In Exercises 19–20, a few steps in the process of simplifying the...

Question

In Exercises 19–20, a few steps in the process of simplifying the given matrix to row-echelon form, with 1s down the diagonal from upper left to lower right, and 0s below the 1s, are shown. Fill in the missing numbers in the steps that are shown.

Matrix simplification steps for row-echelon form with missing numbers.

Accepted Answer

hey everyone here are asked to fill in the missing numbers of the matrices which are undergoing the process of reducing the original matrix to row echelon form. So the key to solving this problem here is to look case by case and identify the row operations which occurred. So beginning with our first case we have the provided matrix with three rows and the first row has the values one, 82, negative six. The second row is negative 32 negative one and five. And our third row is 706, negative six. So now we need to compare this original matrix with the second matrix with missing values. So we can first note that the 2nd and 3rd row both undergo changes and the first row remains the same. And next we want to notice that we have a negative three in the original matrix which becomes a zero in the second matrix and similarly the seven in the original matrix also becomes a zero. So now beginning with row two, if we were to make negative 30, we need to use row one. So we'll have row to undergo the operation where we take row one and multiply it by three and then add it to row two. And now similarly looking at row three, we need to use row one. So we'll have row three become the result of negative seven times row one plus row three. So now we can write our second matrix here but have our filled in values. So again our first row remains the same with the values 182 and negative six. And now our second row becomes 0 5 and negative 13. And then our third row becomes zero, negative 56 negative eight and negative 48. So now we can see that from this first case here, our second row was missing the values five and negative 13 and our third row was missing the values negative AIDS and negative 48. So now that we have filled in the second matrix, we can move on to our second case where we compare this second matrix with the third and final matrix. So once again we can see that the 2nd and 3rd rows undergo changes while the first row remains the same. So looking closely we see that we have from the second matrix which becomes a one in the third matrix. Additionally the negative 56 from the second matrix also becomes a one. So beginning with our second row to get 26 to become one, we have to divide by 26. So that means we'll have wrote to become the results of row two divided by 20 six. Similarly with row three to get negative 56 to become one, we need to divide by negative 56. Therefore row three will become the result of row three divided by negative 56. And so with this we can write our final reduced matrix. So once again our first row remains the same and has the values 182 and negative six. And now our second row becomes 015 divided by negative one half. And finally our 3rd row takes deep values 01 seven and 7th. And with this we have finished filling in all of the missing numbers and we can see that from our final matrix. The second case here we were missing five x 26 and negative one half in row two And in row three we were missing the values 1/ and 6/7. And with this we are finished with our problem and left with answer choice D. Thanks for watching. I hope you found this video helpful and I'll see you again next time.

In Exercises 19–20, a few steps in the process of simplifying the given matrix to row-echelon form, with 1s down the diagonal from upper left to lower right, and 0s below the 1s, are shown. Fill in the missing numbers in the steps that are shown.

Key Concepts

Row-Echelon Form

Gaussian Elimination

Leading Coefficient

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