If possible, solve the following linear systems by Cramer's rule.
2x1 + 4x2 + 6x3 = 14
x1 + 2x3 = 0
2x1 + 3x2 − x3 = 30
I will give you 2 nice youtubes to show how this is done
https://www.youtube.com/watch?v=taBHTo8sviM
She sets it up nicely, but does not actually show how to find the determinant of a 3by3.
This one shows a nice short-cut method to find the determinant, but I must point out that this only works for 3by3 matrix
https://www.youtube.com/watch?v=ROFcVgehEYA
To solve the given linear system using Cramer's rule, we need to find the values of x1, x2, and x3 by calculating determinants.
Step 1: Write the system in matrix form:
2 4 6 x1 14
1 0 2 x2 0
2 3 -1 x3 30
Step 2: Calculate the determinant of the coefficient matrix, A.
|A| = |2 4 6|
|1 0 2|
|2 3 -1|
To calculate the determinant, we can use the Laplace expansion method. Let's calculate the determinant of matrix A:
|A| = 2(det|0 2|) - 4(det|1 2|) + 6(det|1 0|)
|-1 -1| |3 -1| |2 3|
Now, calculate the determinants:
|A| = 2(0 - 2) - 4(1 * 3 + 2 * (-1)) + 6(1 * 3 - 0 * 2)
= -4 - 4(-1) + 6(3)
= -4 + 4 + 18
= 18
So, |A| = 18.
Step 3: Calculate the determinant of the x1 matrix, A1.
Replace the first column of matrix A with the column of constants. Calculate the determinant of this new matrix.
A1 = |14 4 6|
|0 0 2|
|30 3 -1|
Using the same Laplace expansion method, we can calculate A1:
A1 = 14(det|0 2|) - 4(det|0 2|) + 6(det|0 0|)
|-1 -1| |3 -1| |3 3|
Now, calculate the determinants:
A1 = 14(0 - 2) - 4(0 * (-1) + 2 * (-1)) + 6(0 * (-1) - 0 * 3)
= -28 - 4(-2) + 6(0)
= -28 + 8 + 0
= -20
So, A1 = -20.
Step 4: Calculate the determinant of the x2 matrix, A2.
Replace the second column of matrix A with the column of constants. Calculate the determinant of this new matrix.
A2 = |2 14 6|
|1 0 2|
|2 30 -1|
Using the same Laplace expansion method, we can calculate A2:
A2 = 2(det|0 2|) - 14(det|1 2|) + 6(det|1 0|)
|-1 -1| |3 -1| |3 3|
Now, calculate the determinants:
A2 = 2(0 - 2) - 14(1 * 3 + 2 * (-1)) + 6(1 * 3 - 0 * 2)
= -4 - 14(-1) + 6(3)
= -4 + 14 + 18
= 28
So, A2 = 28.
Step 5: Calculate the determinant of the x3 matrix, A3.
Replace the third column of matrix A with the column of constants. Calculate the determinant of this new matrix.
A3 = |2 4 14|
|1 0 0|
|2 3 30|
Using the same Laplace expansion method, we can calculate A3:
A3 = 2(det|0 0|) - 4(det|1 0|) + 14(det|1 0|)
|-1 3| |2 30| |2 3|
Now, calculate the determinants:
A3 = 2(0 - 0) - 4(1 * 30 - 0 * 2) + 14(1 * 3 - 0 * 2)
= 0 - 4(30) + 14(3)
= 0 - 120 + 42
= -78
So, A3 = -78.
Step 6: Calculate the values of x1, x2, and x3.
Using Cramer's rule, we can find the values of the variables:
x1 = A1 / |A| = -20 / 18 = -10/9
x2 = A2 / |A| = 28 / 18 = 14/9
x3 = A3 / |A| = -78 / 18 = -13/3
Therefore, the solution to the given linear system is:
x1 = -10/9
x2 = 14/9
x3 = -13/3