Let S be the set of all lambda in mathbb{R} f | vedclass

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

Question

(C) For a system of linear equations to have no solution,the determinant of the coefficient matrix $D$ must be $0$,and at least one of the Cramer's de

Vedclass · Accepted Answer

contains exactly two elements.

Vedclass · Answer

(C) For a system of linear equations to have no solution,the determinant of the coefficient matrix $D$ must be $0$,and at least one of the Cramer's determinants $(D_x, D_y, D_z)$ must be non-zero.The determinant $D$ is given by:$D = \begin{vmatrix} 2 & -1 & 2 \ 1 & -2 & \lambda \ 1 & \lambda & 1 \end{vmatrix} = 2(-2 - \lambda^2) + 1(1 - \lambda) + 2(\lambda + 2) = -4 - 2\lambda^2 + 1 - \lambda + 2\lambda + 4 = -2\lambda^2 + \lambda + 1$Setting $D = 0$:$-2\lambda^2 + \lambda + 1 = 0 \Rightarrow 2\lambda^2 - \lambda - 1 = 0 \Rightarrow (2\lambda + 1)(\lambda - 1) = 0$Thus,$\lambda = 1$ or $\lambda = -\frac{1}{2}$.Now,check $D_x$ for these values:$D_x = \begin{vmatrix} 2 & -1 & 2 \ -4 & -2 & \lambda \ 4 & \lambda & 1 \end{vmatrix} = 2(-2 - \lambda^2) + 1(-4 - 4\lambda) + 2(-4\lambda + 8) = -4 - 2\lambda^2 - 4 - 4\lambda - 8\lambda + 16 = -2\lambda^2 - 12\lambda + 8$For $\lambda = 1$,$D_x = -2 - 12 + 8 = -6 
eq 0$.For $\lambda = -\frac{1}{2}$,$D_x = -2(\frac{1}{4}) - 12(-\frac{1}{2}) + 8 = -0.5 + 6 + 8 = 13.5 
eq 0$.Since $D=0$ and $D_x 
eq 0$ for both values,the system has no solution for both $\lambda = 1$ and $\lambda = -\frac{1}{2}$.Therefore,$S = \{1, -\frac{1}{2}\}$,which contains exactly two elements.

Let $S$ be the set of all $\lambda \in \mathbb{R}$ for which the system of linear equations
$2x - y + 2z = 2$
$x - 2y + \lambda z = -4$
$x + \lambda y + z = 4$
has no solution. Then the set $S$

Similar Questions

Consider the following system of equations in matrix form $\begin{bmatrix} 1 \\ 2 \\ \lambda \end{bmatrix} \begin{bmatrix} 1 & 2 & \lambda \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = 0$. Then which one of the following statements is true?

If the system of linear equations $2x + 2y + 3z = a$,$3x - y + 5z = b$,and $x - 3y + 2z = c$,where $a, b, c$ are non-zero real numbers,has more than one solution,then:

If the system of simultaneous linear equations $3x - 4y + kz + 13 = 0$,$x + 2y - z - 9 = 0$,and $kx - y + 3z + 7 = 0$ has a unique solution $x = \alpha, y = \beta, z = \gamma$ for $k \neq m$ and $2\beta - \gamma = 8$,then $\alpha + m =$

The system of equations $x + y + z = 6$,$x + 2y + 3z = 10$,and $x + 2y + \lambda z = \mu$ has no solution for:

The system of equations $\lambda x - y + (\cos\theta) z = 0$,$3x + y + 2z = 0$,and $(\cos\theta) x + y + 2z = 0$ for $0 < \theta < 2\pi$ has non-trivial solution$(s)$:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Let $S$ be the set of all $\lambda \in \mathbb{R}$ for which the system of linear equations$2x - y + 2z = 2$$x - 2y + \lambda z = -4$$x + \lambda y + z = 4$has no solution. Then the set $S$