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
This is linear algebra by the way;
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To find the solution for the given system of linear equations, you can use the method of Gaussian elimination or matrix operations. In this case, we can start by setting up the augmented matrix and perform row operations to bring it to row-echelon form or row-reduced echelon form.
The given system of equations can be written in matrix form as:
[(-3-i) (1-2i)] [x1] = [0]
[(-7-9i) (7-4i)] [x2] = [0]
Let's perform the row operations to find the row-echelon form of the augmented matrix:
1. Multiply the first row by (-7 - 9i)/(-3 - i) and subtract it from the second row to eliminate x1 in the second equation:
[(-3-i) (1-2i) ] [x1] [0]
[0 (-12 - 4i)] [x2] = [0]
2. Multiply the second row by (-3 - i)/(-12 - 4i) to simplify the coefficient of x2:
[(-3-i) (1-2i) ] [x1] [0]
[0 1 ] [x2] = [0]
3. Multiply the second equation by (1 - 2i) and subtract it from the first equation to eliminate x2 in the first equation:
[(-3-i) 0 ] [x1] [0]
[0 1 ] [x2] = [0]
The resulting matrix is now in row-echelon form. Now, we can solve the equations recursively:
From the second equation, we know that x2 = 0.
Substituting this into the first equation:
(-3 - i)x1 = 0
Simplifying further, we get:
(-3 - i)x1 = 0
x1 = 0
Therefore, the solution for [x1] is [0], and [x2] can be any value since it is multiplied by t.
Now, let's analyze the given answer choices:
A: [5] [1+7i] * t
B: [5] [1-7i] * t
C: [5] [-1-7i] * t
D: [5] [-1+7i] * t
Since we have found that x1 = 0, the only correct answer choice is:
[x1] = [0]
[x2] = [1 + 7i] * t, which corresponds to answer choice A.
So, the correct solution is:
[x1] = [0]
[x2] = [1 + 7i] * t, where t is any value.