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

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SUMMARY

This discussion focuses on solving a linear system with a variable constant, specifically identifying values of k for which the system has no solutions, one solution, or infinitely many solutions. The system is represented by the equations: x + 2y + z = 2, 2x − 2y + 3z = 1, and x + 3y + (k^2 − 3)z = k. Through Gaussian elimination, it is established that infinitely many solutions occur when both 6k^2 - 23 = 0 and 6k - 15 = 0, while no solutions arise when only one of these conditions holds true. The discussion emphasizes the importance of the last non-zero row in determining the nature of the solutions.

PREREQUISITES
  • Understanding of linear algebra concepts, particularly linear systems.
  • Proficiency in Gaussian elimination techniques.
  • Familiarity with the implications of row echelon form in matrices.
  • Knowledge of polynomial equations and their roots.
NEXT STEPS
  • Study the implications of the rank of a matrix in relation to the number of solutions in linear systems.
  • Explore the method of substitution for solving linear equations with parameters.
  • Learn about the conditions for consistency in linear systems, particularly in relation to augmented matrices.
  • Investigate the use of determinants in determining the uniqueness of solutions in linear algebra.
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Students and professionals in mathematics, particularly those studying linear algebra, as well as educators seeking to clarify concepts related to solving linear systems with variable constants.

delc1
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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?
 
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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.
 
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
 

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