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

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(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?

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