Solving Linear Systems with a Variable: How to Handle Unknown Constants?

In summary, the linear system has no solutions when the last non-zero row of [U|c] ends with c, where c != 0.
  • #1
delc1
9
0
Hello,

I am stuck on a question for some time now and I am unsure how to solve this. I have tried the substitution and gaussian elimination methods but have had no luck at all.

Identify the value(s) of k for which the following linear system
x + 2y + z = 2
2x − 2y + 3z = 1
x + 3y + (k^2 − 3)z = k
has:
(i) no solutions
(ii) one solution
(iii) infinitely many solutions

I am particularly having trouble with the variable 'k'. I am not sure how to handle it?
 
Physics news on Phys.org
  • #2
We have the following:

$$\begin{pmatrix}
1 & 2 & 1\\
2 & -2 & 3\\
1 & 3 & k^2-3
\end{pmatrix}\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}=\begin{pmatrix}
2\\
1\\
k
\end{pmatrix}
$$

Applying gaussian elimination we have the following:

(Let $(a)$ be the first line, $(b)$ the second and $(c)$ the third one.)

$$\begin{pmatrix}
1 & 2 & 1 & | & 2\\
2 & -2 & 3 & | & 1\\
1 & 3 & k^2-3 & | & k
\end{pmatrix} \overset{(b)'=(b)-2(a), (c)'=(c)-(a)}{ \longrightarrow } \begin{pmatrix}
1 & 2 & 1 & | & 2\\
0 & -6 & 1 & | & -3\\
0 & 1 & k^2-4 & | & k-2
\end{pmatrix} \overset{(c)''=6(c)'+(b)}{ \longrightarrow } \begin{pmatrix}
1 & 2 & 1 & | & 2\\
0 & -6 & 1 & | & -3\\
0 & 0 & 6k^2-23 & | & 6k-15
\end{pmatrix} $$

When we get from the last line $0=0$ we have infinitely many solutions, so when $6k^2-23=0$ AND $6k-15=0$.

When only one of the two is zero ($6k^2-23=0$ OR $6k-15=0$ but not both) we have no solutions.

In any other case we have one solution.
 
  • #3
mathmari said:
We have the following:
When only one of the two is zero ($6k^2-23=0$ OR $6k-15=0$ but not both) we have no solutions.

Wouldn't there only be no solutions when the second last column in the bottom row of this matrix equeled 0?

If we found k such that $6k-15=0$, then one of the equations would be:

0x + 0y + 6k2-23z = 0

and since z can be 0, this would result in a solution?

Every text I've read says it must be when the last non-zero row of [U|c] ends with c, where c != 0, that there is no solutions and nothing about when there is another row which also has an unknown constant, for instance k, which is in both the final column in the last non-zero row and another column in the same row.

Please correct me if I am wrong
 

FAQ: Solving Linear Systems with a Variable: How to Handle Unknown Constants?

What is a 3-Equation Linear System?

A 3-Equation Linear System is a set of three linear equations that involve three unknown variables. These equations can be solved simultaneously to find the values of the variables that satisfy all three equations.

How do you solve a 3-Equation Linear System?

To solve a 3-Equation Linear System, you can use different methods such as substitution, elimination, or matrix operations. These methods involve manipulating the equations to eliminate one variable at a time until the values of all three variables are found.

What is the importance of solving a 3-Equation Linear System?

Solving a 3-Equation Linear System is important in many fields of science and engineering. It can help in solving real-life problems involving multiple variables, such as finding the optimal solution in economics or predicting the trajectory of a projectile in physics.

What are the possible outcomes when solving a 3-Equation Linear System?

There are three possible outcomes when solving a 3-Equation Linear System: a unique solution, no solution, or infinite solutions. A unique solution means that there is one set of values that satisfies all three equations, no solution means that there is no set of values that satisfies all three equations, and infinite solutions means that there are infinite sets of values that satisfy all three equations.

Can a 3-Equation Linear System have more than three unknown variables?

No, a 3-Equation Linear System can only have three unknown variables. If there are more than three unknown variables, then the system becomes overdetermined and cannot be solved using three equations.

Similar threads

Replies
11
Views
2K
Replies
3
Views
1K
Replies
2
Views
1K
Replies
1
Views
1K
Replies
5
Views
1K
Replies
6
Views
2K
Replies
1
Views
3K
Replies
9
Views
2K
Back
Top