Question

If $A=\left[\begin{array}{ccc}1& 2& 3\\ 2& 3& 4\\ 0& 5& 6\end{array}\right]$, then $2A=$

• A. $\left[\begin{array}{ccc}2& 4& 6\\ 2& 3& 4\\ 0& 5& 6\end{array}\right]$
• B. $\left[\begin{array}{ccc}1& 2& 3\\ 4& 6& 8\\ 0& 5& 6\end{array}\right]$
• C. $\left[\begin{array}{ccc}1& 2& 3\\ 2& 3& 4\\ 0& 10& 12\end{array}\right]$
• D. $\left[\begin{array}{ccc}2& 4& 6\\ 4& 6& 8\\ 0& 10& 12\end{array}\right]$

Answer 1

The correct option is D $\left[\begin{array}{ccc}2& 4& 6\\ 4& 6& 8\\ 0& 10& 12\end{array}\right]$

Since, $A=\left[\begin{array}{ccc}1& 2& 3\\ 2& 3& 4\\ 0& 5& 6\end{array}\right]$

$\therefore 2A=\left[\begin{array}{ccc}2\times 1& 2\times 2& 2\times 3\\ 2\times 2& 2\times 3& 2\times 4\\ 2\times 0& 2\times 5& 2\times 6\end{array}\right]=\left[\begin{array}{ccc}2& 4& 6\\ 4& 6& 8\\ 0& 10& 12\end{array}\right]$
Ans: D