Question

Which elementary row operation will reduce the x–element in row 2 to 0 in this matrix?

Answer 1

Multiply the FIRST row by -2:  -2  -2  -2  |  -24

Now add this result to the 2nd row.  Voila!  You'll end up with a 0 x-element in row 2.

Answer 2

Answer:

$\text{The operation is }R_2\rightarrow R_2-R_3$            

Step-by-step explanation:

Given the matrix

$\begin{bmatrix}1&1&1\\2&1&0 \\2&0&3\end{bmatrix}$

An elementary row operations are the operations apply in matrix which includes multiply each element in a row by a non-zero number, Multiply a row by a non-zero number and add the result to another row.

These are used to convert the matrix in echelon form.

Here we have to reduce the x–element in row 2 to 0 in the matrix.

$\begin{bmatrix}1&1&1\\2&1&0 \\2&0&3\end{bmatrix}$

$R_2\rightarrow R_2-R_3$

$\begin{bmatrix}1&1&1\\0&1&-3 \\2&0&3\end{bmatrix}$

The operation will be

$R_2\rightarrow R_2-R_3$

$2\rightarrow 2-2=0$

$\text{Hence, the operation is }R_2\rightarrow R_2-R_3$