The set of all values of lambda for which the | vedclass

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

Question

(D) The given system of linear equations is:$(2-\lambda)x_1 - 2x_2 + x_3 = 0$$2x_1 - (3+\lambda)x_2 + 2x_3 = 0$$-x_1 + 2x_2 - \lambda x_3 = 0$For a no

Vedclass · Accepted Answer

contains two elements

Vedclass · Answer

(D) The given system of linear equations is:$(2-\lambda)x_1 - 2x_2 + x_3 = 0$$2x_1 - (3+\lambda)x_2 + 2x_3 = 0$$-x_1 + 2x_2 - \lambda x_3 = 0$For a non-trivial solution,the determinant of the coefficient matrix must be zero:$\begin{vmatrix} 2-\lambda & -2 & 1 \ 2 & -(3+\lambda) & 2 \ -1 & 2 & -\lambda \end{vmatrix} = 0$Applying $R_1 	o R_1 + R_3$:$\begin{vmatrix} 1-\lambda & 0 & 1-\lambda \ 2 & -(3+\lambda) & 2 \ -1 & 2 & -\lambda \end{vmatrix} = 0$Taking $(1-\lambda)$ common from $R_1$:$(1-\lambda) \begin{vmatrix} 1 & 0 & 1 \ 2 & -(3+\lambda) & 2 \ -1 & 2 & -\lambda \end{vmatrix} = 0$Expanding along $R_1$:$(1-\lambda) [1(\lambda(3+\lambda) - 4) + 1(4 - (3+\lambda))] = 0$$(1-\lambda) [3\lambda + \lambda^2 - 4 + 4 - 3 - \lambda] = 0$$(1-\lambda) [\lambda^2 + 2\lambda - 3] = 0$$(1-\lambda)(\lambda+3)(\lambda-1) = 0$$-(1-\lambda)^2(\lambda+3) = 0$Thus,$\lambda = 1$ and $\lambda = -3$. The set of values is $\{1, -3\}$,which contains two elements.

The set of all values of $\lambda$ for which the system of linear equations $2x_1 - 2x_2 + x_3 = \lambda x_1$,$2x_1 - 3x_2 + 2x_3 = \lambda x_2$,and $-x_1 + 2x_2 = \lambda x_3$ has a non-trivial solution:

Similar Questions

The values of $x$ in the following determinant equation,$\left| \begin{array}{ccc} a+x & a-x & a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x \end{array} \right| = 0$ are

If $A_{\lambda} = \begin{bmatrix} \lambda & \lambda - 1 \\ \lambda - 1 & \lambda \end{bmatrix}; \lambda \in N$,then $|A_1| + |A_2| + \dots + |A_{300}|$ is equal to

The solutions of the equation $\left| \begin{array}{ccc} 1 & 1 & x \\ p+1 & p+1 & p+x \\ 3 & x+1 & x+2 \end{array} \right| = 0$ are:

If $\Delta=\left|\begin{array}{lll}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{array}\right|=K(a-b)(b-c)(c-a)$,then $K=$

If $A_\alpha = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$,then the determinant of $A_{\pi / 5} A_{\pi / 4} A_{3 \pi / 10}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module