(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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(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 \times 10 - 8 \times 5) - 3(2 \times 10 - 8 \times 3) + (\lambda + 2)(2 \times 5 - 4 \times 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$.

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

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

A
$-2$
B
$4$
C
$2$
D
$-4$

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3 and 4 .Determinants and Matrices

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

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Let $S = \left\{ A = \begin{bmatrix} 0 & 1 & c \\ 1 & a & d \\ 1 & b & e \end{bmatrix} : a, b, c, d, e \in \{0, 1\} \text{ and } |A| \in \{-1, 1\} \right\}$,where $|A|$ denotes the determinant of $A$. Then the number of elements in $S$ is:

If the system of equations $ (k+1)^3 x + (k+2)^3 y = (k+3)^3 $, $ (k+1) x + (k+2) y = k+3 $, and $ x + y = 1 $ is consistent, then the value of $ k $ is:

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