Consider the system of linear equations x+y+z=4 ,x+2y+2 z=10 ,x+3y+4 ^2z= ^2+15 where , R.

Question

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

Accepted Answer

$x+y+z=4\mu ,x+2y+2\lambda z=10\mu ,x+3y+4\lambda {}^{2}z={\mu }^{2}+15$,

$\Delta=\begin{vmatrix} 1 \quad  1 \quad  1 \ 1 \quad  2 \quad  2\lambda \ 1 \quad  3 \quad  4{\lambda }^{2} \end{vmatrix}=(2\lambda -1{)}^{2}$

For unique solution $\Delta
e 0,2\lambda -1
e 0,(\lambda 
e \frac{1}{2})$

Let $\Delta=0,\lambda =\frac{1}{2}$

$\begin{gathered} {\Delta}_{y}=0,{\Delta}_{x}={\Delta}_{z}=|\begin{array}{ccc}4\mu  \quad  1 \quad  1 \ 10\mu  \quad  2 \quad  1 \ {\mu }^{2}+15 \quad  3 \quad  1\end{array}| \ =(\mu -15)(\mu -1) \end{gathered}$

For infinite solution $\lambda =\frac{1}{2},\mu =1$ or 15

Solutions · 1