# Linear algebra question

1. Dec 9, 2008

### transgalactic

i got a set of linear equations with x,y,z,t variables.

x=4,y=-2,z=-2,t=4 is a solution
x=-2 y=4 z=4 t=-2 is a solution

x=2,y=2,z=2,t=2 is not a solution.

prove that this set is not homogeneous ?
prove that x=0,y=2,z=2,t=0 is a solution?

??

2. Dec 9, 2008

### Defennder

I'm not clear as to what homogenous here means. On the other hand, if a set of linear equations has more than one solutions, then it has infinite number of solutions. That allows you to come up with a general solution given just two specific possible solutions. This should help you with part 2.

3. Dec 9, 2008

### HallsofIvy

Staff Emeritus
I believe that by "homogeneous" here, you mean that the equations are of the form
$a_1x+ b_1y+ c_1z+ d_1t= 0$
$a_2x+ b_2y+ c_2z+ d_2t= 0$
$a_3x+ b_3y+ c_3z+ d_3t= 0$
$a_4x+ b_4y+ c_4z+ d_4t= 0$
That is the same as the matrix equation
$$\left[\begin{array}{cccc}a_1 & b_1 & c_1 & d_1 \\a_2 & b_2 & c_2 & d_2 \\a_3 & b_3 & c_3 & d_3 \\a_4 & b_4 & c_4 & d_4 \end{array}\right]\left[\begin{array}{c} x \\ y \\ z \\ t\end{array}\right]= \left[\begin{array}{c}0 \\ 0 \\ 0 \\ 0\end{array}\right]$$

Such an equation will have a unique solution if and only if the coefficient matrix is invertible. Since this has at least two solutions, the coefficient matrix is not invertible. Since there are at least two solutions, the coefficient matrix is not invertible. If it were homogeneous, then any linear combination of solutions would also be a solution: if $Ax_1= 0$ and $Ax_2= 0$ then $A(cx_1+ dx_2)= cAx_1+ dAx_2= 0$

4. Dec 9, 2008

### transgalactic

how to prove that
x=0,y=2,z=2,t=0 is not a solution?

5. Dec 9, 2008

### Defennder

You're supposed to prove it's a solution, you mean.

6. Dec 9, 2008

### transgalactic

i am supposed
but i dont know how??

7. Dec 9, 2008

### Staff: Mentor

You said you have a set of linear equations (which you didn't show). Replace x, y, z, and t with 0, 2, 2, and 0, respectively, on the expressions on the left side of the equations you have. You should get 0 on the right side of all of your equations.

8. Dec 9, 2008

### transgalactic

i got a crazy idea
if i use this given sentences and transform them into equations:
x=4,y=-2,z=-2,t=4 is a solution
x=-2 y=4 z=4 t=-2 is a solution

(4,-2,-2,4)+2*(-2,4,4,-2)=(0,6,6,0)=3*(0,2,2,0)

so i got my solution that i suppose to prove from algebraic manipulation

is that a proove??
if it is what formal words do i need to say in order to confirm this method?

9. Dec 9, 2008

### Staff: Mentor

Sort of.
Here's the situation as I see it (necessariy sketchy, since you didn't provide many details):
You have a system of equations which I will represent as a matrix equation:
Ax = 0

Any vector x in R^4 is a solution to the equation above if A times x equals the zero vector.

You are given that for x_1 = (4, -2, -2, 4)^T and x_2 = (-2, 4, 4, -2),
A*x_1 = 0 and A*x_2 = 0.

For x_3 = (0, 2, 2, 0), you have shown that x_3 is a linear combination of x_1 and x_2, namely x_3 = 1/3*x_1 + 2/3*x_2.
By the linearity of matrix multiplication A*x_3 = A*(1/3*x_1 + 2/3*x_2)
You should be able to fill in what's missing at the end to show that x_3 is a solution to the homogeneous equation Ax = 0.