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.

Accepted Answer

Welcome back. I am so glad you're here. We are given a system of equations and we need to solve for it. So when we've got a system of equations, we recall from previous lessons that we can create a matrix using the coefficients and constant terms from our system of equations. The first equation becomes the first row of our matrix using the coefficients two negative 71. And the constant term 24. And then the second equation becomes the second row of our matrix 153 21. And the third equation becomes the third row seven, negative 81 50. And now that we've got this matrix, we can use Gaussian elimination or Gauss Jordan elimination. I'm going to demonstrate Gauss Jordan elimination and we are going to transform this matrix into a matrix that has for the first three columns, a diagonal of one's surrounded by zeros. So the first column will be 100, the 2nd 010 and the 3rd 001. And then the fourth column. That's where it's awesome because that will give us our x, y and Z solutions. So how do we get from the one on the left to the one on the right? Well we do a series of transformations and we're going to start with column one. We want to get that one in position. So we're going to transform that too. We're going to work on row one and to get that to to become a one, we're going to multiply everything in that row by the twos reciprocal. So we're multiplying everything by one half in row one. Because we are transforming row one. We can copy down rows two and three. So 153 7 negative 81 50. Now doing the work to replace row 12 times one half is the one that we want negative seven times one half is negative seven halves, one times one half is one half and 24 times one half is 12. Alright, we've got one spot done. Now we want to transform the rest of column one. We want to get zeros into place. So now let's work on row two. We'll get that one into a zero. We're going to use the row that has the one in the desired position. That's row one. And we're going to multiply that in this case by negative one and add that to row two to get a zero in the desired position. Because we're replacing road to we can copy over rose one and negative seven halves one half, 12 and seven negative 50. Now replacing road to negative one times one is negative one plus one is the zero that we want. He negative one times negative seven halves is a positive seven halves plus five is 17 halves negative one times negative or negative one times one half is negative one half plus three gives us five halves and negative one times 12 is negative plus 21 equals nine. Okay, one more position in column one. We want the seven to be a zero and we're going to use the road that has the one in the desired position that's row one. In this case we'll multiply row one times negative seven And add that to row three. Because we're replacing row three we can copy down rows one and negative seven halves one half, 12 and 0 17 halves, five halves and nine. Now for row three negative seven times one is negative seven plus seven is the zero that we want. Ye negative seven times negative seven halves is 49 halves plus negative eight is 33 halves negative seven times one half is negative seven halves plus one is negative five halves. The negative seven times 12 is negative 84 plus 50 is a negative 34. Okay the first column is done. Yay now we want to get a one in position for column to that is row two of column two in order to get that 17 halves to become a one. We're going to multiply by its reciprocal multiply row two times two seventeenth's because we're replacing road too we can copy over rose one and 31 negative seven halves one half 12 and 33 halves negative five halves and negative 34. And now to replace rho to zero times 2 17. 0 17 halves times 2 17 is the one that we want. Ye five halves times two, seventeenth's is five, seventeenth's and nine times to 17th is 18 17th. Sorry for the pause. Okay now we've got the one in place and we can use that one in order to get zeros into the other positions of column two. So let's work on row one replacing that negative seven halves. In order to replace that. We will use the road that has the one in position that's row two. And in this case we're going to multiply road two times a positive seven halves And add that to real one. Because we are replacing row one we can copy down Rose two and three. Let's give ourselves a little more space. Okay Rose two and three are 01, 5/17 and 18 seventeenth's. Rose 30 33 halves, negative five halves and negative 34. Are you doing the work to replace row 17 house times 00 plus one is still 17 halves, times one is seven halves plus negative seven houses. The zero that we want by seven half times five seventeenth's is 35 34th plus one half is 26 seventeenth's and seven halves times 18 17th is 63 seventeenth's plus 12 is seventeenth's. I did do these ahead of time wrote them on a sheet of paper. I do not have these memorized nor can I do them in my head that quickly. Okay now we've got one of our zeros in position for column two. Now let's get this last zero in position for column two, that's in row three. How do we get that? 33 halves to become a zero? We use the row that has the one in the desired position and we'll multiply it by negative 33 halves And we will add that to row three because we are replacing row three we can copy over rows one and 210 17th and 267 seventeenth's, 015, seventeenth's and 18 seventeenth's. Right now replacing row three, negative 33 halftime 00 plus 00 negative 33 half times one is negative 33 halves plus 33 halves is the zero that we want. You negative 33 halves times 5/17 is negative 165. 34th plus the negative five halves that we have. There is negative 125 seventeenth's. This is so fraction E. Alright negative 33 halves times 18 seventeenth's is negative 297 seventeenth's plus negative 34 is negative seventeenth's. Oh my Alright we're almost there though. We've got one column left. We can do this. Okay we need the one in the desired position for column three. So that's row three of column three. So we're replacing row three again. And in this case we will multiply it by the negative reciprocal of that target number. So it will become a positive one. So multiplying row three times negative 17, 125th. So And we can copy over rows one and 2 1, 0, 26, seventeenth's and 267 seventeenth's 01, 5/17 and 18 seventeenth's. And now replacing row 30 times negative 17, 125th is zero, both times negative. 125 17th times negative, 17th, 125th is the one that we want in negative 17, 125th times negative. 875 17th equals seven. Alright, that third row is done now let's just get zeros into place above it. Let's work on row one. And to get a zero into place in row one, we're going to use the row that has the one in the desired spot. So that's row three. And in this case will multiply row three times negative seventeenth's And then add that to row one. Give ourselves a little bit more space. And because we're replacing row one we can copy down Rose two and three. So we've got 015, seventeenth's and 18 seventeenth's. And then this beautiful row 0017. Now replacing row one negative 26 17 times zero is zero plus one is one, negative, 26 17 times 00 plus zero is the zero negative, 26 seventeenth's times one is negative 26 seventeenth's plus positive. 26 17th is the zero that we want ye and negative seventeenth's times seven is negative seventeenth's plus 267 17th equals five. Alright, one more spot. We've got this. Let's get a zero into place in row two will use row three to do that. And this case will multiply row three times negative five seventeenth's and we'll add that to row two and we can copy over rose one and 31005. So nice. 0017. So nice. Now let's replace rho to zero times negative 17 0 plus 000 times negative 5 17 0 plus one is one negative 5/17 times one is negative 5/17 plus 5/17 is the zero that we want there and negative 5/17 times seven is negative 35 seventeenth's plus 18, 17th is negative one and there we have it we have this matrix in the desired form where we have diagonal ones on the left hand side. And the fourth column corresponds with our x, y and Z solutions. So our solution set is five negative one and seven. Now we look at our answer choices and this matches with answer choice. A Well done. We'll catch you on the next one

Struggling with College Algebra?

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

Watch next

Introduction to Matrices practice set