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If $A^{\prime}=\left[\begin{array}{cc}-2 & 3 \\ 1 & 2\end{array}\right]$ and $B=\left[\begin{array}{cc}-1 & 0 \\ 1 & 2\end{array}\right],$ then find $(\mathrm{A}+2 \mathrm{B})^{\prime}$
$\left[\begin{array}{cc}-4 & 5 \\ 1 & 6\end{array}\right]$
$\left[\begin{array}{cc}-4 & 5 \\ 1 & 6\end{array}\right]$
$\left[\begin{array}{cc}-4 & 5 \\ 1 & 6\end{array}\right]$
$\left[\begin{array}{cc}-4 & 5 \\ 1 & 6\end{array}\right]$
Solution
We know that $A=\left(A^{\prime}\right)^{\prime}$
$\therefore A=\left[\begin{array}{cc}-2 & 1 \\ 3 & 2\end{array}\right]$
$\therefore \,\,A + 2B = \left[ {\begin{array}{*{20}{c}}
{ – 2}&1 \\
3&2
\end{array}} \right]$ $ + 2\left[ {\begin{array}{*{20}{c}}
{ – 1}&0 \\
1&2
\end{array}} \right] = $ $\left[ {\begin{array}{*{20}{c}}
{ – 2}&1 \\
3&2
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
{ – 2}&0 \\
2&4
\end{array}} \right]$ $ = \left[ {\begin{array}{*{20}{c}}
{ – 4}&1 \\
5&6
\end{array}} \right]$
$\therefore(A+2 B)^{\prime}=\left[\begin{array}{cc}-4 & 5 \\ 1 & 6\end{array}\right]$