Matrices and Determinants 4 Question 29

30. Show that the system of equations, $3 x-y+4 z=3$, $x+2 y-3 z=-2$ and $6 x+5 y+\lambda z=-3$ has atleast one solution for any real number $\lambda \neq-5$. Find the set of solutions, if $\lambda=-5$.

$(1983,5 \mathrm{M})$

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Answer:

Correct Answer: 30. ( $k=0$, the given system has infinitely many solutions)

Solution:

  1. The given system of equations

$ \begin{gathered} 3 x-y+4 z=3 \\ x+2 y-3 z=-2 \\ 6 x+5 y+\lambda z=-3 \end{gathered} $

has atleast one solution, if $\Delta \neq 0$

$ \therefore \quad \Delta=\left|\begin{array}{ccc} 3 & -1 & 4 \\ 1 & 2 & -3 \\ 6 & 5 & \lambda \end{array}\right| \neq 0 $

$ \begin{array}{lc} \Rightarrow & 3(2 \lambda+15)+1(\lambda+18)+4(5-12) \neq 0 \\ \Rightarrow & 7(\lambda+5) \neq 0 \\ \Rightarrow & \lambda \neq-5 \end{array} $

Let $z=-k$, then equations become

and $\quad \begin{gathered}3 x-y=3-4 k \\ x+2 y=3 k-2\end{gathered}$

On solving, we get

$ x=\frac{4-5 k}{7}, y=\frac{13 k-9}{7}, z=k $



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