Question

$A=\left[\begin{array}{ccc}2& 2& 1\\ 0& 1& 4\\ 0& 2& 6\end{array}\right]$, $B=\left[\begin{array}{ccc}2& 2& 1\\ 0& 1& 4\\ 0& 0& 1\end{array}\right]$

To obtain B from the matrix A, order of operations would be

• A. $R_3\to R_3-3R_1$, $R_3\to R_2-R_1$
• B. $R_3\to R_1-2R_2$, $R_3\to (R_3\times -2)$
• C. $R_2\to R_2-2R_2$, $R_3\to (R_3÷2)$
• D. $R_3\to R_3-2R_2$, $R_3\to (R_3÷-2)$

Answer 1

The correct option is D $R_3\to R_3-2R_2$, $R_3\to (R_3÷-2)$

$A=\left[\begin{array}{ccc}2& 2& 1\\ 0& 1& 4\\ 0& 2& 6\end{array}\right]B=\left[\begin{array}{ccc}2& 2& 1\\ 0& 1& 4\\ 0& 0& 1\end{array}\right]$

From A to B

$R_1$ & $R_2$ are same

changing $R_3$ w.r.t. $R_2$

Clearly the term ${}^{\prime }{2}^{\prime }$ in $R_3{C}_2$ in A is change to ${}^{\prime }{0}^{\prime }$ in comparing them we find

$R_3\to R_3-2R_2$

We get the matrix

$\left[\begin{array}{ccc}2& 2& 1\\ 0& 1& 4\\ 0& 0& -2\end{array}\right]$

Now if we divide $R_3$ w.r.t. ${}^{\prime }-2^{\prime }$

We get the derived term $1$ in $R_3{C}_3$ of B

$\left[\begin{array}{ccc}2& 2& 1\\ 0& 1& 4\\ 0& 0& 1\end{array}\right]=B$

$\therefore$Operations are

$R_3\to R_3-2R_2$

and $R_3\to R_3-(-2)$

Option D