If possible, solve the following linear systems by Cramer's rule.
2x1 + 4x2 + 6x3 = 14
x1 + 2x3 = 0
2x1 + 3x2 − x3 = 30
i am having trouble with this one
To solve the given linear system using Cramer's rule, we need to find the determinant of the coefficient matrix and the determinants of the matrices formed by replacing each column with the constant terms. Here's a step-by-step solution:
Step 1: Write the system of equations in matrix form. Let's call the coefficient matrix A and the constant matrix B.
A = | 2 4 6 |
| 1 0 2 |
| 2 3 -1 |
B = | 14 |
| 0 |
| 30 |
Step 2: Calculate the determinant of the coefficient matrix A, denoted as det(A).
det(A) = | 2 4 6 |
| 1 0 2 |
| 2 3 -1 |
To compute the determinant, you can use various methods such as the expansion by minors or using the row or column operations. I will use the expansion by minors method.
det(A) = 2(0(-1) - 2(3)) - 4(1(-1) - 2(2)) + 6(1(3) - 0(2))
= 2(-6) - 4(-5) + 6(3)
= -12 + 20 + 18
= 26
Therefore, det(A) = 26.
Step 3: Replace the first column of the coefficient matrix A with the constant matrix B to form a new matrix A1.
A1 = | 14 4 6 |
| 0 0 2 |
| 30 3 -1 |
Step 4: Calculate the determinant of matrix A1, denoted as det(A1).
det(A1) = | 14 4 6 |
| 0 0 2 |
| 30 3 -1 |
Using the same method as before,
det(A1) = 14(0(-1) - 2(3)) - 4(0(-1) - 2(-1)) + 6(0(3) - 0(-1))
= 14(-6) - 4(2) + 6(0)
= -84 - 8 + 0
= -92
Therefore, det(A1) = -92.
Step 5: Replace the second column of the coefficient matrix A with the constant matrix B to form a new matrix A2.
A2 = | 2 14 6 |
| 1 0 2 |
| 2 30 -1 |
Step 6: Calculate the determinant of matrix A2, denoted as det(A2).
det(A2) = | 2 14 6 |
| 1 0 2 |
| 2 30 -1 |
Again, using the same method,
det(A2) = 2(0(-1) - 2(30)) - 14(1(-1) - 2(-1)) + 6(1(30) - 0(-1))
= 2(-60) - 14(3) + 6(30)
= -120 - 42 + 180
= 18
Therefore, det(A2) = 18.
Step 7: Replace the third column of the coefficient matrix A with the constant matrix B to form a new matrix A3.
A3 = | 2 4 14 |
| 1 0 0 |
| 2 3 30 |
Step 8: Calculate the determinant of matrix A3, denoted as det(A3).
det(A3) = | 2 4 14 |
| 1 0 0 |
| 2 3 30 |
Using the same method as before,
det(A3) = 2(0(30) - 0(3)) - 4(1(30) - 0(2)) + 14(1(3) - 0(2))
= 2(0) - 4(30) + 14(3)
= 0 - 120 + 42
= -78
Therefore, det(A3) = -78.
Step 9: Use Cramer's rule to find the values of x1, x2, and x3.
x1 = det(A1) / det(A) = (-92) / 26 = -92/26 = -46/13
x2 = det(A2) / det(A) = 18 / 26 = 9 / 13
x3 = det(A3) / det(A) = -78 / 26 = -3
So, the solutions to the given linear system are:
x1 = -46/13
x2 = 9/13
x3 = -3