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 21 in college algebra, chapter on matrices.

Accepted Answer

hey everyone here we are asked to use Gaussian elimination with back substitution or Gauss Jordan elimination to solve the given system of equations. So the first step here is to write our given system of equations into an augmented matrix. So if we recall an augmented matrix has this vertical line which represents the equal sign from the equations. And to the left hand side we have the coefficients for each given variable. So going from left to right, we have variables X, y and D. For our columns. And to the right hand side of this line, we have all of the constant values. So beginning with our first row, we look at our first given equation which gives us the values 51, negative two and 37 from our second equation. We get our second row with the values to negative and negative eight for our third row. We look at our third equation and we get the values eight negative 23 and five. And now from here we want to begin using elementary row operations to get this matrix into reduced row echelon form. So starting with our first column, we can look at this first value here which is five and we want this value to be become one. So we'll have row one be replaced by row one divided by five. And now doing this operation we can rewrite this tricks. So our role one now has the values 1, 1/ Rose are unchanged. So they have the values to negative 31 and negative eight. And lastly eight negative 23 and five. So now we can continue down our first column to the second term which is to we want this term to become zero. So we'll have row to undergo the operation where we use row one. So we have negative two times row one plus wrote doing this. We once again rewrite our matrix. Our first row remains as 1, 1/5 negative 2/5 and 5th. And now our second row takes the values zero negative 17 5th, 9/5 and negative fits. And our last row remains the same with the values eight negative 23 and five. So now we can move on to the last value in column one which is eight, we want this eight to become zero. So again using ro one, we'll have row three undergo the operation where we take negative eight Row one Plus Row three. Again we rewrite our matrix. Our first two rows are the same. Our first row has the values 1, 1/5 negative 2/5 and 5th. Our second row has the values zero negative 17 5th 9/5 and negative 114 5th. And now our third row becomes 0, 18 5th, 31/5 and -271/5. 71 5th. And now from here to our second column and we can see that we can easily make this second term here. This negative 17 5th into a positive one. So to do this will have row to become the result of this. Reciprocal here negative 5/17 times row two. And now once again doing this operation here will rewrite our matrix, our first row remains the same with the values 1, 1/5 negative 2/5 and 5th. Our second row now becomes 01, negative 9/17 And 114 17th. And now our third role remains the same with the values zero negative 18 5th, 31 5th and negative 5th. So now that we have this positive one in column two, we can use it to get the other two terms to become zero. So we can start with this first term, the 1/5 in column two. And we can again use row too. So we'll have row two or sorry row one become the result of the operation where we take negative 1/5 times row two plus row one. Now doing this, we once again rewrite our matrix Values 1, 0 -5/17 and 103/17. And then our last two rows remain the same. We have the values 01, negative 9/17 and 114 17th for row two. And then our third row has the values zero negative 5th, 31 5th and negative 200 and 5th. So now we can move to the last column to which is this negative 18 5th, we want this term to become zero. So we'll have row three undergo the operation where we have the reciprocal here 18 5th times two plus row three. Once again we rewrite our matrix with our new values, row one in the same. So we have the values 10, negative 5/17 and 17th. Our second row remains as 01, negative 9/17 and 17th. And now our third row takes the values 00 73 17th and negative 17th. So now from here we can move on to our third column and we can go ahead and get this first term here to become zero. But before we can move to that we first have to make this last term here, one in row three so that we can use it to get the other terms to become zero. So doing this, We will have this last term become one by having row three. Take on the operation where we take the reciprocal 17 divided by 73 times row three. And now once again we rewrite the matrix with the new values. So our row one has the values 10 negative 5/17 and 103 17th. Our second row is still 01, negative 9/17 and 17th. And then our third row now has the values 001 and negative seven. And now once again we can use this one value to make the other two terms zero. So we'll start with this first term here with the negative 5/17. And so we'll have undergo the operation of having 5/17 times row three plus row one. Once again we rewrite our matrix with our new values row one. Now the values 1004. And then our last two rows are still the same with the values 01, negative 9/17 and 100 14 17th. And our last row has the values 001 and negative seven. So now we can make this last term here. This negative 9 17 in 20 using row three. So we'll have row to become the result of seven, or sorry, 9/17 times row three plus row two. And with this our final matrix can be formed. So our first row remains as 1004. Our second row now becomes 0103 and our last row is still 001, negative seven. So with this we have our final reduced matrix. And if we recall these terms on the left hand side of the vertical line or the X, Y and z coefficients. So from here we get three equations from each of the rows, from our first role. We see that X is equal to four, from our second row, we see that y is equal to three and from our third, third row we see that Z is equal to negative seven. And so from this we've gotten direct solutions for each of our variables, so all that is left is to write this as a solution set. So our final answer is the solution set with values for three and negative seven, which leaves us with answer choice. D. Thanks for watching. I hope you found this video helpful and I'll see you again next time.

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

Key Concepts

Matrices

Gaussian Elimination

Back-Substitution

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