- Home
- Standard 12
- Mathematics
If $A = \left( {\begin{array}{*{20}{c}}1&{ - 2}&1\\2&1&3\end{array}} \right)$ and $B = \left( {\begin{array}{*{20}{c}}2&1\\3&2\\1&1\end{array}} \right)$, then ${(AB)^T}$ is equal to
$\left( {\begin{array}{*{20}{c}}{ - 3}&{ - 2}\\{10}&7\end{array}} \right)$
$\left( {\begin{array}{*{20}{c}}{ - 3}&{10}\\{ - 2}&7\end{array}} \right)$
$\left( {\begin{array}{*{20}{c}}{ - 3}&7\\{10}&2\end{array}} \right)$
None of these
Solution
(b) $AB = \left( {\begin{array}{*{20}{c}}
1&{ – 2}&1 \\
2&1&3
\end{array}} \right)$ $\left( {\begin{array}{*{20}{c}}
2&1 \\
3&2 \\
1&1
\end{array}} \right)$= $\left( {\begin{array}{*{20}{c}}
{ – 3}&{ – 2} \\
{10}&7
\end{array}} \right)$
${(AB)^T} = \left( {\begin{array}{*{20}{c}}{ – 3}&{10}\\{ – 2}&7\end{array}} \right)$.