Question

A= $\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right]$.

B is matrix obtained by subtracting $4times{1}^{st}rowfrom{2}^{nd}row$ of A.

C is matrix obtained by subtracting $7times{1}^{st}rowfrom{3}^{rd}row$, then $C$ is

• A. $\left[\begin{array}{ccc}1& 2& 3\\ 0& 3& 6\\ 0& 6& 12\end{array}\right]$
• B. $\left[\begin{array}{ccc}1& 2& 3\\ 7& 0& 0\\ 4& 5& 6\end{array}\right]$
• C. $\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 2\\ 3& 4& 5\end{array}\right]$
• D. $\left[\begin{array}{ccc}1& 2& 3\\ 0& -3& -6\\ 0& -6& -12\end{array}\right]$

Answer 1

The correct option is D $\left[\begin{array}{ccc}1& 2& 3\\ 0& -3& -6\\ 0& -6& -12\end{array}\right]$

Given $A=\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right]$

B is obtained by substracting $4R_1$ from $R_2$

$\Rightarrow R_2\leftrightarrow R_2-4R_1$$\to B\sim \left[\begin{array}{ccc}1& 2& 3\\ 0& -3& -6\\ 7& 8& 9\end{array}\right]$

C is obtained by subtracting $7R_1$ from $R_3$

$\Rightarrow R_3\leftrightarrow R_3-7R_1$

$C\sim \left[\begin{array}{ccc}1& 2& 3\\ 0& -3& -6\\ 0& -6& -12\end{array}\right]$

$\therefore$option D is correct