Question

The system of equations $x+y+z=6$, $x+2y+3z=10$,$x+2y+\lambda z=\mu$ has no solution if;

• A. $\lambda =3$,$\mu =10$
• B. $\lambda \ne 3$,$\mu =10$
• C. $\lambda \ne 3$,$\mu \ne 10$
• D. $\lambda =3$,$\mu \ne 10$

Answer 1

The correct option is D

$\lambda =3$,$\mu \ne 10$

Step$1$ : Write the given information

Given equations :

$x+y+z=6$ . . .(i)

$x+2y+3z=10$ . . .(ii)

$x+2y+\lambda z=\mu$ . . .(iii)

Step2: finding ' $\lambda$ ' value, $\Delta =0$

For no solution, $\Delta =0$

$\Rightarrow$ $\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 2& \lambda \end{array}\right|=0$

$\Rightarrow$ $1(2\lambda -6)–1(\lambda -3)+1(2-2)=0$

$\Rightarrow$ $2\lambda -6-\lambda +3=0$

$\Rightarrow$ $\lambda =3$

Step3: To find the value of ' $\mu$ '

Consider, ${\Delta }_1\ne 0$

$\Rightarrow$ $\left|\begin{array}{ccc}6& 1& 1\\ 10& 2& 3\\ \mu & 2& 3\end{array}\right|\ne 0$

$\Rightarrow$ $6(6-6)–1(30-3\mu )+1(20-2\mu )\ne 0$

$\Rightarrow$ $-30+3\mu +20-2\mu \ne 0$

$\Rightarrow$ $-10+\mu \ne 0$

$\Rightarrow$ $\mu \ne 10$

Hence, option (D) is the answer.