SOLUTION: 1] . Using matrix method, solve the following system of linear equations : 2x – y = 4, 2y + z = 5, z + 2x = 7.

Algebra ->  Matrices-and-determiminant -> SOLUTION: 1] . Using matrix method, solve the following system of linear equations : 2x – y = 4, 2y + z = 5, z + 2x = 7.      Log On


   

Question 868933: 1] . Using matrix method, solve the following system of linear equations :
2x – y = 4, 2y + z = 5, z + 2x = 7.
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
2 –1 +0 4
0 +2 +1 5
2 +0 +1 7
gauss jordan method
Divide row1 by 2
1 -1/2 0 2
0 2 1 5
2 0 1 7
Add (-2 * row1) to row3
1 -1/2 0 2
0 2 1 5
0 1 1 3
Divide row2 by 2
1 -1/2 0 2
0 1 1/2 5/2
0 1 1 3
Add (-1 * row2) to row3
1 -1/2 0 2
0 1 1/2 5/2
0 0 1/2 1/2
Divide row3 by 1/2
1 -1/2 0 2
0 1 1/2 5/2
0 0 1 1
Add (-1/2 * row3) to row2
1 -1/2 0 2
0 1 0 2
0 0 1 1
Add (1/2 * row2) to row1
1 0 0 3
0 1 0 2
0 0 1 1
x=3 y=2 z=1
Cramers rule
Solved by pluggable solver: Using Cramer's Rule to Solve Systems with 3 variables



system%282%2Ax%2B-1%2Ay%2B0%2Az=4%2C0%2Ax%2B2%2Ay%2B1%2Az=5%2C2%2Ax%2B0%2Ay%2B1%2Az=7%29



First let A=%28matrix%283%2C3%2C2%2C-1%2C0%2C0%2C2%2C1%2C2%2C0%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 4, 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=2. 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=6. 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=%286%29%2F%282%29=3



So the first solution is x=3




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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%2C2%2C-1%2C0%2C0%2C2%2C1%2C2%2C0%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=4.



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=%284%29%2F%282%29=2



So the second solution is y=2




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



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=%282%29%2F%282%29=1



So the third solution is z=1




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Final Answer:




So the three solutions are x=3, y=2, and z=1 giving the ordered triple (3, 2, 1)




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.