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3 and 4 .Determinants and Matrices
easy
જો $A=\left[\begin{array}{lll}3 & \sqrt{3} & 2 \\ 4 & 2 & 0\end{array}\right]$ અને $B=\left[\begin{array}{rrr}2 & -1 & 2 \\ 1 & 2 & 4\end{array}\right]$ , તો ચકાસો કે $\left(\mathrm{A}^{\prime}\right)^{\prime}=\mathrm{A}$.
Option A
Option B
Option C
Option D
Solution
We have
$A=\left[\begin{array}{lll}3 & \sqrt{3} & 2 \\ 4 & 2 & 0\end{array}\right]$ $\Rightarrow A^{\prime}=\left[\begin{array}{ll}3 & 4 \\ \sqrt{3} & 2 \\ 2 & 0\end{array}\right] \Rightarrow$ $\left(\mathrm{A}^{\prime}\right)^{\prime}=\left[\begin{array}{lll}3 & \sqrt{3} & 2 \\ 4 & 2 & 0\end{array}\right]=\mathrm{A}$
Thus $\left(A^{\prime}\right)^{\prime}=A$
Standard 12
Mathematics
