Question

Assertion :Consider the system 'S' of equations: $x+y+z=6,x+2y+3z=10,x+2y+\lambda z=\mu$, where $\lambda ,\mu ϵ$ {1,2,3,....., 10}. Reason: Number of ways of choosing $\lambda$ & $\mu$ so that system 'S' has a unique solution is 100.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is D Assertion is incorrect but Reason is correct

writing in matrix form

$\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 2& \lambda \end{array}\right|\left|\begin{array}{c}x\\ y\\ z\end{array}\right|\left|\begin{array}{c}6\\ 10\\ \mu \end{array}\right|△=\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 2& \lambda \end{array}\right|\left(1\right)(2x-6)-1(\lambda -3)+1(2-2)=2\lambda -6-\lambda +3=(\lambda -3){△}_x=\left|\begin{array}{ccc}6& 1& 1\\ 10& 2& 3\\ \mu & 2& \lambda \end{array}\right|=(2\lambda -6)(10\lambda -3\mu )(20-2\mu )=12\lambda -36-10\lambda +3\mu +20-2\mu =2\lambda +\mu -16{△}_y=\left|\begin{array}{ccc}1& 6& 1\\ 1& 10& 3\\ 1& \mu & \lambda \end{array}\right|=1(10\lambda -3\mu )-6(\lambda -3)+1(\mu -10)=10\lambda -3\mu -6\lambda +18+\mu -10=7\lambda -24+8{△}_z=\left|\begin{array}{ccc}1& 1& 6\\ 1& 2& 10\\ 1& 2& \mu \end{array}\right|=1(2\mu -20)-1(\mu -10)+6(2-2)=2\mu -20-\mu +10=(\mu -10)$

Assertion is correct and reason is incorrect

For infinite solution$△=0{△}_x{=△}_y{=△}_z$

i.e.$\lambda =3,\mu =10$

For unique solution $△\ne 0$ i.e.$\lambda \ne 3$

Total number of $(\lambda +\mu )$ pairs are $10x10=100$

neglecting $\lambda$=$3$ case we have=$100-10=90$