Question

Which of the following are correct in respect of the system of equations $x+y+z=8,x-y+2z=6$ and $3x-y+5z=k$?
1. They have no solution, if $k=15$.
2. They have infinitely many solutions, if $k=20$.
3. They have unique solution, if $k=25$.
Select the correct answer using the code given below

• A. 1 and 2 only
• B. 2 and 3 only
• C. 1 and 3 only
• D. 1, 2 and 3

Answer 1

Answer: (A)

1 and 2 only

First we need to know if the above equations are linearly dependent or no,

In order to figure that out lets find the determinant $D$ which is;

$D=\left|\begin{array}{ccc}1& 1& 1\\ 1& -1& 2\\ 3& -1& 5\end{array}\right|=0$

as $D=0$ the equations are linearly dependent, which means the system of equations can either be inconsistent or have infinitely many solutions. That shall depend on the value of k.

For there to be inifinitely many solutions , values of $D_1,D_2,D_3$ should be $0$ , defined as;

$D_1=\left|\begin{array}{ccc}8& 1& 1\\ 6& -1& 2\\ k& -1& 5\end{array}\right|D_2=\left|\begin{array}{ccc}1& 8& 1\\ 1& 6& 2\\ 3& k& 5\end{array}\right|D_3=\left|\begin{array}{ccc}1& 1& 8\\ 1& -1& 6\\ 3& -1& k\end{array}\right|$

which on solving you get;

$D_1=\left|\begin{array}{ccc}8& 1& 1\\ 6& -1& 2\\ k& -1& 5\end{array}\right|=3k-60=3(k-20)D_2=\left|\begin{array}{ccc}1& 8& 1\\ 1& 6& 2\\ 3& k& 5\end{array}\right|=k-20D_3=\left|\begin{array}{ccc}1& 1& 8\\ 1& -1& 6\\ 3& -1& k\end{array}\right|=40-2k=-2(k-20)$

Now for all the Determinants to be zero , its evident that $k=20$

Hence statement 2 is correct.

Subsequently for there to be no solution $k\ne 20$, hence statement 1 is correct as well,

As $D$ is zero the system of equations CANNOT have a unique solution hence, statement 3 is WRONG.