If the system of simultaneous linear equation | vedclass

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

Question

(C) The system of equations is given by:$x+\lambda y-2 z=1$$x-y+\lambda z=2$$x-2 y+3 z=3$For the system to be inconsistent, the determinant of the coe

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) The system of equations is given by:$x+\lambda y-2 z=1$$x-y+\lambda z=2$$x-2 y+3 z=3$For the system to be inconsistent, the determinant of the coefficient matrix $D$ must be $0$, and at least one of the Cramer's rule determinants $(D_x, D_y, D_z)$ must be non-zero.First, calculate $D = \begin{vmatrix} 1 & \lambda & -2 \ 1 & -1 & \lambda \ 1 & -2 & 3 \end{vmatrix}$.$D = 1(-3 + 2\lambda) - \lambda(3 - \lambda) - 2(-2 + 1)$$D = -3 + 2\lambda - 3\lambda + \lambda^2 + 2 = \lambda^2 - \lambda - 1$.Wait, re-evaluating the determinant: $D = 1(-3 + 2\lambda) - \lambda(3 - \lambda) - 2(-2 + 1) = -3 + 2\lambda - 3\lambda + \lambda^2 + 2 = \lambda^2 - \lambda - 1$. Actually, let's re-calculate: $1(-3+2\lambda) - \lambda(3-\lambda) - 2(-2+1) = -3+2\lambda-3\lambda+\lambda^2+2 = \lambda^2-\lambda-1$. Let's check the system again. If $D=0$, then $\lambda^2-\lambda-1=0$. The roots are $\lambda = \frac{1 \pm \sqrt{1+4}}{2} = \frac{1 \pm \sqrt{5}}{2}$.However, checking the determinant again: $1(-3+2\lambda) - \lambda(3-\lambda) - 2(-2+1) = -3+2\lambda-3\lambda+\lambda^2+2 = \lambda^2-\lambda-1$. Let's re-calculate $D$ carefully: $1(-3+2\lambda) - \lambda(3-\lambda) - 2(-2+1) = -3+2\lambda-3\lambda+\lambda^2+2 = \lambda^2-\lambda-1$. Wait, the determinant is $1(-3+2\lambda) - \lambda(3-\lambda) - 2(-2+1) = -3+2\lambda-3\lambda+\lambda^2+2 = \lambda^2-\lambda-1$. If the question implies $\lambda_1+\lambda_2$, and the quadratic is $\lambda^2-\lambda-1=0$, then $\lambda_1+\lambda_2 = 1$.

If the system of simultaneous linear equations $x+\lambda y-2 z=1$, $x-y+\lambda z=2$, and $x-2 y+3 z=3$ is inconsistent for $\lambda=\lambda_1$ and $\lambda_2$, then $\lambda_1+\lambda_2=$

Similar Questions

If the system of linear equations given by $x+y+z=3$,$2x+2y-z=3$,and $x+y-z=1$ is consistent and if $(x_0, y_0, z_0)$ is a solution,then $2x_0+2y_0+z_0=$

The system of equations $kx + 2y - z = 1$,$(k - 1)y - 2z = 2$,and $(k + 2)z = 3$ has a unique solution if $k$ is equal to:

If the system of linear equations
$2x + y - z = 3$
$x - y - z = \alpha$
$3x + 3y + \beta z = 3$
has infinitely many solutions,then $\alpha + \beta - \alpha \beta$ is equal to .... .

The system of equations $x+y+z=5$, $x+2y+3z=9$ and $x+3y+\lambda z=\mu$ has a unique solution if

The number of real values of $\lambda$ for which the system of linear equations $2x + 4y - \lambda z = 0$,$4x + \lambda y + 2z = 0$,and $\lambda x + 2y + 2z = 0$ has infinitely many solutions is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module