MHB Solve for k in System of Equations

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To solve for k in the given system of equations, the determinant of the matrix must be analyzed. A unique solution exists if the determinant is nonzero, while a determinant of zero indicates either no solutions or infinitely many solutions. The discussion suggests two approaches: determining k for an invertible matrix or finding k that results in infinitely many solutions. Participants emphasize the importance of understanding the role of the determinant in these scenarios. Ultimately, the solution hinges on the value of k that satisfies these conditions.
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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