Solve for k in System of Equations

[wonguyen1995]

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

 

[Dustinsfl]

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?

 

[wonguyen1995]

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.

 

[Dustinsfl]

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.

 

[CellCoree]

To solve for k in this system of equations, we can use the method of elimination. First, we can eliminate the variable x by adding the two equations together, resulting in:

x-3y+x+3z=6-3
2x+kx+(3-k)z=1

This simplifies to:

2x+kx=3

Next, we can factor out the common variable x:

x(2+k)=3

To find the value of k, we need to isolate it on one side of the equation. We can do this by dividing both sides by x:

(2+k)=3/x

Finally, we can subtract 2 from both sides to get the value of k:

k=3/x-2

Therefore, to have a solution for this system of equations, k must be equal to 3/x-2.