SOLUTION: x+y+z=1 x+2y+3z=2 x+4y+9z=4 solve with Cramer method

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Question 1103596: x+y+z=1
x+2y+3z=2
x+4y+9z=4 solve with Cramer method

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
Before we begin, you should note that the coefficient of y is equal to the right hand value. This means that y=1 and x=0 and z=0
Solved by pluggable solver: Using Cramer's Rule to Solve Systems with 3 variables



system%281%2Ax%2B1%2Ay%2B1%2Az=1%2C1%2Ax%2B2%2Ay%2B3%2Az=2%2C1%2Ax%2B4%2Ay%2B9%2Az=4%29



First let A=%28matrix%283%2C3%2C1%2C1%2C1%2C1%2C2%2C3%2C1%2C4%2C9%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 1, 2, and 4 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=0. 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=%280%29%2F%282%29=0



So the first solution is x=0




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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%2C1%2C1%2C1%2C2%2C3%2C1%2C4%2C9%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%282%29=1



So the second solution is y=1




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Let's reset again by letting A=%28matrix%283%2C3%2C1%2C1%2C1%2C1%2C2%2C3%2C1%2C4%2C9%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=0.



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=%280%29%2F%282%29=0



So the third solution is z=0




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Final Answer:




So the three solutions are x=0, y=1, and z=0 giving the ordered triple (0, 1, 0)




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.