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:
- 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 $
