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 31 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 from previous lessons that when we have a system of equations we can create a matrix using the equations, coefficients and constant terms. So the first row of our matrix will correspond with the first equation and its coefficients are one negative negative two. And then row two will correspond with equation too. So that will be four negative 11 negative and row three equation 31 negative negative four. And what we want to do using either Gaussian or Gauss Jordan elimination is to create to transform this into a new matrix. And I'm going to demonstrate Gauss Jordan elimination. So what we will be transforming it into will be with a diagonal of one's starting in the upper left hand corner surrounded by zeros for the first three columns. So the first column will be 100, the second column will be 010, and the third will be 001. And then the fourth column is where it's just wonderful and amazing because in this case it will be our a B and C solutions. So we just have to go through a series of transformations. What we need to do first is get the one in position for column one and we already have it. You see the first step is done for us. So now we want the rest of column one to be zeros. So let's work on transforming this four will be replacing row two. And in order to get that four into zero. We're going to work with the row that has the one in the desired position which is row one for column one. And we are going to multiply row one in this case, times negative four And then add that to row two because we're replacing road to we can copy down rose one and three. So that'll be one negative 33, negative two and one negative negative four. And road two will be now taking negative four times row one and adding that to road to negative four times one is negative four plus four is the zero that we want negative four times negative three is positive 12 plus negative one is a positive 11 negative four times three is negative 12 plus one is negative 11 and negative four times negative two is positive, eight plus negative 19 is negative 11. Alright, we've got row two replaced now to finish off column one, we want a zero in the third row. How do we get a zero there? We're going to work with the row that has the one in the desired position which is still row one. In this case we'll multiply row one times a negative one and add that to row three, copying over rows one and two. We've got one negative 33 negative 2011, negative 11 and negative 11. Now transforming row three negative one times one is zero plus one. Excuse me negative one times one is negative one plus one is the zero that we want there negative one times negative three is positive three plus negative seven is negative four negative one times three is negative three plus one is negative two and negative one. Times negative two is positive two plus negative four is negative two. Alright, column one is done. You. Now let's work on getting the one in the desired position for column to which is right here where we've got this 11 in order to get that to be a one. We are going to multiply row two times the reciprocal of that target number. We're going to multiply it by 1/11. So row two times 1/11. We can copy down rose one and 31, negative 33, negative two and zero negative four negative two negative two. Now replacing road to zero times 1/11 that is 0 11 times 1/11 is the one that we want. A negative 11 times 1/11 is negative one both times. Alright, now let's work on getting zeros into the rest of the positions for column two. Let's work on this negative three in row one. What we'll do is work with the row that has the one in the desired position in this case road to and we'll multiply it by three and add that to row one, replacing row one. We can copy over rows two and 301. Negative one. Negative 10, negative four negative two negative two. Now replacing row 10 times three is zero plus one is one. Three times one is three plus negative three is the zero that we want. Three times negative one is negative three plus positive three is another zero. Alright. And three times negative one is negative three plus negative two is negative five. Look at that row one is done. Now let's get a zero into the other position for column to where this negative four is. So we'll be replacing row three will work with road to to do it and this time we'll multiply row two times four And add that to row three. Because we're replacing row three we can copy down rows one and two. So we'll have 100, negative five and 01 negative one negative one. Now to replace row 34 times 00 plus zero is still 04 times one is four plus negative four. Gives us the zero that we want four times negative one is negative four plus negative two is negative six and four times negative one is negative four plus negative two is again negative six. And now we want to get our one in the desired position for ro excuse me for column three which is row three and we need To multiply this time by the negative reciprocal of that target number. So we're going to multiply everything times a negative 1/6. So that will be a positive one in our targeted position. So we can copy over rows one and negative 501 negative one negative one. And now four row 30 times negative 1/ is zero both times and negative six times negative 1/ is one both times. Alright, so row three is done. We're almost done with this. We need to now convert the last zero in row two and we'll work with row three to do that because it is the one in the desired position. And in this case we're just going to add row three to row two and we'll get our zero where we need it. So copying over rose one and -50011. And now doing the work to replace road to zero plus zero is 01 plus zero is one negative one plus one is zero and negative one plus one is another zero. And look at this. This matrix looks just like our target matrix where we have the diagonal row of ones and now column four corresponds with our A B and C. Solutions. So that means our solution set is negative 501. We look at our answer choices and this matches with answer choice B. 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.

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