Difference between revisions of "031 Review Part 2, Problem 11"

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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 1:    
 
!Step 1:    
 +
|-
 +
|To begin with, we turn this system into an augmented matrix.
 +
|-
 +
|Hence, we get
 
|-
 
|-
 
|
 
|
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::<math>B=   
 +
    \begin{bmatrix}
 +
          1 & k & 1 \\
 +
          3 & 5  & 2k 
 +
        \end{bmatrix}.</math>
 
|}
 
|}
  

Revision as of 10:53, 10 October 2017

Consider the following system of equations.

Find all real values of    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
Step 2:  


Final Answer:  
      

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