(B) To evaluate the determinant $\left|\begin{array}{cc}x & x+1 \\ x-1 & x\end{array}\right|$,we use the formula for a $2 \times 2$ determinant: $\lef

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(B) To evaluate the determinant $\left|\begin{array}{cc}x & x+1 \\ x-1 & x\end{array}\right|$,we use the formula for a $2 \times 2$ determinant: $\left|\begin{array}{cc}a & b \\ c & d\end{array}\right| = ad - bc$.Applying this to our expression:$= x(x) - (x+1)(x-1)$$= x^2 - (x^2 - 1)$$= x^2 - x^2 + 1$$= 1$

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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Evaluate $\left|\begin{array}{cc}x & x+1 \\ x-1 & x\end{array}\right|$

A
$0$
B
$1$
C
$2$
D
$3$

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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

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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The value of $x$ obtained from the equation $\left| \begin{array}{ccc} x + \alpha & \beta & \gamma \\ \gamma & x + \beta & \alpha \\ \alpha & \beta & x + \gamma \end{array} \right| = 0$ is:

Medium

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$f(x)$ is an $n^{\text{th}}$ degree polynomial satisfying $f(x) = \frac{1}{2} \begin{vmatrix} f(x) & f(\frac{1}{x}) - f(x) \\ 1 & f(\frac{1}{x}) \end{vmatrix}$. If $f(2) = 33$,then the value of $f(3)$ is

DifficultAP EAMCET 2025

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If $A_{\lambda} = \begin{bmatrix} \lambda & \lambda - 1 \\ \lambda - 1 & \lambda \end{bmatrix}; \lambda \in N$,then $|A_1| + |A_2| + \dots + |A_{300}|$ is equal to

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Evaluate $\left|\begin{array}{rr}2 & 4 \\ -1 & 2\end{array}\right|$

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Evaluate $\left|\begin{array}{cc}x & x+1 \\ x-1 & x\end{array}\right|$

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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:

The value of $x$ obtained from the equation $\left| \begin{array}{ccc} x + \alpha & \beta & \gamma \\ \gamma & x + \beta & \alpha \\ \alpha & \beta & x + \gamma \end{array} \right| = 0$ is:

$f(x)$ is an $n^{\text{th}}$ degree polynomial satisfying $f(x) = \frac{1}{2} \begin{vmatrix} f(x) & f(\frac{1}{x}) - f(x) \\ 1 & f(\frac{1}{x}) \end{vmatrix}$. If $f(2) = 33$,then the value of $f(3)$ is

If $A_{\lambda} = \begin{bmatrix} \lambda & \lambda - 1 \\ \lambda - 1 & \lambda \end{bmatrix}; \lambda \in N$,then $|A_1| + |A_2| + \dots + |A_{300}|$ is equal to

Evaluate $\left|\begin{array}{rr}2 & 4 \\ -1 & 2\end{array}\right|$

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