Homogeneous Matrix help *Linear Algebra

[VP1]

Hello- I'm stuck on a linear algebra problem.
The problem is to find the solutions using a matrix.

x + 4y -2z = 0
3x +10y -z = 0
-2x -6y - z = 0

Using elementary matrix operations I get to

x +4y - 2z = 0
0 +2y -5z = 0
0 -2y + 5z = 0

The last two rows cancel each other out. Does this mean I need to parametrize the remaining x +4y -2z = 0 ?
Thanks!

 

[daon2]

If one or more of your rows do not "zero out" then your matrix of coefficients is invertible and has only the trivial solution x=y=z=0. That is the simple case. One more row reduction will tell you your free variable is z, and you can set up a parameterization. The rows do not cancel each-other out. One row may be cancelled by adding it to the other. One thing though, always reduce to the reduced-row echelon form before doing this. It will save you time later.

 

[HallsofIvy]

Hello- I'm stuck on a linear algebra problem.
The problem is to find the solutions using a matrix.

x + 4y -2z = 0
3x +10y -z = 0
-2x -6y - z = 0

Using elementary matrix operations I get to

x +4y - 2z = 0
0 +2y -5z = 0
0 -2y + 5z = 0

The last two rows cancel each other out. Does this mean I need to parametrize the remaining x +4y -2z = 0 ?
Thanks!

As Daon2 said, it is not the two rows "cancel each other out", leaving only the first equation. It is, rather, that one of the rows cancels the the other leaving you with two equations instead of three. That second equation tells you that 2y= 5z or z= (5/2)y. If we replace z in the first equation by (5/2)y, we have x+ 4y- 5y= x- y= 0 so that x= y. Now you could use y itself as parameter: x= y. z= (5/2)y. Or, if you prefer, just "rename" y to t: x= t, y= t, z= (5/2)t gives the solution. If you don't like fractions, replace t with 2s to get x= 2s, y= 2s, z= 5s. Note that s= 0 gives the "obvious" solution x= y= z= 0.

(The coefficient matrix of your last set of equations is \(\displaystyle \begin{bmatrix}1 & 4 & -2 \\ 0 & 2 & -5 \\ 0 & -2 & 5\end{bmatrix}\). You can get the "reduced row echelon" form that Daon2 refers to by
1) Add the second row to the third
2) Subtract twice the second row from the first
3) Divide the second row by 2

Doing those gives \(\displaystyle \begin{bmatrix}1 & 2 & 3 \\ 0 & 1 & -\frac{5}{2} \\ 0 & 0 & 0 \end{bmatrix}\) which gives the two equations x+ 2y+ 3z= 0, y- (5/2)z= 0 leading to the solution above.

 

[soroban]

Hello, VP1!

\(\displaystyle \text{Solve, using a matrix: }\;\begin{array}{ccc} x + 4y -2z &=& 0 \\ 3x +10y -z &=& 0 \\ \text{-}2x -6y - z &=& 0 \end{array}\)

We have: .\(\displaystyle \left|\begin{array}{ccc|c}1&4&\text{-}2&0 \\ 3&10&\text{-}1&0 \\ \text{-}2&\text{-}6&\text{-}1&0 \end{array}\right| \)

\(\displaystyle \begin{array}{c} \\ \\ R_2-3R_1 \\ R_3+2R_1 \end{array}\:\left|\begin{array}{ccc|c} 1&4&\text{-}2 & 0 \\ 0 & \text{-}2 & 5 & 0 \\ 0 & 2 &\text{-}5 & 0 \end{array}\right|\)

\(\displaystyle \begin{array}{c} \\ \\ (\text{-}1)R_2 \\ R_3+R_2 \end{array}\:\left|\begin{array}{ccc|c} 1&4&\text{-}2 & 0 \\ 0&2&\text{-}5 & 0 \\ 0&0&0&0 \end{array}\right| \)

\(\displaystyle \begin{array}{c}R_1-2R_2 \\ \frac{1}{2}R_2 \\ \\ \end{array}\:\left|\begin{array}{ccc|c}1&0&8&0 \\ 0&1&\text{-}\frac{5}{2} & 0 \\ 0&0&0&0 \end{array}\right| \)

We have: .\(\displaystyle \begin{Bmatrix}x + 8z\;=\; 0 &\Rightarrow& x \;=\; \text{-}8t \\ y -\frac{5}{2}z \;=\; 0 &\Rightarrow& y \;=\; \frac{5}{2}t \\ z \;=\; z & \Rightarrow & z\;=\;z\end{Bmatrix}\)


On the right, replace \(\displaystyle z\) with \(\displaystyle 2t\):

. . \(\displaystyle \begin{Bmatrix}x &=& \text{-}16z \\ y &=& 5z \\ z &=& 2t \end{Bmatrix}\)

These are the parametric equations for the solution.