SOLUTION: I am stuck on a project, where I have to solve a matrix ( [2 5] [3 1] with the 2 being over the 3) by using cramer's rule.

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Question 442543: I am stuck on a project, where I have to solve a matrix ( [2 5] [3 1] with the 2 being over the 3) by using cramer's rule.
Found 2 solutions by MathLover1, swincher4391:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

Solved by pluggable solver: Finding the Determinant of a 2x2 Matrix

If you have the general 2x2 matrix:

%28matrix%282%2C2%2Ca%2Cb%2Cc%2Cd%29%29

the determinant is: D=a%2Ad-c%2Ab

So this means that

abs%28matrix%282%2C2%2Ca%2Cb%2Cc%2Cd%29%29=a%2Ad-c%2Ab

Note: the vertical bars denote a determinant.


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So in this case the determinant of %28matrix%282%2C2%2C2%2C5%2C3%2C1%29%29 is:


abs%28matrix%282%2C2%2C2%2C5%2C3%2C1%29%29=%282%29%281%29-%283%29%285%29=2-15=-13


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Answer:

So abs%28matrix%282%2C2%2C2%2C5%2C3%2C1%29%29=-13 which means that the determinant of the matrix %28matrix%282%2C2%2C2%2C5%2C3%2C1%29%29 is -13

Answer by swincher4391(1107) About Me  (Show Source):
You can put this solution on YOUR website!
Given a matrix:
matrix%282%2C2%2C2%2C5%2C3%2C1%29
Cramer's rule states that the xi = det(Ai)/det(A)
So what is the det(A)? det(A) = 2*1 - 3*5 = -13
You do not specify the solution matrix in this problem, but I will act like you did and you can apply it to your problem.
Let's say the solution matrix is matrix%282%2C1%2C1%2C2%29.
Then to solve for x1 you would create a matrix where the solution matrix is substituted in for column 1.
In this made up example, you'd find the det(A1) where A1 is:
matrix%282%2C2%2C1%2C5%2C2%2C1%29
This determinant is 1 -10 = -9.
So the solution to x1 = -9/-13 = (9/13)
To find x2 create an A2 matrix.
The matrix would be:
matrix%282%2C2%2C2%2C1%2C3%2C2%29
The det(A2) = 4 - 3 = 1
So then x2 = 1 / -13 = (-1/13)
To show that Cramer's rule indeed works:
2x1 + 5x2 = 1
3x1 + 1x2 = 2
Where as I said matrix%281%2C2%2C1%2C2%29 is my solution matrix and matrix%282%2C2%2C2%2C5%2C3%2C1%29 was the matrix you gave me.
If I plug in x1=9/13 and x2=-1/13
2(9/13) + 5(-1/13) = 13/13 = 1. Check.
3(9/13) + 1(-1/13) = 26/13 = 2. Check.
Now for whatever your solution matrix is, apply the same steps.