SOLUTION: use cramer's 8X1- X2 =16 2X2+5X3 = 5 2X1+3X2 =7

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Question 859143: use cramer's
8X1- X2 =16
2X2+5X3 = 5
2X1+3X2 =7
Answer by ewatrrr(24785) 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%288%2Ax%2B1%2Ay%2B0%2Az=16%2C0%2Ax%2B2%2Ay%2B5%2Az=5%2C2%2Ax%2B3%2Ay%2B0%2Az=7%29



First let A=%28matrix%283%2C3%2C8%2C1%2C0%2C0%2C2%2C5%2C2%2C3%2C0%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 16, 5, and 7 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=-110. 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=-205. 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=%28-205%29%2F%28-110%29=41%2F22



So the first solution is x=41%2F22




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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%2C8%2C1%2C0%2C0%2C2%2C5%2C2%2C3%2C0%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=-120.



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=%28-120%29%2F%28-110%29=12%2F11



So the second solution is y=12%2F11




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Let's reset again by letting A=%28matrix%283%2C3%2C8%2C1%2C0%2C0%2C2%2C5%2C2%2C3%2C0%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=-62.



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=%28-62%29%2F%28-110%29=31%2F55



So the third solution is z=31%2F55




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Final Answer:




So the three solutions are x=41%2F22, y=12%2F11, and z=31%2F55 giving the ordered triple (41/22, 12/11, 31/55)




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.