I previously posted this question under the wrong topic.
Which of the following choices is a solution for:
(-3-i)x1 + (1-2i)x2 = 0
(-7-9i)x1 + (7-4i)x2 = 0
for any value of the variable t
[x1] = ?
[x2]
A: [5]
[1+7i] * t
B: [5]
[1-7i] * t
C: [5]
[-1-7i] * t
D: [5]
[-1+7i] * t
(-3-i)x1 + (1-2i)x2 = 0.....(1)
(-7-9i)x1 + (7-4i)x2 = 0 ....(2)
or
Ax=0
The system is a homogeneous equation. A trivial solution is x1=0 and x2=0.
For the system to have non-trivial solutions, the determinant of A must vanish, or |A|=0.
Calculate
|A|
=(-3-i)(7-4i) - (-7-9i)(1-2i)
= -21-4+5i - (-25+5i)
=0
So non-trivial solutions exist, because the two equations are linearly dependent.
Prove that the two equations are linearly dependent by applying Gauss elimination which results in two identical equations. Eliminate the second equation.
Take equation (1),
(-3-i)x1 + (1-2i)x2 = 0
We multiply by the conjugate of (1-2i) to get
(1+2i)(-3-i)x1 + (1+2i)(1-2i)x2 = 0
-(1+7i)x1 + 5x2 = 0
Let x1=5t, where t is a variable parameter, then
-(1+7i)5t + 5x2 = 0
Solving, x2=1+7i
Therefore
x1=5t, x2=1+7i
Make your pick for the answer.
Thanks a lot!
You're welcome!
To find the solution for the given system of equations, we can use the method of elimination or substitution. Let's solve it step-by-step:
Step 1: Multiply the first equation by (-1-7i) and the second equation by (1+7i) to eliminate the imaginary terms.
(-1-7i)[(-3-i)x1 + (1-2i)x2] = 0
(1+7i)[(-7-9i)x1 + (7-4i)x2] = 0
Simplifying these equations:
(10 + 32i)x1 + (1+7i)x2 = 0
(-100 - 32i)x1 + (1+7i)x2 = 0
Step 2: Add the two equations together to eliminate x1.
(10 + 32i)x1 + (1+7i)x2 + (-100 - 32i)x1 + (1+7i)x2 = 0 + 0
Simplifying the equation:
(-90 + 39i)x2 = 0
Step 3: Since any value of the variable t can be a solution, we can set x2 = t.
(-90 + 39i)t = 0
Step 4: Solve for t:
-90t + 39it = 0
Dividing both sides by i:
-90t/i + 39it/i = 0/i
-90t - 39t = 0
-129t = 0
t = 0
Step 5: Substitute t = 0 into the original equations to find x1:
(-3-i)x1 + (1-2i)x2 = 0
(-7-9i)x1 + (7-4i)x2 = 0
Substituting t = 0:
(-3-i)x1 + (1-2i)(0) = 0
(-7-9i)x1 + (7-4i)(0) = 0
Simplifying the equations:
(-3-i)x1 = 0
(-7-9i)x1 = 0
Since the coefficient of x1 is non-zero, the only solution for x1 is 0.
Therefore, the solution for [x1] and [x2] is [0] and [t] respectively, for any value of the variable t.
So, the correct choice would be:
[x1] = [0]
[x2] = [t]