If k in R and operatorname{det} A = begin{vma | vedclass

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(C) Given that $\det A = \begin{vmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3 \end{vmatrix} = K$.We are given $\det B = \begin{vmatri

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$K$

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(C) Given that $\det A = \begin{vmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3 \end{vmatrix} = K$.We are given $\det B = \begin{vmatrix} a_1 & b_1 & c_1 \ a_2 + 2a_1 & b_2 + 2b_1 & c_2 + 2c_1 \ a_3 & b_3 & c_3 \end{vmatrix}$.Using the property of determinants,we apply the row operation $R_2 	o R_2 - 2R_1$.This operation does not change the value of the determinant.$\det B = \begin{vmatrix} a_1 & b_1 & c_1 \ (a_2 + 2a_1) - 2a_1 & (b_2 + 2b_1) - 2b_1 & (c_2 + 2c_1) - 2c_1 \ a_3 & b_3 & c_3 \end{vmatrix}$$\det B = \begin{vmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3 \end{vmatrix}$Since the resulting determinant is identical to $\det A$,we have $\det B = K$.

If $k \in R$ and $\operatorname{det} A = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = K$,then $\operatorname{det} B = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 + 2a_1 & b_2 + 2b_1 & c_2 + 2c_1 \\ a_3 & b_3 & c_3 \end{vmatrix}$ is equal to

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