(D) Let $\Delta = \left|\begin{array}{ccc}x+2 & x+3 & x+5 \\ x+4 & x+6 & x+9 \\ x+8 & x+11 & x+15\end{array}\right|$.Apply operations $R_2 \rightarrow

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(D) Let $\Delta = \left|\begin{array}{ccc}x+2 & x+3 & x+5 \\ x+4 & x+6 & x+9 \\ x+8 & x+11 & x+15\end{array}\right|$.Apply operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$,we get:$\Delta = \left|\begin{array}{ccc}x+2 & x+3 & x+5 \\ 2 & 3 & 4 \\ 6 & 8 & 10\end{array}\right|$.Apply operation $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1$,we get:$\Delta = \left|\begin{array}{ccc}x+2 & 1 & 3 \\ 2 & 1 & 2 \\ 6 & 2 & 4\end{array}\right|$.Expand along $R_1$:$\Delta = (x+2)(4-4) - 1(8-12) + 3(4-6)$$\Delta = (x+2)(0) - 1(-4) + 3(-2)$$\Delta = 0 + 4 - 6 = -2$.

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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$\left|\begin{array}{ccc}x+2 & x+3 & x+5 \\ x+4 & x+6 & x+9 \\ x+8 & x+11 & x+15\end{array}\right|$ is equal to

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A
$3x^2+4x+5$
B
$x^3+8x+2$
C
$0$
D
$-2$

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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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Let $A(a, 0)$,$B(b, 2b+1)$,and $C(0, b)$,where $b \neq 0$ and $|b| \neq 1$,be points such that the area of triangle $ABC$ is $1 \, \text{sq. unit}$. Then,the sum of all possible values of $a$ is:

DifficultJEE MAIN 2021

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DifficultJEE MAIN 2024

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If $A_n = \begin{bmatrix} 1-n & n \\ n & 1-n \end{bmatrix}$,then $|A_1| + |A_2| + \dots + |A_{2021}| = $

MediumKCET 2022

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If $a, b, c$ are distinct positive real numbers,then the value of the determinant $\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ is

EasyAP EAMCET 2016

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If the area of the triangle with vertices $(x, 0), (1, 1)$ and $(0, 2)$ is $4$ square units,then a value of $x$ is

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$\left|\begin{array}{ccc}x+2 & x+3 & x+5 \\ x+4 & x+6 & x+9 \\ x+8 & x+11 & x+15\end{array}\right|$ is equal to

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Let $A(a, 0)$,$B(b, 2b+1)$,and $C(0, b)$,where $b \neq 0$ and $|b| \neq 1$,be points such that the area of triangle $ABC$ is $1 \, \text{sq. unit}$. Then,the sum of all possible values of $a$ is:

If $\alpha \neq a, \beta \neq b, \gamma \neq c$ and $\left|\begin{array}{lll}\alpha & b & c \\ a & \beta & c \\ a & b & \gamma\end{array}\right|=0$,then $\frac{a}{\alpha-a}+\frac{b}{\beta-b}+\frac{\gamma}{\gamma-c}$ is equal to :

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