Solve for k in System of Equations

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SUMMARY

The discussion focuses on determining the value of k in the system of equations represented by the matrix \[ \begin{bmatrix} 1&-3&0\\ 1&0&3\\ 2&k&3-k \end{bmatrix} \] to ensure a solution exists. The key insight is that the determinant of the matrix must be analyzed; if it is non-zero, the system has a unique solution, while a zero determinant indicates either no solutions or infinitely many solutions. Participants suggest two approaches: finding k for an invertible matrix or identifying k for infinite solutions.

PREREQUISITES
  • Understanding of linear algebra concepts, particularly determinants.
  • Familiarity with matrix representation of systems of equations.
  • Knowledge of conditions for unique and infinite solutions in linear systems.
  • Basic skills in manipulating algebraic expressions and solving equations.
NEXT STEPS
  • Research how to calculate the determinant of a 3x3 matrix.
  • Learn about conditions for unique solutions in linear algebra.
  • Explore methods for finding values that yield infinite solutions in systems of equations.
  • Study examples of systems of equations with varying determinants.
USEFUL FOR

Students and professionals in mathematics, particularly those studying linear algebra, as well as educators seeking to enhance their understanding of systems of equations and determinants.

wonguyen1995
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Find k to have a solution?
x-3y=6
x+3z=-3
2x+kx+(3-k)z=1LATEX
 
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wonguyen1995 said:
Find k to have a solution?
x-3y=6
x+3z=-3
2x+kx+(3-k)z=1LATEX

Have you thought about what role the determinant would play?
 
dwsmith said:
Have you thought about what role the determinant would play?

of course I think it should better if i have sample of solution. i will research carefully on this.
 
wonguyen1995 said:
of course I think it should better if i have sample of solution. i will research carefully on this.

We know the matrix
\[
\begin{bmatrix}
1&-3&0\\
1&0&3\\
2&k&3-k
\end{bmatrix}
\]
has unique solution if the determinant is what?
Second, if the determinant is zero, we have no solutions or infinitely many solutions.

We have two approaches. One assume the determinant is nonzero and find k that makes it invertible or assume the determinant is zero and try to find a k such that we have infinitely many solutions.
 

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