SOLUTION: 1a + 2b + 1c = 5 1a - 1b - 1c = 3 0a + 1b + 1c = 2

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Question 499824: 1a + 2b + 1c = 5
1a - 1b - 1c = 3
0a + 1b + 1c = 2
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

Solved by pluggable solver: Using Cramer's Rule to Solve Systems with 3 variables



system%281%2Ax%2B2%2Ay%2B1%2Az=5%2C1%2Ax%2B-1%2Ay%2B-1%2Az=3%2C0%2Ax%2B1%2Ay%2B1%2Az=2%29



First let A=%28matrix%283%2C3%2C1%2C2%2C1%2C1%2C-1%2C-1%2C0%2C1%2C1%29%29. This is the matrix formed by the coefficients of the given system of equations.


Take note that the right hand values of the system are 5, 3, and 2 and they are highlighted here:




These values are important as they will be used to replace the columns of the matrix A.




Now let's calculate the the determinant of the matrix A to get abs%28A%29=-1. To save space, I'm not showing the calculations for the determinant. However, if you need help with calculating the determinant of the matrix A, check out this solver.



Notation note: abs%28A%29 denotes the determinant of the matrix A.



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Now replace the first column of A (that corresponds to the variable 'x') with the values that form the right hand side of the system of equations. We will denote this new matrix A%5Bx%5D (since we're replacing the 'x' column so to speak).






Now compute the determinant of A%5Bx%5D to get abs%28A%5Bx%5D%29=-5. Again, as a space saver, I didn't include the calculations of the determinant. Check out this solver to see how to find this determinant.



To find the first solution, simply divide the determinant of A%5Bx%5D by the determinant of A to get: x=%28abs%28A%5Bx%5D%29%29%2F%28abs%28A%29%29=%28-5%29%2F%28-1%29=5



So the first solution is x=5




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We'll follow the same basic idea to find the other two solutions. Let's reset by letting A=%28matrix%283%2C3%2C1%2C2%2C1%2C1%2C-1%2C-1%2C0%2C1%2C1%29%29 again (this is the coefficient matrix).




Now replace the second column of A (that corresponds to the variable 'y') with the values that form the right hand side of the system of equations. We will denote this new matrix A%5By%5D (since we're replacing the 'y' column in a way).






Now compute the determinant of A%5By%5D to get abs%28A%5By%5D%29=2.



To find the second solution, divide the determinant of A%5By%5D by the determinant of A to get: y=%28abs%28A%5By%5D%29%29%2F%28abs%28A%29%29=%282%29%2F%28-1%29=-2



So the second solution is y=-2




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Let's reset again by letting A=%28matrix%283%2C3%2C1%2C2%2C1%2C1%2C-1%2C-1%2C0%2C1%2C1%29%29 which is the coefficient matrix.



Replace the third column of A (that corresponds to the variable 'z') with the values that form the right hand side of the system of equations. We will denote this new matrix A%5Bz%5D






Now compute the determinant of A%5Bz%5D to get abs%28A%5Bz%5D%29=-4.



To find the third solution, divide the determinant of A%5Bz%5D by the determinant of A to get: z=%28abs%28A%5Bz%5D%29%29%2F%28abs%28A%29%29=%28-4%29%2F%28-1%29=4



So the third solution is z=4




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Final Answer:




So the three solutions are x=5, y=-2, and z=4 giving the ordered triple (5, -2, 4)




Note: there is a lot of work that is hidden in finding the determinants. Take a look at this 3x3 Determinant Solver to see how to get each determinant.