Question

For what values of \[k,\]the system of linear equations
$\begin{align}
  & x+y+z=2 \\
 & 2x+y-z=3 \\
 & 3x+2y+kz=4 \\
\end{align}$
has a unique solution?

Answer 1

Hint: In the given question we have given three linear equations in three variables viz. $x,y\text{, and }z$and the question demand the value of $k$, so that that system of the given equation has a unique solution. So, in order to find the unique solution, we have to transform the equation in matrix form as we write

Complete step by step answer:
$A=\left[ \begin{matrix}
   1 & 1 & 1 \\
   2 & 1 & -1 \\
   3 & 2 & k \\
\end{matrix} \right],X=\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]\text{ and }B=\left[ \begin{matrix}
   2 \\
   3 \\
   4 \\
\end{matrix} \right]$ so, we get the solution by comparing $X={{A}^{-1}}B$. As inverse of matrix $\left( {{A}^{-1}} \right)$ as ${{A}^{-1}}=\dfrac{1}{\left| A \right|}adjA$. Hence system of equation has unique solution is $\left| A \right|\ne 0.$
So, we have to find those values of $k,$ for which the determinant of matrix $A$ does not equal to zero.
Step by step solution:
We have given the system of linear equations as
$\begin{align}
  & x+y+z=2 \\
 & 2x+y-z=3 \\
 & 3x+2y+kz=4 \\
\end{align}$
Let the determinant of coefficient of the variables is $\left| A \right|$, So we can write
$\left| A \right|=\left| \begin{matrix}
   1 & 1 & 1 \\
   2 & 1 & -1 \\
   3 & 2 & k \\
\end{matrix} \right|$--------------------------------(1)
Now in the above determinant we do the operation ${{C}_{1}}\to {{C}_{1}}+{{C}_{2}}$, So we can write
$\left| \begin{matrix}
   1 & 1 & 1 \\
   2 & 1 & -1 \\
   3 & 2 & k \\
\end{matrix} \right|=\left| \begin{matrix}
   3 & 2 & 0 \\
   2 & 1 & -1 \\
   3 & 2 & k \\
\end{matrix} \right|$----------------------------(2)
Again, we do the operation ${{C}_{1}}\to {{C}_{1}}-{{C}_{2}}$, we can write
$\left| \begin{matrix}
   3 & 2 & 0 \\
   2 & 1 & -1 \\
   3 & 2 & k \\
\end{matrix} \right|=\left| \begin{matrix}
   0 & 0 & -k \\
   2 & 1 & -1 \\
   3 & 2 & k \\
\end{matrix} \right|$ ---------------------------(3)
Now on expanding we can write
$\begin{align}
  & \left| \begin{matrix}
   0 & 0 & -k \\
   2 & 1 & -1 \\
   3 & 2 & k \\
\end{matrix} \right|=0\left| \begin{matrix}
   1 & -1 \\
   2 & k \\
\end{matrix} \right|-0\left| \begin{matrix}
   2 & -1 \\
   3 & k \\
\end{matrix} \right|+(-k)\left| \begin{matrix}
   2 & 1 \\
   3 & 2 \\
\end{matrix} \right| \\
 & \left| \begin{matrix}
   0 & 0 & -k \\
   2 & 1 & -1 \\
   3 & 2 & k \\
\end{matrix} \right|=(-k)\left\{ 2(2)-1(3) \right\} \\
 & \left| \begin{matrix}
   0 & 0 & -k \\
   2 & 1 & -1 \\
   3 & 2 & k \\
\end{matrix} \right|=(-k)(1) \\
\end{align}$------------------(4)
Hence, we see here from (1), (2), (3), (4) the value of determinant $\left| A \right|=-k$.
As we know that system of the given equation has a unique solution if $\left| A \right|\ne 0$
So $k\ne 0$
Hence, we conclude that the given system of the equation has a unique solution for all values of $k$except zero, so we can write the solution as
$k\in R-\left\{ 0 \right\}$

Note:
There is a system of equations
A consistent system of the equation: A given system of equation is said to be consistent if it has one or more solutions. In this case if $\left| A \right|\ne 0,$ then the given system has unique solution.
Inconsistent system of the equation: A given system of equation is said to be inconsistent if it has no solution. In this case $\left| A \right|=0$, and $\left( adjA \right)B\ne O$
If $\left| A \right|=0$, and $\left( adjA \right)B=O,$ then system has infinitely many solutions.
Here O is the null matrix and (adj) is the adjoint of a matrix.