The system of equations lambda x + y + z = 0, | vedclass

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

Question

(A) For a system of homogeneous linear equations to have a non-zero solution,the determinant of the coefficient matrix must be equal to zero.The coeff

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) For a system of homogeneous linear equations to have a non-zero solution,the determinant of the coefficient matrix must be equal to zero.The coefficient matrix is given by:$A = \begin{bmatrix} \lambda & 1 & 1 \ -1 & \lambda & 1 \ -1 & -1 & \lambda \end{bmatrix}$Setting the determinant to zero:$\begin{vmatrix} \lambda & 1 & 1 \ -1 & \lambda & 1 \ -1 & -1 & \lambda \end{vmatrix} = 0$Expanding along the first row:$\lambda(\lambda^2 - (-1)) - 1(-\lambda - (-1)) + 1(1 - (-\lambda)) = 0$$\lambda(\lambda^2 + 1) - 1(-\lambda + 1) + 1(1 + \lambda) = 0$$\lambda^3 + \lambda + \lambda - 1 + 1 + \lambda = 0$$\lambda^3 + 3\lambda = 0$$\lambda(\lambda^2 + 3) = 0$This gives $\lambda = 0$ or $\lambda^2 = -3$. Since $\lambda$ must be a real value,$\lambda^2 = -3$ has no real solutions.Therefore,the only real value for $\lambda$ is $0$.

The system of equations $\lambda x + y + z = 0, -x + \lambda y + z = 0, -x - y + \lambda z = 0$ will have a non-zero solution if real values of $\lambda$ are given by

Similar Questions

The number of real values of $t$ such that the system of homogeneous equations
$\begin{aligned}
t x+(t+1) y+(t-1) z &=0 \\
(t+1) x+t y+(t+2) z &=0 \\
(t-1) x+(t+2) y+t z &=0
\end{aligned}$
has non-trivial solutions is

If $x=k$ satisfies the equation $\left|\begin{array}{ccc}x-2 & 3x-3 & 5x-5 \\ x-4 & 3x-9 & 5x-25 \\ x-8 & 3x-27 & 5x-125\end{array}\right|=0$, then $x=k$ also satisfies the equation

The number of positive integral solutions of the equation $\left| \begin{array}{ccc} 1 - \lambda & 2 & 1 \\ -3 & \lambda & -2 \\ 2 & -2 & 1 + \lambda \end{array} \right| = 0$ is

The set of all values of $\lambda$ for which the system of linear equations $x - 2y - 2z = \lambda x$,$x + 2y + z = \lambda y$,and $-x - y = \lambda z$ has non-zero solutions.

Let $A = \begin{bmatrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \end{bmatrix}$. If $|A|^2 = 25$, then $|\alpha|$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module