(D) system of linear homogeneous equations has a non-trivial solution if and only if the determinant of the coefficient matrix is equal to $0$.The coe

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(D) system of linear homogeneous equations has a non-trivial solution if and only if the determinant of the coefficient matrix is equal to $0$.The coefficient matrix is given by:$D = \begin{vmatrix} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 3 & -4 \end{vmatrix} = 0$Expanding the determinant along the first row:$1(k(-4) - 3(-2)) - k(3(-4) - 2(-2)) + 3(3(3) - 2(k)) = 0$$1(-4k + 6) - k(-12 + 4) + 3(9 - 2k) = 0$$-4k + 6 + 8k + 27 - 6k = 0$$-2k + 33 = 0$$2k = 33$$k = \frac{33}{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

Medium

The value of $k$ for which the set of equations $x + ky + 3z = 0, 3x + ky - 2z = 0, 2x + 3y - 4z = 0$ has a non-trivial solution over the set of rationals is

A
$15$
B
$31/2$
C
$16$
D
$33/2$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Mathematics Questions

Chapter

3 and 4 .Determinants and Matrices

Subtopic

Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

The value of the determinant $\left| \begin{array}{ccc} 4 & -6 & 1 \\ -1 & -1 & 1 \\ -4 & 11 & -1 \end{array} \right|$ is

Easy

View Solution

The value of $x$ such that the matrix $\begin{bmatrix} x & 2 & 3 \\ 4 & 5 & 6 \\ 2 & 3 & 5 \end{bmatrix}$ is not invertible is

EasyMHT CET 2020

View Solution

If $k > 1$ and the determinant of the matrix $A^2$,where $A = \begin{bmatrix} k & k\alpha & \alpha \\ 0 & \alpha & k\alpha \\ 0 & 0 & k \end{bmatrix}$,is $k^2$,then $|\alpha|$ is equal to

MediumTS EAMCET 2014

View Solution

If $a, b$ and $c$ are real numbers,and $\Delta=\begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix}=0$,show that either $a+b+c=0$ or $a=b=c$.

Difficult

View Solution

If $f(x) = \begin{vmatrix} 1 & x & x+1 \\ 2x & x(x-1) & (x+1)x \\ 3x(x-1) & x(x-1)(x-2) & (x+1)x(x-1) \end{vmatrix}$,then $f(100)$ is equal to:

MediumWBJEE 2015

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

The value of $k$ for which the set of equations $x + ky + 3z = 0, 3x + ky - 2z = 0, 2x + 3y - 4z = 0$ has a non-trivial solution over the set of rationals is

Similar Questions

The value of the determinant $\left| \begin{array}{ccc} 4 & -6 & 1 \\ -1 & -1 & 1 \\ -4 & 11 & -1 \end{array} \right|$ is

The value of $x$ such that the matrix $\begin{bmatrix} x & 2 & 3 \\ 4 & 5 & 6 \\ 2 & 3 & 5 \end{bmatrix}$ is not invertible is

If $k > 1$ and the determinant of the matrix $A^2$,where $A = \begin{bmatrix} k & k\alpha & \alpha \\ 0 & \alpha & k\alpha \\ 0 & 0 & k \end{bmatrix}$,is $k^2$,then $|\alpha|$ is equal to

If $a, b$ and $c$ are real numbers,and $\Delta=\begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix}=0$,show that either $a+b+c=0$ or $a=b=c$.

If $f(x) = \begin{vmatrix} 1 & x & x+1 \\ 2x & x(x-1) & (x+1)x \\ 3x(x-1) & x(x-1)(x-2) & (x+1)x(x-1) \end{vmatrix}$,then $f(100)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module