The system of linear equations x + lambda y - | vedclass

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

Question

(B) For a system of homogeneous linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero,i.e.,$\Delta =

Vedclass · Accepted Answer

exactly three values of $\lambda$

Vedclass · Answer

(B) For a system of homogeneous linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero,i.e.,$\Delta = 0$.The coefficient matrix is:$\Delta = \begin{vmatrix} 1 & \lambda & -1 \ \lambda & -1 & -1 \ 1 & 1 & -\lambda \end{vmatrix} = 0$Expanding the determinant along the first row:$1((-1)(-\lambda) - (-1)(1)) - \lambda((\lambda)(-\lambda) - (-1)(1)) - 1((\lambda)(1) - (-1)(1)) = 0$$1(\lambda + 1) - \lambda(-\lambda^2 + 1) - 1(\lambda + 1) = 0$$(\lambda + 1) - \lambda(1 - \lambda^2) - (\lambda + 1) = 0$$-\lambda(1 - \lambda)(1 + \lambda) = 0$$\lambda(1 - \lambda)(1 + \lambda) = 0$Solving for $\lambda$,we get $\lambda = 0, 1, -1$.Thus,there are exactly three values of $\lambda$ for which the system has a non-trivial solution.

The system of linear equations $x + \lambda y - z = 0, \lambda x - y - z = 0, x + y - \lambda z = 0$ has a non-trivial solution for:

Similar Questions

Find the equation of the line joining $A(1, 3)$ and $B(0, 0)$ using determinants and find $k$ if $D(k, 0)$ is a point such that the area of triangle $ABD$ is $3 \, \text{sq units}$.

Find the values of $k$ if the area of the triangle is $4$ square units and the vertices are $(k, 0), (4, 0), (0, 2)$.

The equation whose roots are also the roots of the equation $\left|\begin{array}{ccc}1 & -3 & 1 \\ 1 & 6 & 4 \\ 1 & 3x & x^2\end{array}\right|=0$ is

If $\left|\begin{array}{ccc}1+x & 1 & 1 \\ 1+y & 1+2 y & 1 \\ 1+z & 1+z & 1+3 z\end{array}\right| = 10 k x y z \left(3+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$,then $k = \text{ . . . . . . }$ (where $x, y, z \neq 0$ and $3+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \neq 0$).

The value of $\left|\begin{array}{lll}(a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1\end{array}\right|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module