3 and 4 .Determinants and Matrices
easy

यदि $\left[ {\begin{array}{*{20}{c}}2&{ - 3}\\4&0\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}a&c\\b&d\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&4\\2&{ - 5}\end{array}} \right]$, तो $(a,b,c,d) = $

A

$(1,\,6,\,2,\,5)$

B

$(1, 2, 7, 5)$

C

$(1, 2, -7, 5)$

D

$(-1, -2, 7, -5)$

Solution

(c) $\left[ {\begin{array}{*{20}{c}}2&{ – 3}\\4&0\end{array}} \right] – \left[ {\begin{array}{*{20}{c}}a&c\\b&d\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&4\\2&{ – 5}\end{array}} \right]$

$\Rightarrow$ $\left[ {\begin{array}{*{20}{c}}a&c\\b&d\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2&{ – 3}\\4&0\end{array}} \right] – \left[ {\begin{array}{*{20}{c}}1&4\\2&{ – 5}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ – 7}\\2&5\end{array}} \right]$

$\Rightarrow$ $(a,b,c,d) = (1,\,\,2,\,\, – 7,\,\,5)$.

Standard 12
Mathematics

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