Difference between revisions of "031 Review Part 2, Problem 11"
Jump to navigation
Jump to search
Kayla Murray (talk | contribs) |
Kayla Murray (talk | contribs) |
||
Line 24: | Line 24: | ||
{| 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 | ||
|- | |- | ||
| | | | ||
+ | ::<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 |
|
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: |
---|