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3 and 4 .Determinants and Matrices
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જો $A = \left[ {\begin{array}{*{20}{c}}3&1\\{ - 1}&2\end{array}} \right]$, તો ${A^2} = $
A
$\left[ {\begin{array}{*{20}{c}}8&{ - 5}\\{ - 5}&3\end{array}} \right]$
B
$\left[ {\begin{array}{*{20}{c}}8&{ - 5}\\5&3\end{array}} \right]$
C
$\left[ {\begin{array}{*{20}{c}}8&{ - 5}\\{ - 5}&{ - 3}\end{array}} \right]$
D
$\left[ {\begin{array}{*{20}{c}}8&5\\{ - 5}&3\end{array}} \right]$
Solution
(d) $A = \left[ {\begin{array}{*{20}{c}}3&1\\{ – 1}&2\end{array}} \right]$
${A^2} = A.A = \left[ {\begin{array}{*{20}{c}}3&1\\{ – 1}&2\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}3&1\\{ – 1}&2\end{array}} \right]$
${A^2} = \left[ {\begin{array}{*{20}{c}}8&5\\{ – 5}&3\end{array}} \right]$.
Standard 12
Mathematics
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