Solving a Matrix: Finding Positive & Negative Solutions

[helpm3pl3ase]

have a matrix:

X1+X2+X3+X4 = 0
X1+X2-X3-X4 = 0
X1-X2+X3-X4 = 0

the solutions that are formed are:
t
-t
-t
t

Now i understand that they just used t to label the free variables.. but Iam not sure how they were able to tell which were negative and positive.. I tried to do it out, but just confused myself more.. Help anyone?

 

[wildman]

The answer means:
X1 = t
X2 = -t
X3 = -t
X4 = t

The reason for the signs is they have to add to zero in all three cases.

For instance:
Take X1+X2+X3+X4

If you substitute the answer in you get:

t+ (-t) + (-t)+t

t+(-t) is zero and (-t)+t is zero so they add to zero.

In the second case:

X1+X2-X3-X4

In this case substituting in the answer:

t + (-t) - (-t) - t

t + (-t) is zero and -(-t)-t is zero

The third case is left as an exercise for the student.

 

[tacman]

The solutions in this matrix are formed by using the method of elimination. In this case, since all the variables are equal to zero, we can set up a system of equations to solve for the values of X1, X2, X3, and X4.

Starting with the first equation, X1+X2+X3+X4 = 0, we can rearrange it to solve for X1. This gives us X1 = -X2-X3-X4. We can then substitute this value for X1 into the second and third equations to eliminate the X1 variable.

In the second equation, we now have (-X2-X3-X4)+X2-X3-X4 = 0. Simplifying this gives us -2X3-2X4 = 0. Dividing both sides by -2, we get X3+X4 = 0. Similarly, in the third equation, we now have (-X2-X3-X4)+X2+X3-X4 = 0. Simplifying this gives us -2X4 = 0. Dividing both sides by -2, we get X4 = 0.

Now, we can substitute the value of X4 = 0 into the second equation to solve for X2. This gives us X2-X3 = 0, or X2 = X3. Finally, we can substitute the values of X4 = 0 and X2 = X3 into the first equation to solve for X1. This gives us X1 = -2X3. So, our solutions are X1 = -2t, X2 = t, X3 = t, and X4 = 0.

Since t is just a placeholder for any real number, we can see that there are infinitely many solutions to this matrix. However, we can determine whether each solution is positive or negative by plugging in different values for t. For example, if we plug in t = 1, we get the solution X1 = -2, X2 = 1, X3 = 1, X4 = 0. This solution has both positive and negative values. But if we plug in t = -1, we get the solution X1 = 2, X2 = -1, X3 = -1, X4 = 0. This solution has only negative values.

In summary, the method of elimination allows