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3 and 4 .Determinants and Matrices
medium
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 કોઈ પણ અચળ $k$ માટે$(k B)^{\prime}=k B^{\prime}.$
Option A
Option B
Option C
Option D
Solution
We have
$k B=k\left[\begin{array}{rrr}2 & -1 & 2 \\ 1 & 2 & 4\end{array}\right]=\left[\begin{array}{lll}2 k & -k & 2 k \\ k & 2 k & 4 k\end{array}\right]$
Then $(k \mathrm{B})^{\prime}=\left[\begin{array}{cc}2 k & k \\ -k & 2 k \\ 2 k & 4 k\end{array}\right]=k\left[\begin{array}{cc}2 & 1 \\ -1 & 2 \\ 2 & 4\end{array}\right]=k \mathrm{B}^{\prime}$
Thus $(k \mathrm{B})^{\prime}=k \mathrm{B}^{\prime}$
Standard 12
Mathematics
