, which is row one. So we're going to take row one times a negative one And add that to row two. And because we're replacing road to we can copy over. Call Rose. Excuse me, Rose one and three. So that'll be 10061, negative two negative 2 16. And now doing the work to replace row two negative one times one is negative one plus one is the zero that we want negative one times 00 plus one is one negative one times zero. Still zero plus negative one is negative one negative six negative one times six is negative six plus one is negative five. Okay, now we need a zero in this third row of column one. So we'll replace row three will use the row that has the one in the desired position to do that, which is row one again, we'll multiply row one times a negative one and add that to row three. Because we're replacing row three, we can copy down rows one and negative. This one is just a one and negative one and negative five. And replacing row three negative one times one is negative one plus one is zero, negative one times zero is zero plus negative two is negative two both times and negative one times six is negative six plus is 10. Okay, now we work on column two and we want to have a one in the second row, second column and we already do ye And now we want a zero above that. Well there's already one there. Yeah, So now we just need to work on row threes, column two. We're going to replace row three again. We want to get a zero there. So we work with the road that has a one in the desired position, which is row two. And we're going to multiply row two times two And add that to row three. We can copy over rows one and 2100601, negative one, negative five. And now to replace row three, so two times zero is zero plus zero is still 02 times one is two plus negative two is the zero. That we want, two times negative one is negative two plus negative two is negative four and two times negative five is negative 10 plus 10 is zero. Alright, we're getting close. Now We want to get the one in the desired position of column three, which is row three. So again, we're working on row three and in order to do that, we are going to take row three and multiply it by the negative reciprocal of that number so that it will become a positive ones. We're going to multiply by negative 1/ negative 1/4 times row three. We can copy down rows one and -1 -5. And now doing the work to replace row 30 times negative 1/4. Well all three times is zero and negative four times negative 1/4 is the one that we want. Ye look at that Rose one and three are looking good and row two is almost there. We just need to get a zero into this position here in row two. And to do that will work with the row that has the one in the desired position which is row three. And this time we can just add row three plus, row two can copy over Rose one and 310060010. And now replacing road to we're adding ro plus 000 plus one is 11 plus negative one is zero and five plus zero plus negative five is negative five. And look at that. We have our matrix in the desired format with the diagonal ones in the first three columns, and then we have a corresponding X, Y and Z solutions in the fourth column. So our solution set is six negative 50. And we look at our answer choices and that matches with answer choice a well done, we'll catch you on the next one.

In Exercises 21–38, solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.

Key Concepts

Systems of Equations

Matrices

Gaussian Elimination

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