Perform each matrix row operation and write the new matrix.

Question

$\begin{bmatrix}1 & -1 & 1 & 1 & \vert & 3 \0 & 1 & -2 & -1 & \vert & 0 \2 & 0 & 3 & 4 & \vert & 11 \5 & 1 & 2 & 4 & \vert & 6\end{bmatrix}\quad\begin{array}{l}-2R_1 + R_3 \-5R_1 + R_4\end{array}$

Accepted Answer

hey everyone here we are asked to write the new matrix after performing the matrix row operations. So first here we can just write out the borders of the matrix along with the vertical line. So now beginning with our first given operation here we have to take negative seven times row three plus row one. And then we need to replace this day data that we get with the data already in row one. So with this we can begin with row one which now we know will be replaced by this operation. So we can go ahead and start with our first column. So again we have negative seven times the value in column one of row three which is two. And then we add this value that's also in column one of row one. So we'll add one. Now we can move to our second column we have negative seven again times the value in column two of row three which is eight, so negative seven times eight. And then we add the value in row one which is to moving on to our third column we have again negative seven times the value of row three here which is zero plus the value in row one which is also column three which is adding negative one. And finally with our third or sorry, fourth column we have negative seven times the value in row three which is negative four plus the value in row one which is two. And now we can move to our final column on the right hand side of the vertical line. Once again we have negative seven times the value in row three which is just three plus the value in row one which is simply zero. And then our last two rows just stayed the same here. So our second row has the values, sorry, last three rows remain the same. So our second row has the values 03, negative 51 and six. Our third row has the values to 80, negative four and three. And our final role has the values nine, 08, -7 and eight. So now we can just go ahead and simplify our matrix here. So our first row simplifying, we have the values negative seven times two plus one which is negative 13. We then have negative 54 we then have negative one and lastly 30 and then on the right hand side we have negative 21 and then again our last three rows remain the same with the values 03 negative 51 and six. Then we have 280, negative 43. And lastly we have 908, negative seven and eight. And so now we can go ahead and move on to our second operation So we can go ahead and rewrite the matrix here. So our role one will remain the same with the values negative 13, negative and negative one 30 and negative 21. But now we our second operation, we have two times row four and then we add row two and we're going to replace row two with the result of this operation. So just like before we can write out this operation for each column. So starting here we'll have to in the first column we'll have two times the value of column one of row four which is nine. We then add the value of Row two which is zero. Moving on to our second column, we again have two times the value in column two, row four which is zero. Then adding the value of row two which is three from our third column we get two times row four which is eight plus row two Which is -5. And now we can move on to our 4th row here. Let me just move over these values here. So with row four we again have to or sorry, column four we have two times row four which is negative seven. Then adding row two, we have plus one and then moving on to our final column on the right hand side of the vertical line, we have two times eight plus six. And then our last two rows remain the same. We have the values to 80, negative four and three. Then we have 908, negative seven and eight. So now up here I'm just going to rewrite the matrix and we're going to simplify the second row to get our final answer here. So again our first row has the values negative 13 negative at 54 negative 1, 30 And 20-21. Then our simplified second row we get the values 18 3, 11, negative 13 and 22. And our last two rows are still unchanged with the values to 80 negative 43 and lastly 908 negative seven and eight. And with this we have finished our row operations and we are left with answer choice. C. Thanks for watching. I hope you found this video helpful and I'll see you again next time.

Perform each matrix row operation and write the new matrix.
$\begin{bmatrix}1 & -1 & 1 & 1 & \vert & 3 \\0 & 1 & -2 & -1 & \vert & 0 \\2 & 0 & 3 & 4 & \vert & 11 \\5 & 1 & 2 & 4 & \vert & 6\end{bmatrix}\quad\begin{array}{l}-2R_1 + R_3 \\-5R_1 + R_4\end{array}$

Key Concepts

Matrix Row Operations

Performing Linear Combinations of Rows

Augmented Matrices and Systems of Equations

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