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(B) matrix is singular if its determinant is equal to $0$.Given the matrix $A = \begin{bmatrix} 1 & 3 & \lambda + 2 \ 2 & 4 & 8 \ 3 & 5 & 10 \end{bm

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$4$

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(B) matrix is singular if its determinant is equal to $0$.Given the matrix $A = \begin{bmatrix} 1 & 3 & \lambda + 2 \ 2 & 4 & 8 \ 3 & 5 & 10 \end{bmatrix}$.Setting the determinant $|A| = 0$:$|A| = 1(4 	imes 10 - 8 	imes 5) - 3(2 	imes 10 - 8 	imes 3) + (\lambda + 2)(2 	imes 5 - 4 	imes 3) = 0$$|A| = 1(40 - 40) - 3(20 - 24) + (\lambda + 2)(10 - 12) = 0$$|A| = 1(0) - 3(-4) + (\lambda + 2)(-2) = 0$$0 + 12 - 2(\lambda + 2) = 0$$12 - 2\lambda - 4 = 0$$8 - 2\lambda = 0$$2\lambda = 8$$\lambda = 4$.

If the matrix $\begin{bmatrix} 1 & 3 & \lambda + 2 \\ 2 & 4 & 8 \\ 3 & 5 & 10 \end{bmatrix}$ is singular,then $\lambda = $

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