Question

One ordered triple (x,y,z) which satisfies the system is (0, 0, 0). What is the only numerical value of "k" for which this system has a solution other than (0,0,0)?
x-2y+kz=0
2x-y+3z=0
3x-3y-2z=0

Answer 1

Let
\[A=\left[\begin{matrix}1&-2&k\\
2&-1&3\\
3&-3&-2\end{matrix}\right]\]
\[A\left[\begin{matrix}x\\y\\z\end{matrix}\right]=\left[\begin{matrix}0\\0\\0\end{matrix}\right]\\
(A^{-1}\cdot A)\left[\begin{matrix}x\\y\\z\end{matrix}\right]=A^{-1}\left[\begin{matrix}0\\0\\0\end{matrix}\right]\\
\left[\begin{matrix}x\\y\\z\end{matrix}\right]=A^{-1}\left[\begin{matrix}0\\0\\0\end{matrix}\right]\]
Do you know how to find the inverse of a 3x3 matrix?
Also, you might need to know the inverse exists if the determinant of A is non-zero.

Answer 2

I haven't learn matrix yet

Answer 3

Is there an easier approach to these problems?

Answer 4

I think the matrix approach is easier, but that's just my opinion. I'm pretty sure there's something you can do using what you learn about elimination when solving a system of equations. Just a sec...

Answer 5

\[\begin{cases}x-2y+kz=0\\
2x-y+3z=0\\
3x-3y-2z=0\end{cases}\]
Multiply the first equation by -2 and add it to the second to eliminate x. Then, multiply the first equation by -3 and add it to the third to eliminate x.
\[\begin{cases}-2x+4y-2kz=0\\
2x-y+3z=0\end{cases}\\
3y+(3-2k)z=0\ldots(A)\]
\[\begin{cases}-3x+6y-3kz=0\\
3x-3y-2z=0\end{cases}\\
3y-(2+3k)z=0\ldots(B)\]
Rewriting A and B, you get
\[3y = -(3-2k)z\\
3y=(2k-3)z\\\text{ and}\\
3y=(2+3k)z\]
Setting the equations equal to each other, you have
\[(2k-3)z=(2+3k)z\\
2k-3=2+3k\\
k=-5\]
So now you just have to solve the system
\[\begin{cases}x-2y-5z=0\\
2x-y+3z=0\\
3x-3y-2z=0\end{cases}\]
Think you can handle that? If I'm not mistaken, I think you should get "many solutions".

Answer 6

The variables will be cancelled out...

x-2y-5z=0
2x-y+3z=0
3x-3y-2z=0

6x-12y-30z=0
6x-3y+9z=0
6x-6y-4z=0

Subtract the first two: -9y-39z=0
subtract the second and third: 3y+13z=0
9y+39z=0

Add those two: (-9y-39z=0) + (9y+39z=0) = 0

Answer 7

Looks like it's all right. However, I think the last step takes it a bit too far. Right before that, you have
\[3y+13z=0, \text{ or}\\
y=-\frac{13}{3}z\]
If you plug that into the original first equation, you get
\[x=-3z\]
This means that if you let z be any (non-zero) number, x and y will also be non-zero because they both depend on z. You just have to be able to write x, y, and z in terms of another variable.

Answer 8

Oh, I see. Thank you so much!!