3 and 4 .Determinants and Matrices
easy

यदि $A = \left[ {\begin{array}{*{20}{c}}3&{ - 5}\\{ - 4}&2\end{array}} \right],$ तो ${A^2} - 5A = $  

A

$I$

B

$14\ I$

C

$0$

D

इनमें से कोई नहीं

Solution

(b) ${A^2} = A.A = \left[ {\begin{array}{*{20}{c}}3&{ – 5}\\{ – 4}&2\end{array}} \right]{\rm{ }}\left[ {\begin{array}{*{20}{c}}3&{ – 5}\\{ – 4}&2\end{array}} \right]$

$ \Rightarrow \,{A^2} = \left[ {\begin{array}{*{20}{c}}{29}&{ – 25}\\{ – 20}&{24}\end{array}} \right]$ 

तथा $5A = \left[ {\begin{array}{*{20}{c}}{15}&{ – 25}\\{ – 20}&{10}\end{array}} \right]$

$\therefore$ ${A^2} – 5A = 14\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = 14I$.

Standard 12
Mathematics

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