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3 and 4 .Determinants and Matrices
easy
જો $2X + \left[ {\begin{array}{*{20}{c}}1&2\\3&4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3&8\\7&2\end{array}} \right]$ તો શ્રેણિક $X$ મેળવો.
A
$\left[ {\begin{array}{*{20}{c}}1&3\\2&{ - 1}\end{array}} \right]$
B
$\left[ {\begin{array}{*{20}{c}}1&{ - 3}\\2&{ - 1}\end{array}} \right]$
C
$\left[ {\begin{array}{*{20}{c}}2&6\\4&{ - 2}\end{array}} \right]$
D
$\left[ {\begin{array}{*{20}{c}}2&{ - 6}\\4&{ - 2}\end{array}} \right]$
Solution
(a) $2X\, = \left[ {\begin{array}{*{20}{c}}3&8\\7&2\end{array}} \right] – \left[ {\begin{array}{*{20}{c}}1&2\\3&4\end{array}} \right]$
$2X = \left[ {\begin{array}{*{20}{c}}2&6\\4&{ – 2}\end{array}} \right] \Rightarrow X = \left[ {\begin{array}{*{20}{c}}1&3\\2&{ – 1}\end{array}} \right]$.
Standard 12
Mathematics
