- Home
- Standard 12
- Mathematics
यदि $A = \left[ {\begin{array}{*{20}{c}}1&{ - 2}\\3&0\end{array}} \right],$ $B = \left[ {\begin{array}{*{20}{c}}{ - 1}&4\\2&3\end{array}} \right]$, $C = \left[ {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right]$, तो $5A - 3B - 2C$=
$\left[ {\begin{array}{*{20}{c}}8&{20}\\7&9\end{array}} \right]$
$\left[ {\begin{array}{*{20}{c}}8&{ - 20}\\7&{ - 9}\end{array}} \right]$
$\left[ {\begin{array}{*{20}{c}}{ - 8}&{20}\\{ - 7}&9\end{array}} \right]$
$\left[ {\begin{array}{*{20}{c}}8&7\\{ - 20}&{ - 9}\end{array}} \right]$
Solution
(b) $5A – 3B + 2C = \left[ {\begin{array}{*{20}{c}}5&{ – 10}\\{15}&0\end{array}} \right] – \left[ {\begin{array}{*{20}{c}}{ – 3}&{12}\\6&9\end{array}} \right] – \left[ {\begin{array}{*{20}{c}}0&{ – 2}\\2&0\end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}}5&{ – 10}\\{15}&0\end{array}} \right] – \left[ {\begin{array}{*{20}{c}}{ – 3}&{10}\\8&9\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}8&{ – 20}\\7&{ – 9}\end{array}} \right]$.