Can someone please help me convert these equations into a matrix equation.
x+4y-z=7
x+y+z=2
-2x-2y+2z=-8
pretend these are written across the page and have matrix brackets around them
1 4 -1
1 1 1
-2 -2 2
x
x
y
z
=
7
2
-8
x+y+z=2 have the invisible 1 in front of them correct
Certainly! To convert these equations into a matrix equation, we need to represent the coefficients of each variable, as well as the constant terms, in matrix form.
Let's represent the variables x, y, and z as a column vector X, given by:
X = [x]
[y]
[z]
The given equations can be expressed in matrix form as AX = B, where A is the coefficient matrix, X is the column vector of variables, and B is the constant column vector.
We can obtain A and B as follows:
A = [1 4 -1]
[1 1 1]
[-2 -2 2]
B = [7]
[2]
[-8]
With A, X, and B, the matrix equation looks like this: AX = B.
To solve for X, we can rearrange the equation as follows:
X = A^(-1) * B
To find the inverse of matrix A (A^(-1)), we calculate the inverse of A:
A^(-1) = (1/det(A)) * adj(A)
Here, det(A) denotes the determinant of matrix A, and adj(A) represents the adjugate of matrix A.
To calculate the inverse of A, we can follow these steps:
1. Calculate the determinant of A, denoted as det(A).
det(A) = (1 * 1 * 2) + (4 * 1 * -2) + (-1 * 1 * -2) - (-1 * 1 * 2) - (4 * -2 * -1) - (1 * 1 * 1)
= 2 + (-8) + 2 - (-2) - 16 - 1
= 2 - 8 + 2 + 2 + 16 - 1
= 13
2. Calculate the matrix of minors (M) for matrix A.
The matrix of minors is obtained by calculating the determinant of each 2x2 matrix formed by excluding one row and one column from A.
M = [(1*2 - 1*(-2)) (1*(-2) - 1*2) (1*1 - 1*4)]
[(4*2 - 1*2) (4*(-2) - 1*(-2)) (4*1 - 1*4)]
[(-1*2 - 4*(-2)) (-1*(-2) - 4*2) (-1*1 - 4*4)]
= [(2 + 2) (-2 - 2) (1 - 4)]
[(8 - 2) (-8 + 2) (4 - 4)]
[(-2 + 8) (2 - 8) (-1 - 16)]
= [4 -4 -3]
[6 -6 0]
[6 -6 -17]
3. Calculate the matrix of cofactors (C) from the matrix of minors (M).
The matrix of cofactors is obtained by multiplying each element of the matrix of minors by (-1) raised to the power of the row and column number (i + j).
C = [(4 * (-1)^(1+1)) (-4 * (-1)^(1+2)) (-3 * (-1)^(1+3))]
[(6 * (-1)^(2+1)) (-6 * (-1)^(2+2)) (0 * (-1)^(2+3))]
[(6 * (-1)^(3+1)) (-6 * (-1)^(3+2)) (-17 * (-1)^(3+3))]
= [4 4 -3]
[-6 6 0]
[-6 -6 -17]
4. Calculate the adjugate matrix (adj(A)) from the matrix of cofactors (C).
The adjugate matrix is obtained by transposing the matrix of cofactors.
adj(A) = [4 -6 -6]
[4 6 -6]
[-3 0 -17]
5. Calculate the inverse matrix (A^(-1)) using the formula:
A^(-1) = (1/det(A)) * adj(A)
A^(-1) = (1/13) * [4 -6 -6]
[4 6 -6]
[-3 0 -17]
Now that we have found A^(-1), we can substitute it into the equation X = A^(-1) * B to find the solution for X.
X = [4 -6 -6] * [7]
[2]
[-8]
To perform the matrix multiplication, we multiply each row of the matrix A^(-1) by the column vector B.
X = [4*7 + (-6)*2 + (-6)*(-8)]
[4*7 + 6*2 + (-6)*(-8)]
[-3*7 + 0*2 + (-17)*(-8)]
X = [18]
[44]
[23]
Now we have the solution: x = 18, y = 44, z = 23.