2x-y+3z=9,x+y+z=6x-y+z=2 solve equation on matrix inversion method
To solve the system of equations using the matrix inversion method, we need to represent the system in matrix form. Let's call the coefficient matrix A, the variable matrix X, and the constant vector B.
The original system of equations can be written as:
2x - y + 3z = 9 (equation 1)
x + y + z = 6 (equation 2)
x - y + z = 2 (equation 3)
First, we need to organize the coefficients of the variables into the coefficient matrix A:
A = | 2 -1 3 |
| 1 1 1 |
| 1 -1 1 |
Next, we organize the variables into the variable matrix X:
X = | x |
| y |
| z |
Finally, we organize the constants into the constant vector B:
B = | 9 |
| 6 |
| 2 |
Now we can solve for X by using the matrix inversion:
X = A^(-1) * B
To find the inverse of A, we can use the formula for a 3x3 matrix:
A^(-1) = (1/det(A)) * adj(A)
where det(A) is the determinant of A and adj(A) is the adjugate of A.
Before we calculate the inverse, let's find det(A):
det(A) = (2 * 1 * 1) + (-1 * 1 * 1) + (3 * 1 * -1) - (3 * 1 * 1) - (2 * 1 * -1) - (-1 * 1 * 1)
= 2 - 1 - 3 - 3 + 2 + 1
= -2
Now, let's calculate the adjugate of A:
adj(A) = | (1 * 1 - 1 * -1) (-1 * 1 - 1 * 1) (1 * -1 - 1 * 1) |
| (-1 * 1 - 1 * -1) (2 * 1 - 1 * 1) (-1 * 1 - 2 * -1) |
| (1 * -1 - 3 * 1) (-1 * -1 - 2 * 1) (2 * 1 - 1 * -1) |
= | 0 -2 -2 |
| 0 1 1 |
| -4 1 3 |
Now, we can calculate A^(-1):
A^(-1) = (1/det(A)) * adj(A)
= (1/-2) * | 0 -2 -2 |
| 0 1 1 |
| -4 1 3 |
= | 0 1 1 |
| 0.5 -0.5 -0.5 |
| 2 -0.5 -1.5 |
Finally, we can solve for X:
X = A^(-1) * B
= | 0 1 1 | * | 9 |
| 6 |
| 2 |
= | 0*9 + 1*6 + 1*2 |
| 0*9 + 0.5*6 + (-0.5)*2 |
| 2*9 + (-0.5)*6 + (-1.5)*2 |
= | 6 + 2 |
| 0 + 3 - 1 |
| 18 - 3 - 3 |
= | 8 |
| 2 |
| 12 |
So the solution to the system of equations is x = 8, y = 2, and z = 12.
To solve the system of linear equations using the matrix inversion method, we need to represent the system of equations in matrix form and then find the inverse of the coefficient matrix.
Let's represent the system of equations in matrix form:
[A] [X] = [B]
where:
[A] is the coefficient matrix,
[X] is the unknown matrix containing variables x, y, and z,
[B] is the matrix containing constants on the right side of each equation.
From the given system of equations:
2x - y + 3z = 9 ---> equation 1
x + y + z = 6 ---> equation 2
x - y + z = 2 ---> equation 3
We can create the coefficient matrix [A]:
[A] = | 2 -1 3 |
| 1 1 1 |
| 1 -1 1 |
And the matrix [B] containing the constants:
[B] = | 9 |
| 6 |
| 2 |
Now, let's find the inverse of matrix [A] using the matrix inversion method.
1. Calculate the determinant of [A], denoted as |A|:
|A| = 2[(1)(1) - (-1)(1)] - (-1)[(1)(1) - (-1)(3)] + 3[(1)(-1) - (1)(3)]
= 2(2) - (-1)(-2) + 3(-4)
= 4 + 2 - 12
= -6
2. Calculate the adjoint of [A], denoted as adj(A):
adj(A) = | 1(1) - (-1)(1) -1(1) - (-1)(3) 1(1) - (1)(-1) |
| -1(1) - (1)(1) 2(1) - (1)(3) -2(1) - (-1)(1) |
| -1(-1) - (1)(1) 2(-1) - (-1)(3) -2(-1) - (-1)(-1) |
adj(A) = | 2 -2 2 |
| -2 -1 -1 |
| 1 2 -1 |
3. Calculate the inverse of [A], denoted as [A]^-1:
[A]^-1 = (1 / |A|) * adj(A)
= (1 / -6) * | 2 -2 2 |
| -2 -1 -1 |
| 1 2 -1 |
[A]^-1 = | -2/3 1/3 -1/3 |
| -1/3 -1/6 1/6 |
| -1/6 1/3 -1/6 |
Now that we have the inverse matrix [A]^-1, we can solve for [X] by multiplying both sides of the equation by [A]^-1:
[A]^-1[A][X] = [A]^-1[B]
[X] = [A]^-1[B]
Substituting the values we derived, we have:
| x | | -2/3 1/3 -1/3 | | 9 |
| y | = | -1/3 -1/6 1/6 | * | 6 |
| z | | -1/6 1/3 -1/6 | | 2 |
Multiplying the matrices, we get:
|x| | -2/3 * 9 + 1/3 * 6 - 1/3 * 2 |
|y| = | -1/3 * 9 - 1/6 * 6 + 1/6 * 2 |
|z| | -1/6 * 9 + 1/3 * 6 - 1/6 * 2 |
Simplifying:
|x| | -6 + 2 - 2/3 |
|y| = | -3/3 - 1 + 1/3 |
|z| | -3/6 + 2 - 1/6 |
|x| | -6 + 2 - 2/3 |
|y| = | -1 - 1/3 |
|z| | -1/2 + 2 - 1/6 |
|x| | -22/3 |
|y| = | -4/3 |
|z| | 5/6 |
Therefore, the solution to the system of linear equations is:
x = -22/3
y = -4/3
z = 5/6
If you have Ax=B, then x = A^-1*B
Here, A^-1 =
http://www.wolframalpha.com/input/?i=inverse+%7B%7B2,-1,3%7D,%7B1,1,1%7D,%7B1,-1,1%7D%7D