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 23 in college algebra, chapter on matrices.

Accepted Answer

Welcome back. I am so glad you're here. We are given this system of equations and we're asked to solve for it. Well, we recall that when we have a system of equations, we can convert it to a matrix. So the first equation will become the first row. We're going to use the coefficients and constant terms. But we note that for this first equation we have an X term which will go in the first column of the first row, we have a wide term which will go in the second column of the first row, but there is no Z term. So the coefficient for the Z term here is zero. So the first row is reading three negative 80. And then we have our constant term 19. And the second equation becomes the second row, so the coefficients are 1, -1 and the constant term -16. And then for the third row we will grab from the third equations, coefficients and constant terms, so six negative 12 and 22. And now that we've set up our matrix for the system of equations, we get to choose either Gaussian elimination or Gauss Jordan elimination to solve for it, I'm going to demonstrate Gauss Jordan elimination, which means that we are going to go through a series of transformations to get this matrix looking like one that has a diagonal of one's surrounded by zeros for the first three columns. So the first column will be 100, the 2nd 010 and the 3rd 001. And then the fourth column, that's where it's at because in this case that's going to give us our X, Y and Z solutions. So how do we get from our matrix to the one with the ones and zeros? While we're going to start off by getting the one in the first column. That's where this three is currently. Excuse me, How do we transform that three into one? While we're going to take the first row and we are going to multiply it by the reciprocal of that three. So we're going to multiply everything by one third because we're replacing real one. We can copy over Rose two and three. So I'm going to do that 15 negative one, negative 16 and six negative 1 to 22. And now multiplying the first row by one third three times one. Third is the one that we want negative eight times one. Third is negative eight thirds, zero times one third is zero and 19 times one third is 19 3rd. All right. We've got that one in position. Now we're going to use that one to help us get zeros in the rest of column one. So let's work on getting zero in row two of column one. Using the One From Row one. We're going to multiply row one in this case, times negative one and then add that to row two. Because we're replacing road to we can copy over a Rose one and three. So one negative eight thirds 0, 19 3rd and six negative 12 and 22. Now replacing road to negative one times one is negative one plus one is the zero that we want. Ye negative one times negative eight thirds is positive, eight thirds plus five is 23 3rd negative one times 00 plus negative one is still negative one and negative one times 19 3rd is negative 19 3rd plus negative 16 is negative 67 3rd. Okay, we've got 10 in place. Now let's get the rest of column one finish. We want this six in row three to become a zero. We're going to use the road that has the one in the desired position. That's row one and in this case we'll multiply row one times negative six And then add that to row three because we're replacing row three we can copy down rows one and 21 negative eight thirds, zero and 19 3rd and 0 23 3rd negative one and negative 67 3rd. Now replacing row three negative six times one is negative six plus six is the zero that we want. Ye negative six times negative eight thirds is 16 plus negative one is 15, negative six times 00 plus two is still too and negative six times 19 3rd is negative 38 plus 22 is negative 16. All right, column one is done. Yay now we need to get the one in place for column to which is in row two Where this 23 3rd is And to get that to become a one we're going to multiply by its reciprocal by three 23rd. So because we're replacing road to we can copy over rose one and negative eight thirds 0 19 3rd and 0 15 to negative 16. Now replacing road to zero times three 23rd is still 23 3rd times three. 23rd is the one that we want negative one times three. 23rd is negative three 23rd and negative 67 3rd times three 23rd is negative 67 23rd. Right now we can use that one to get zeros in the rest of column two. So now let's work on row one. Getting a zero where that eight thirds is. We'll use the road that has the one in the desired position. So road to in this case will multiply row two times eight thirds And then add that to row one. Because we are replacing round one we can copy down rows two and 3. So we'll have negative three 23rd and negative 23rd and 0 15 to negative 16. Now replacing Rule one zero times eight thirds is zero plus one is still 11 times eight thirds is eight thirds plus negative eight thirds is the zero that we want ye negative three 23rd times eight thirds is negative eight 23rd plus zero is negative eight 23rd and negative 67 23rd times eight thirds is negative 536 69th plus 19 3rd is negative 33 23rd. Okay, one more zero for column to. We want to work on row three. Get that 15 to become a zero. We'll do that by multiplying row two times a negative And adding that to row three. We can copy over rows one and 2. So we'll have 10 negative eight 23rd and negative 33 23rd and 01 negative three 23rd negative 67 23rd. This is so fraction E now four row three negative 15 times 00 plus 00, negative 15 times one is negative, 15 plus 15 is the zero that we want. Ye now negative three 23rd times negative 15 is 45 23rd plus two equals 23rd and negative 15 times negative 67 23rd is 1005, 23rd plus negative 16 is a mere 637. 23rd. Okay, well now we've just got to work on column three. We need to get our one in our desired position and that's row three of column three. So we're going to transform row three again and we will multiply it by its reciprocal to get that 91 23rd to be a one. So multiplying row three times 23 firsts. We can copy over rows one and 210 negative eight 23rd negative 33 23rd and 01 negative three 23rd negative 67 23rd. And now transforming row three. Well zero times 23 91st zero, both times 91 23rd times 23 91st is one. Yay And 637, 23rd Times 23/91 is seven. Cool. All right now we are going to use row three to get zeros in the rest of column one. So let's start with row one or excuse me. The rest of column three. So getting row one, getting that negative eight 23rd to become a zero. We're going to multiply row three times positive. Eight 23rd And we're going to add that to row one Because we're replacing row one. We can copy down rows two and 3. So we've got 01 negative three 23rd negative 67 23rd. And this lovely 0017. Now replacing row 10 times eight, 23rd, zero plus one is 10 times eight, 23rd, zero plus 001 times eight, 23rd is eight 23rd plus negative eight 23rd is the zero that we want and seven times eight 23rd is 56 23rd plus negative 33 23rd is one. Alright, one more. We're almost done. We've got to get this negative three 23rd to become a zero. So we're going to transform row two. We'll use the row that has the one in the desired spot to do that. That's row three. In this case we'll multiply row three times three 23rd And then add that to row two. We can copy over Rose one and three which are now very lovely. 1001 and 0017. Now let's work on road 23 23rd times 00 plus zero is 03 20 third's time. 00 plus one is 13 23rd times one is three 23rd plus negative three 23rd is the zero that we want and three 23rd times seven is 21 23rd plus negative 67. 23rd is negative two. And hey look at this, we did it. The first three columns are our diagonal ones plus surrounded by zeros and our fourth column is our x, y and Z solutions. So our solution set in this case is one negative +27, scroll back up a little so we can see all the answer choices and this matches with answer choice. C. 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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