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Investigate for what values of \[\lambda ,\mu \] simultaneous equations \[x + y + z = 6,x + 2y + 3z = 10,x + 2y + \lambda z = \mu \] have
(i) No solution (ii) a unique solution (iii) an infinite solution

Answer Verified Verified
Hint: We solve this question by writing the system of linear of equations in matrix form and then solving the determinant, then using the definition of system of linear equation having unique solution, no solution and infinite solution evaluate for the values of \[\lambda ,\mu \].
* A system of linear equations has no solution if its determinant is zero and one of the \[{D_x},{D_y},{D_z}\] are not equal to zero.
* A system of linear equations has a unique solution if its determinant is zero.
* A system of linear equations has infinite solutions if \[{D_x},{D_y},{D_z},D\]are all equal to zero.

Complete step-by-step answer:
Given set of three linear equations
\[x + y + z = 6\]
\[x + 2y + 3z = 10\]
\[x + 2y + \lambda z = \mu \]
We write the set of linear equations in the matrix form such that \[AX = B\]
\[\left[ {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&2&3 \\
  1&2&\lambda
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  6 \\
  {10} \\
  \mu
\end{array}} \right]\]
Where \[A = \left[ {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&2&3 \\
  1&2&\lambda
\end{array}} \right],X = \left[ {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right],B = \left[ {\begin{array}{*{20}{c}}
  6 \\
  {10} \\
  \mu
\end{array}} \right]\]
Now we take determinant of the matrix \[A\] and denote it by \[D\].
\[D = \left| {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&2&3 \\
  1&2&\lambda
\end{array}} \right|\]
To solve the determinant
\[1(2\lambda - 6) - 1(\lambda - 3) + 1(2 - 2) = 2\lambda - 6 - \lambda + 3\]
                                                       \[\begin{gathered}
   = (2\lambda - \lambda ) + ( - 6 + 3) \\
   = \lambda - 3 \\
\end{gathered} \]
Also, we take determinants \[{D_x},{D_y},{D_z}\] where we write x coefficients in the first column, y coefficients in the second column and z coefficients in the third column respectively.
i.e. \[{D_x} = \left| {\begin{array}{*{20}{c}}
  6&1&1 \\
  {10}&2&3 \\
  \mu &2&\lambda
\end{array}} \right|,{D_y} = \left| {\begin{array}{*{20}{c}}
  1&6&1 \\
  1&{10}&3 \\
  1&\mu &\lambda
\end{array}} \right|,{D_x} = \left| {\begin{array}{*{20}{c}}
  1&1&6 \\
  1&2&{10} \\
  1&2&\mu
\end{array}} \right|\]
Now, solving the given parts.
(i) No solution
For a system of linear equations to have no solution \[D = 0\] and at least one of \[{D_x},{D_y},{D_z}\] are not equal to zero. Then the system of linear equations is called inconsistent.
Since, we know value of determinant \[D = \lambda - 3\]
Therefore substituting for \[D = \lambda - 3 = 0\]
\[\begin{gathered}
  \lambda - 3 = 0 \\
  \lambda = 3 \\
\end{gathered} \]
Solving for the determinant \[{D_x} = \left| {\begin{array}{*{20}{c}}
  6&1&1 \\
  {10}&2&3 \\
  \mu &2&\lambda
\end{array}} \right|\]
\[{D_x} = 6(2\lambda - 6) - 1(10\lambda - 3\mu ) + 1(20 - 2\mu )\]
       \[\begin{gathered}
   = 12\lambda - 36 - 10\lambda + 3\mu + 20 - 2\mu \\
   = (12 - 10)\lambda + (3 - 2)\mu + (20 - 36) \\
   = 2\lambda + \mu - 16 \\
\end{gathered} \]
Now substitute the value of \[\lambda = 3\] in value of \[{D_x}\]
\[{D_x} = 2(3) + \mu - 16\]
       \[\begin{gathered}
   = 6 - 16 + \mu \\
   = - 10 + \mu \\
\end{gathered} \]
Therefore, for no solution we write \[{D_x} \ne 0\]
Which gives the value of \[\mu \]
 \[\begin{gathered}
   - 10 + \mu \ne 0 \\
  \mu \ne 10 \\
\end{gathered} \]
Therefore, for the values \[\lambda = 3,\mu \ne 10\] the system of linear equations has no solution.
(ii) Unique solution
For a system of linear equations to have unique solution \[D \ne 0\]. Then the system of linear equations is called consistent.
Since, we know value of determinant \[D = \lambda - 3\]
Therefore substituting for \[D = \lambda - 3 \ne 0\]
\[\begin{gathered}
  \lambda - 3 \ne 0 \\
  \lambda \ne 3 \\
\end{gathered} \]
Therefore, for the value \[\lambda \ne 3\] the system of linear equations has unique solution.
(iii) Infinite solution
For a system of linear equations to have infinite solution \[D = {D_x} = {D_y} = {D_z} = 0\]
Since, we know value of determinant \[D = \lambda - 3\]
Therefore substituting for \[D = \lambda - 3 = 0\]
\[\begin{gathered}
  \lambda - 3 = 0 \\
  \lambda = 3 \\
\end{gathered} \]
Solving for the determinant \[{D_x} = \left| {\begin{array}{*{20}{c}}
  6&1&1 \\
  {10}&2&3 \\
  \mu &2&\lambda
\end{array}} \right|\]
\[{D_x} = 6(2\lambda - 6) - 1(10\lambda - 3\mu ) + 1(20 - 2\mu )\]
       \[\begin{gathered}
   = 12\lambda - 36 - 10\lambda + 3\mu + 20 - 2\mu \\
   = (12 - 10)\lambda + (3 - 2)\mu + (20 - 36) \\
   = 2\lambda + \mu - 16 \\
\end{gathered} \]
Now substitute the value of \[\lambda = 3\] in value of \[{D_x}\]
\[{D_x} = 2(3) + \mu - 16\]
       \[\begin{gathered}
   = 6 - 16 + \mu \\
   = - 10 + \mu \\
\end{gathered} \]
Therefore, for no solution we write \[{D_x} = 0\]
Which gives the value of \[\mu \]
 \[\begin{gathered}
   - 10 + \mu = 0 \\
  \mu = 10 \\
\end{gathered} \]
Therefore, for the values \[\lambda = 3,\mu = 10\] the system of linear equations has infinite solutions.

Note: Students are likely to make mistake while solving the determinant as when moving from left to right we change signs from positive to negative and then again to positive. If a matrix \[A = \left[ {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right]\] then, determinant is \[\left| A \right| = a(ei - fh) - b(di - fg) + c(dh - eg)\].