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Systems of Three Equations

Solving using Matrices and Cramer's Rule

Problems

Problems

The Determinant

There is another way to solve systems of equations with three variables. It involves a quantity called the determinant.

Every m×m matrix has a unique determinant. The determinant is a single number. To find the determinant of a 2×2matrix , multiply the numbers on the downward diagonal and subtract the product of the numbers on the upward diagonal:

A =    


detA = a 1 b 2 - a 2 b 1 .
For example,

det = 4(6) - (- 1)(- 2) = 24 - 2 = 22    


To find the determinant of a 3×3 matrix, copy the first two columns of the matrix to the right of the original matrix. Next, multiply the numbers on the three downward diagonals, and add these products together. Multiply the numbers on the upward diagonals, and add these products together. Then subtract the sum of the products of the upward diagonals from the sum of the product of the downward diagonals (subtract the second number from the first number):

A =    

   


Example: Find the determinant of:

   



Solution:

Step 1

   



Step 2


Step 3


Step 4

10 - 80 = -70. detA = - 70 .

Cramer's Rule

Recall the general 3×4 matrix used to solve systems of three equations:

   

This matrix will be used to solve systems by Cramer's Rule. We divide it into four separate 3×3 matrices:

D =    


D x =    

D y =    

D z =    


D is the 3×3 coefficient matrix, and D x , D y , and D z are each the result of substituting the constant column for one of the coefficient columns in D .


Cramer's Rule states that:

x =
y =
z =
Thus, to solve a system of three equations with three variables using Cramer's Rule,
  1. Arrange the system in the following form:
    a 1 x + b 1 y + c 1 z = d 1
    a 2 x + b 2 y + c 2 z = d 2
    a 3 x + b 3 y + c 3 z = d 3
  2. Create D , D x , D y , and D z .
  3. Find detD , detD x , detD y , and detD z .
  4. x = , y = , and z = .
Note: If detD = 0 and detD x , detD y , or detD z≠ 0 , the system is inconsistent. If detD = 0 and detD x = detD y = detD z = 0 , the system has multiple solutions.


Example:Solve the following system:

8x + 10z = 7y + 15
2x + 3y + 8z = 7
5y + 9 = 4x + 2z
  1. Rearrange the system:
    8x - 7y + 10z = 15
    2x + 3y + 8z = 7
    -4x + 5y - 2z = - 9
  2. Create the matrices:

    D =    


    D x =    

    D y =    

    D z =    

  3. Find the determinants:

    detD = (- 48 + 224 + 100) - (- 120 + 320 + 28) = 276 - 228 = 48
    detD x = (- 90 + 504 + 350) - (- 270 + 600 + 98) = 764 - 428 = 336
    detD y = (- 112 - 480 - 180) - (- 280 - 576 - 60) = - 772 - (- 916) = 144
    detD z = (- 216 + 196 + 150) - (- 180 + 280 + 126) = 130 - 226 = - 96

  4. x = = 7 . y = = 3 . z = = - 2 .
Thus, (x, y, z) = (7, 3, - 2) .

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