(C) To find the value of $\lambda$,we expand the determinant along the first column:$\left| \begin{matrix} -6 & 1 & \lambda \\ 0 & 3 & 7 \\ -1 & 0 & 5

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(C) To find the value of $\lambda$,we expand the determinant along the first column:$\left| \begin{matrix} -6 & 1 & \lambda \\ 0 & 3 & 7 \\ -1 & 0 & 5 \end{matrix} \right| = -6(3 \times 5 - 7 \times 0) - 0(1 \times 5 - \lambda \times 0) + (-1)(1 \times 7 - 3 \times \lambda) = 5948$$-6(15) - 0 + (-1)(7 - 3\lambda) = 5948$$-90 - 7 + 3\lambda = 5948$$-97 + 3\lambda = 5948$$3\lambda = 5948 + 97$$3\lambda = 6045$$\lambda = \frac{6045}{3} = 2015$Therefore,the correct option is $C$.

Class 12 Mathematics 3-4. Matrices and Determinants Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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If $\left| \begin{matrix} -6 & 1 & \lambda \\ 0 & 3 & 7 \\ -1 & 0 & 5 \end{matrix} \right| = 5948$,then $\lambda$ is:

A
$2011$
B
$2013$
C
$2015$
D
$2017$

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Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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If $\left| \begin{matrix} -6 & 1 & \lambda \\ 0 & 3 & 7 \\ -1 & 0 & 5 \end{matrix} \right| = 5948$,then $\lambda$ is:

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