Question

The system of linear equations: $3x-2y-kz=10$, $2x-4y-2z=6$ and $x+2y-z=5m$ is inconsistent if:

• A. $k=3$, $m=\frac{4}{5}$
• B. $k\ne 3$, $m\in \mathrm{ℝ}$
• C. $k\ne 3$, $m\ne \frac{4}{5}$
• D. $k=3$, $m\ne \frac{4}{5}$

Answer 1

The correct option is D

$k=3$, $m\ne \frac{4}{5}$

Explanation for the correct option:

Given equations are $3x-2y-kz=10$, $2x-4y-2z=6$ and $x+2y-z=5m$.

For equations $a_{1}x+b_{1}y+c_{1}z=d_1$, $a_{2}x+b_{2}y+c_{2}z=d_2$ and $a_{3}x+b_{3}y+c_{3}z=d_3$, we difine

$\mathrm{\Delta }=\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|,{\mathrm{\Delta }}_1=\left|\begin{array}{ccc}{d}_1& b_1& c_1\\ d_2& b_2& c_2\\ d_3& b_3& c_3\end{array}\right|,{\mathrm{\Delta }}_2=\left|\begin{array}{ccc}{a}_1& d_1& c_1\\ a_2& d_2& c_2\\ a_3& d_3& c_3\end{array}\right|,{\mathrm{\Delta }}_3=\left|\begin{array}{ccc}{a}_1& b_1& d_1\\ a_2& b_2& d_2\\ a_3& b_3& d_3\end{array}\right|$

If, $\mathrm{\Delta }=0$ and ${\mathrm{\Delta }}_1\ne 0$ or ${\mathrm{\Delta }}_2\ne 0$ or ${\mathrm{\Delta }}_3\ne 0$, then the system of equations are inconsistent and have no solutions.

Here,

$$\begin{gathered} \left(a_1,b_1,c_1,d_1\right)=\left(3,-2,-k,10\right) \\ \left(a_2,b_2,c_2,d_2\right)=\left(2,-4,-2,6\right) \\ \left(a_3,b_3,c_3,d_3\right)=\left(1,2,-1,5m\right) \end{gathered}$$

Thus, if

$\begin{array}{cc}\mathrm{\Delta }=& \left|\begin{array}{ccc}3& -2& -\mathrm{k}\\ 2& -4& -2\\ 1& 2& -1\end{array}\right|=0\\ \Rightarrow & 3\left(4+4\right)+2\left(-2+2\right)-k\left(4+4\right)=0\\ \Rightarrow & 24+0-8k=0\\ \Rightarrow & k=3\end{array}$

And if,

$\begin{array}{cc}{\mathrm{\Delta }}_3=& \left|\begin{array}{ccc}3& -2& 10\\ 2& -4& 6\\ 1& 2& 5m\end{array}\right|\ne 0\\ \Rightarrow & 3\left(-20m-12\right)+2\left(10m-6\right)+10\left(4+4\right)\ne 0\\ \Rightarrow & -60m-36+20m-12+80\ne 0\\ \Rightarrow & -40m\ne -32\\ \Rightarrow & m\ne \frac{4}{5}\end{array}$

Therefore, for the given set of equations to be inconsistent, it must be that $k=3$ and $m\ne \frac{4}{5}$

Hence, option D is correct.