Question

The system of linear equations \[3x - 2y - kz = 10,\,2x - 4y - 2z = 6\] and\[x + 2y - z = 5m\] is inconsistent if:
A. \[k = 3,m = \dfrac{4}{5}\]
B. \[k \ne 3,m \in \mathbb{R}\]
C. \[k \ne 3,m \ne \dfrac{4}{5}\]
D. \[k = 3,m \ne \dfrac{4}{5}\]

Answer 1

Hint: Firstly, we write the given system of equations in determinant form and using Cramer’s rule to get the condition of inconsistent. Now we use the condition of inconsistent to find the value of \[k\] and \[m\] .

Formula used:
Cramer’s Rule:
If the system of linear equations are
\[{a_1}x + {b_1}y + {c_1}z = {d_1}\] , where \[{a_1},{b_1},{c_1},{d_1}\] are constant values.
\[{a_2}x + {b_2}y + {c_2}z = {d_2}\], where \[{a_2},{b_2},{c_2},{d_2}\] are constant values.
\[{a_3}x + {b_3}y + {c_3}z = {d_3}\], where \[{a_3},{b_3},{c_3},{d_3}\] are constant values.
 . Then we will get solution as
\[x = \dfrac{{{D_1}}}{D},y = \dfrac{{{D_2}}}{D},z = \dfrac{{{D_3}}}{D}\]
where \[D = \left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}} \right|,{D_1} = \left| {\begin{array}{*{20}{c}}{{d_1}}&{{b_1}}&{{c_1}}\\{{d_2}}&{{b_2}}&{{c_2}}\\{{d_3}}&{{b_3}}&{{c_3}}\end{array}} \right|,{D_2} = \left| {\begin{array}{*{20}{c}}{{a_1}}&{{d_1}}&{{c_1}}\\{{a_2}}&{{d_2}}&{{c_2}}\\{{a_3}}&{{d_3}}&{{c_3}}\end{array}} \right|,{D_3} = \left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{d_1}}\\{{a_2}}&{{b_2}}&{{d_2}}\\{{a_3}}&{{b_3}}&{{d_3}}\end{array}} \right|\]
(i) If \[D \ne 0\], then the system of equations is consistent.
(ii) If \[D = 0\] and at least one of \[{D_1},{D_2},{D_3}\] is non-zero then the given system is inconsistent.
(iii) If \[D = 0\] and \[{D_1} = {D_2} = {D_3} = 0\] then the system is consistent.
Determinant property: If a matrix given \[A = \left[ {\begin{array}{*{20}{c}}a&b&c\\d&e&f\\g&h&i\end{array}} \right]\] then
\[\left| A \right| = a\left| {\begin{array}{*{20}{c}}e&f\\h&i\end{array}} \right| - b\left| {\begin{array}{*{20}{c}}d&f\\g&i\end{array}} \right| + c\left| {\begin{array}{*{20}{c}}d&e\\g&h\end{array}} \right|\]
\[ \Rightarrow \left| A \right| = a\left( {ei - fh} \right) - b\left( {di - gf} \right) + c\left( {dh - eg} \right)\]

Complete step by step solution:
Given system of equations
\[3x - 2y - kz = 10\]
\[2x - 4y - 2z = 6\]
\[x + 2y - z = 5m\]
Using Cramer’s rule, we get
\[D = \left| {\begin{array}{*{20}{c}}3&{ - 2}&{ - k}\\2&{ - 4}&{ - 2}\\1&2&{ - 1}\end{array}} \right|\] , \[{D_1} = \left| {\begin{array}{*{20}{c}}{10}&{ - 2}&{ - k}\\6&{ - 4}&{ - 2}\\{5m}&2&{ - 1}\end{array}} \right|\] , \[{D_2} = \left| {\begin{array}{*{20}{c}}3&{10}&{ - k}\\2&6&{ - 2}\\1&{5m}&{ - 1}\end{array}} \right|\] and \[{D_3} = \left| {\begin{array}{*{20}{c}}3&{ - 2}&{10}\\2&{ - 4}&6\\1&2&{5m}\end{array}} \right|\]
Now finding the determinant of \[D = \left| {\begin{array}{*{20}{c}}3&{ - 2}&{ - k}\\2&{ - 4}&{ - 2}\\1&2&{ - 1}\end{array}} \right|\]
\[D = 3\left| {\begin{array}{*{20}{c}}{ - 4}&{ - 2}\\2&{ - 1}\end{array}} \right| - ( - 2)\left| {\begin{array}{*{20}{c}}2&{ - 2}\\1&{ - 1}\end{array}} \right| + \left( { - k} \right)\left| {\begin{array}{*{20}{c}}2&{ - 4}\\1&2\end{array}} \right|\]
\[ \Rightarrow D = 3\left( {( - 4)( - 1) - 2 \times \left( { - 2} \right)} \right) + 2\left( {2 \times \left( { - 1} \right) - \left( { - 1} \right) \times 2} \right) - k\left( {2 \times 2 - 1 \times \left( { - 4} \right)} \right)\]
\[ \Rightarrow D = 3\left( {4 + 4} \right) + 2\left( { - 2 + 2} \right) - k\left( {4 + 4} \right)\]
\[ \Rightarrow D = 3 \times 8 + 2 \times 0 - k \times 8\]
\[ \Rightarrow D = 24 - k \times 8\] ………………..(1)
If this system of equations are inconsistent then \[D = 0\]
Equation the equation (1) with ā\[D = 0\] and we get
\[ \Rightarrow 24 - 8k = 0\]
\[ \Rightarrow 8k = 24\]
\[ \Rightarrow k = \dfrac{{24}}{8}\]
\[ \Rightarrow k = 3\]
Substitute the value of \[k\] in \[{D_1},{D_2},{D_3}\] and we get
\[{D_1} = \left| {\begin{array}{*{20}{c}}{10}&{ - 2}&{ - 3}\\6&{ - 4}&{ - 2}\\{5m}&2&{ - 1}\end{array}} \right|\] , \[{D_2} = \left| {\begin{array}{*{20}{c}}3&{10}&{ - 3}\\2&6&{ - 2}\\1&{5m}&{ - 1}\end{array}} \right|\] and \[{D_3} = \left| {\begin{array}{*{20}{c}}3&{ - 2}&{10}\\2&{ - 4}&6\\1&2&{5m}\end{array}} \right|\]
Since the system of equations is inconsistent then \[{D_1} \ne 0,{D_2} \ne 0,{D_3} \ne 0\]
\[{D_1} = \left| {\begin{array}{*{20}{c}}{10}&{ - 2}&{ - 3}\\6&{ - 4}&{ - 2}\\{5m}&2&{ - 1}\end{array}} \right|\]
\[\left| {\begin{array}{*{20}{c}}{10}&{ - 2}&{ - 3}\\6&{ - 4}&{ - 2}\\{5m}&2&{ - 1}\end{array}} \right| = 10\left| {\begin{array}{*{20}{c}}{ - 4}&{ - 2}\\2&{ - 1}\end{array}} \right| - \left( { - 2} \right)\left| {\begin{array}{*{20}{c}}6&{ - 2}\\{5m}&{ - 1}\end{array}} \right| + \left( { - 3} \right)\left| {\begin{array}{*{20}{c}}6&{ - 4}\\{5m}&2\end{array}} \right|\]
\[ = 10\left( {\left( { - 4} \right)\left( { - 1} \right) - 2 \times \left( { - 2} \right)} \right) + 2\left( {6 \times \left( { - 1} \right) - 5m \times \left( { - 2} \right)} \right) - 3\left( {6 \times 2 - 5m \times \left( { - 4} \right)} \right)\]
\[ = 10\left( {4 + 4} \right) + 2\left( { - 6 + 10m} \right) - 3\left( {12 + 20m} \right)\]
\[ = 10 \times 8 - 12 + 20m - 36 - 60m\]
\[ = 80 - 12 - 36 - 40m\]
\[ = 32 - 40m\]
\[ = 8(4 - 5m)\]
Now \[{D_1} \ne 0\]
\[ \Rightarrow 8(4 - 5m) \ne 0\]
\[ \Rightarrow 4 - 5m \ne 0\]
\[ \Rightarrow 5m \ne 4\]
\[ \Rightarrow m \ne \dfrac{4}{5}\]
Therefore, the value of \[k = 3\] and \[m \ne \dfrac{4}{5}\] .
Hence the correct option is option D.

Note: The system of linear equations has not always a solution. There may exist unique solutions or infinitely many solutions or no solution. If the system is consistent then the system has either a unique solution or infinitely many solutions. If the system is inconsistent then the system has no solution. So, Cramer’s rule plays an important part for checking inconsistency of the system of linear equations. Cramer’s rule is not applicable if the equations have greater than three variables, applicable for only up to three variables.