Exams › JEE Main › Maths
Let D be the determinant of the matrix | 1 1 1 | | 1 1+x 1 | | 1 1 1+y | where x ≠ 0 and y ≠ 0. Then D is
- divisible by x, but not by y
- divisible by y, but not by x
- divisible by neither x nor y
- divisible by both x and y
Correct answer: divisible by both x and y
Solution
The determinant D can be computed using properties of determinants and shows that it contains terms involving both x and y, indicating that it is divisible by both variables. This is due to the structure of the matrix, where the presence of x and y in the second and third rows leads to terms that include these variables when expanded.
Related JEE Main Maths questions
- In a triangle $ABC$, if the determinant $\begin{vmatrix}\sin A & \sin^2 A\\ \sin B & \sin^2 B\\ \sin C & \sin^2 C\end{vmatrix}=0$, then the triangle must be
- If $[\,\cdot\,]$ represents the greatest integer less than or equal to a real number, and $-1\le x<0$, $0\le y<1$, $1\le z<2$, then the determinant \[ \begin{vmatrix} [x] & [y] & [z] \\ [x+1] & [y+1] & [z] \\ [x] & [y] & [z]+1 \end{vmatrix} \] is equal to:
- Let $x$, $y$, and $z$ be complex numbers. If $\Delta$ denotes the determinant of the matrix \[ \begin{bmatrix} 0 & -y & -z\\ y & 0 & -x\\ z & x & 0 \end{bmatrix}, \] then $\Delta$ is
- For the linear system \[ \begin{aligned} x_1+2x_2+3x_3&=6\\ x_1+3x_2+5x_3&=9\\ 2x_1+5x_2+ax_3&=b \end{aligned} \] if it is consistent and admits infinitely many solutions, then which statement must be true?
- For which value(s) of $x$ does the determinant of the matrix \[ \begin{vmatrix} 1 & x-3 & (x-3)^2\\ 1 & x-4 & (x-4)^2\\ 1 & x-5 & (x-5)^2 \end{vmatrix} \] become zero?
- Let $\Delta_r$ denote the determinant \[ \Delta_r=\begin{vmatrix} \binom{m}{r} & 1 & 1\\ m^2-1 & 2^m & m+1\\ \sin^2(m^2) & \sin^2(m) & \sin^2(m+1) \end{vmatrix}. \] Then the value of $\sum_{r=0}^{m} \Delta_r$ is
⚔️ Practice JEE Main Maths free + battle 1v1 →