Mathematics - Determinant Question with Solution | TestHub

Given equations are

$$\begin{gathered} x+y+z=1 \\ 2x+Ny+2z=2 \\ 3x+3y+Nz=3 \end{gathered}$$

So,

$∆=\left|\begin{array}{ccc}1& 1& 1\\ 2& N& 2\\ 3& 3& N\end{array}\right|$

Applying $C_2\leftrightarrow C_2-C_1$ and $C_3\leftrightarrow C_3-C_1$

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

$\Rightarrow ∆=\left(N-2\right)\left(N-3\right)$

For unique solution $∆\ne 0$, so

$N\ne 2,3$

Also, $N\in \left\{1,2,3,4,5,6\right\}$ and for unique solution $N$ can take values $\left\{1,4,5,6\right\}$.

$\Rightarrow P$(system has unique solution ) $=\frac{4}{6}=\frac{k}{6}$

So, $k=4$

Therefore, required sum is

$=4+1+4+5+6=20$