031 Review Part 2, Problem 11

Consider the following system of equations.

$x_{1}+kx_{2}=1$

$3x_{1}+5x_{2}=2k$

Find all real values of  $k$  such that the system has only one solution.

Foundations:
1. To solve a system of equations, we turn the system into an augmented matrix and
row reduce that matrix to determine the solution.
2. For a system to have a unique solution, we need to have no free variables.

Solution:

Step 1:
To begin with, we turn this system into an augmented matrix.
Hence, we get
$\left[{\begin{array}{cc|c}1&k&1\\3&5&2k\end{array}}\right].$
Now, when we row reduce this matrix, we get
$\left[{\begin{array}{cc|c}1&k&1\\3&5&2k\end{array}}\right]\sim \left[{\begin{array}{cc|c}1&k&1\\0&-3k+5&-3+2k\end{array}}\right].$

Step 2:
To guarantee a unique solution, our matrix must contain two pivots.
So, we must have  $-3k+5\neq 0.$
Hence, we must have
$k\neq {\frac {5}{3}}.$
Therefore,  $k$  can be any real number except  ${\frac {5}{3}}.$

Final Answer:
The system has only one solution when  $k$  is any real number except  ${\frac {5}{3}}.$

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