Which row operation will triangularize this matrix?1 0 1 | 1 0 1 1 | 62 0 1 | 1A) R1-R3B)2R2-R3C)-2R1+R3D)2R1+R3E)3R1+R3

Triangularizing matrix gives the matrix that has only zeroes above or below the main diagonal. To find which option is correct we need to calculate all of them.
In all these options we calculate result and write it into row that is first mentioned:

A)R1-R3
[tex] \left[\begin{array}{ccc}-1&0&0|0\\0&1&1|6\\2&0&1|1\end{array}\right] [/tex]

B)2R2-R3
[tex] \left[\begin{array}{ccc}1&0&1|1\\-2&2&1|4\\2&0&1|1\end{array}\right] [/tex]

C)-2R1+R3
[tex] \left[\begin{array}{ccc}0&0&-1|-1\\0&1&1|6\\2&0&1|1\end{array}\right] [/tex]

D)2R1+R3
[tex] \left[\begin{array}{ccc}4&0&3|3\\0&1&1|6\\2&0&1|1\end{array}\right] [/tex]

E)3R1+R3
[tex] \left[\begin{array}{ccc}5&0&4|4\\0&1&1|6\\2&0&1|1\end{array}\right] [/tex]

None of the options will triangularize this matrix. The only way to triangularize this matrix is
R3-2R1
[tex]\left[\begin{array}{ccc}1&0&1|1\\0&1&1|6\\0&0&-1|-1\end{array}\right]
[/tex]

This equation is similar to C) but in reverse order. Order in which rows are written is important.