Question

Find the value of $k$ for which the simultaneous equation $x+y+z=3;x+2y+3z=4$ and $x+4y+kz=6$ will not have a unique solution

• A. $0$
• B. $5$
• C. $6$
• D. $7$

Answer 1

Answer: (D)

$7$

The given system of equations will be consistent with unique solution, when
$\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 4& k\end{array}\right|\ne 0$
$⟹1(2k-12)+1(3-k)+1(4-2)\ne 0$
$⟹k-12+3+2\ne 0\Rightarrow k-7\ne 0\Rightarrow k\ne 7$

Hence, option D is correct.