Answer:
[tex]\text{The operation is }R_2\rightarrow R_2-R_3[/tex]
Step-by-step explanation:
Given the matrix
[tex]\begin{bmatrix}1&1&1\\2&1&0 \\2&0&3\end{bmatrix}[/tex]
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.
[tex]\begin{bmatrix}1&1&1\\2&1&0 \\2&0&3\end{bmatrix}[/tex]
[tex]R_2\rightarrow R_2-R_3[/tex]
[tex]\begin{bmatrix}1&1&1\\0&1&-3 \\2&0&3\end{bmatrix}[/tex]
The operation will be
[tex]R_2\rightarrow R_2-R_3[/tex]
[tex]2\rightarrow 2-2=0[/tex]
[tex]\text{Hence, the operation is }R_2\rightarrow R_2-R_3[/tex]
