Consider the system of linear equations x+y+z | vedclass

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

Question

(A) The system of equations is:$x+y+z=4\mu$$x+2y+2\lambda z=10\mu$$x+3y+4\lambda^2 z=\mu^2+15$The determinant of the coefficient matrix $\Delta$ is:$\

Vedclass · Accepted Answer

The system has a unique solution if $\lambda 
eq \frac{1}{2}$.

Vedclass · Answer

(A) The system of equations is:$x+y+z=4\mu$$x+2y+2\lambda z=10\mu$$x+3y+4\lambda^2 z=\mu^2+15$The determinant of the coefficient matrix $\Delta$ is:$\Delta = \begin{vmatrix} 1 & 1 & 1 \ 1 & 2 & 2\lambda \ 1 & 3 & 4\lambda^2 \end{vmatrix} = 1(8\lambda^2 - 6\lambda) - 1(4\lambda^2 - 2\lambda) + 1(3 - 2) = 4\lambda^2 - 4\lambda + 1 = (2\lambda - 1)^2$.For a unique solution,$\Delta 
eq 0$,which implies $(2\lambda - 1)^2 
eq 0$,so $\lambda 
eq \frac{1}{2}$. Thus,option $A$ and $D$ are consistent with this.When $\lambda = \frac{1}{2}$,$\Delta = 0$. We check for consistency using Cramer's rule or augmented matrix reduction. The augmented matrix is:$\begin{bmatrix} 1 & 1 & 1 & | & 4\mu \ 1 & 2 & 1 & | & 10\mu \ 1 & 3 & 1 & | & \mu^2+15 \end{bmatrix}$Subtracting $R_1$ from $R_2$ and $R_3$:$\begin{bmatrix} 1 & 1 & 1 & | & 4\mu \ 0 & 1 & 0 & | & 6\mu \ 0 & 2 & 0 & | & \mu^2+15-4\mu \end{bmatrix}$From $R_2$,$y = 6\mu$. Substituting into $R_3$: $2(6\mu) = \mu^2 - 4\mu + 15 \implies \mu^2 - 16\mu + 15 = 0 \implies (\mu-1)(\mu-15) = 0$.So,if $\lambda = \frac{1}{2}$ and $\mu \in \{1, 15\}$,the system is consistent (infinite solutions). If $\mu 
eq 1, 15$,the system is inconsistent.Comparing with the options,option $A$ is the incorrect statement because the condition for a unique solution depends only on $\lambda 
eq \frac{1}{2}$,regardless of $\mu$.

Consider the system of linear equations $x+y+z=4\mu$,$x+2y+2\lambda z=10\mu$,and $x+3y+4\lambda^2 z=\mu^2+15$,where $\lambda, \mu \in \mathbb{R}$. Which one of the following statements is $NOT$ correct?

Similar Questions

The number of solutions of the equations $x + 4y - z = 0,$ $3x - 4y - z = 0,$ and $x - 3y + z = 0$ is

If $A=\begin{bmatrix} x & y & y \\ y & x & y \\ y & y & x \end{bmatrix}$ is a matrix such that $5 A^{-1}=\begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{bmatrix}$,then $A^2-4 A=$

If the solution for the system of equations $x+2y-z=3$,$3x-y+2z=1$ and $2x-2y+3z=2$ is $(\alpha, \beta, \gamma)$,then $\alpha^2+\beta^2+\gamma^2=$

Given the system of linear equations: $2x + 3y + 4z = 9$,$4x + 9y + 3z = 10$,and $5x + 10y + 5z = 11$. The value of $x$ is given by:

Let $\alpha_1, \alpha_2$ be two values of $\alpha$ for which the system $2\alpha x + y = 5$,$x - 6y = \alpha$,and $x + y = 2$ is consistent. Then $|2(\alpha_1 + \alpha_2)|$ is -

Vedclass Test Series

Exam Paper Generator

Online Exam Module