Perform each matrix row operation and write the new matrix.

Question

$\begin{bmatrix}1 & -3 & 2 & \vert & 0 \3 & 1 & -1 & \vert & 7 \2 & -2 & 1 & \vert & 3\end{bmatrix}\quad -3R_1 + R_2$

Accepted Answer

hey everyone here we are asked to write the new matrix after performing the matrix row operation. So here we just have the operation where we take five and multiply it by row three and then add it to row one and then we're going to take the result operation, take this data and replace it with row one. So first we can just write out our matrix here, the borders with the vertical line and then we can go ahead and start with row one and like I just stated our world one will be replaced by this operation so we can begin at our first column for row one and write out this operation here. So again we have five times row three which means we look at the first column of row three. So we have five times negative four and then we add this to row one. So we look at row one, column one which is negative six. So we have plus negative six. Now we can move on to our second column where we have five times row three, looking at the second column, we have a value of zero, so we have five times zero. Then we add row one. Which in this case is to next we can move on to column three where we have five again, times row three which is eight. We then add row one again looking at column three of row one, we have the value of one. So we'll add one. Now we can move to the last column on the right hand side of the vertical line here and we'll repeat the operation. So we again have five times row three which is six plus row one which is negative four. And now we just have to fill in the last two rows which remain unchanged here. So we have the values to 1234 row two. And for our last row we have negative and six. So now all that is left to do is to simplify our first row and rewrite the matrix to our final answer here. So row one simplified as the values negative 26 to 26 then our last two rows once again remain unchanged. So for row two we have the values to and then four row three. We have the values negative 408 and six. And with this we have found our new matrix. And so we are left with answer choice D. 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.
$\begin{bmatrix}1 & -3 & 2 & \vert & 0 \\3 & 1 & -1 & \vert & 7 \\2 & -2 & 1 & \vert & 3\end{bmatrix}\quad -3R_1 + R_2$

Key Concepts

Matrix Row Operations

Augmented Matrices

Applying Scalar Multiplication and Addition to Rows

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