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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