Technology
Solving a System of Linear Equations Using the Matrix Method
Solving a System of Linear Equations Using the Matrix Method
Often, we face systems of linear equations that can be tedious to solve manually. However, the matrix method offers a systematic and efficient approach. Let's explore solving the following system of linear equations using the matrix method:
Problem Statement
Consider the following system of linear equations:
3x 5y - z 13
2x 7y z 28
x 7y 2z 32
Step 1: Representing the System in Matrix Form
We can represent this system of equations in matrix form as ( Amathbf{x} mathbf{b} ), where:
Matrix ( A )
A begin{pmatrix} 3 5 -1 2 7 1 1 7 2 end{pmatrix}
Vector ( mathbf{x} )
mathbf{x} begin{pmatrix} x y z end{pmatrix}
Vector ( mathbf{b} )
mathbf{b} begin{pmatrix} 13 28 32 end{pmatrix}
Step 2: Finding the Inverse of Matrix ( A )
To solve for ( mathbf{x} ), we need to find the inverse of ( A ). The first step is to calculate the determinant of ( A ).
det ( A ) 3*7*2 - 1*(2*7 - 1*7) - 5*(2*2 - 1*1) - (-1)*(2*7 - 1*1) 42 - (14 - 7) - 5*(4 - 1) (14 - 1) 42 - 7 - 15 13 33
Since the determinant is not zero, the matrix ( A ) is invertible. We can find the inverse ( A^{-1} ) using the formula:
( A^{-1} frac{1}{text{det}(A)} text{adj}(A) )
For the sake of brevity, let's use a calculator or software to find the inverse:
Matrix ( A^{-1} )
A^{-1} ≈ begin{pmatrix} 2 -5 1 -1 3 -1 1 -1 1 end{pmatrix}
Step 3: Solving for ( mathbf{x} )
Now, we can compute ( mathbf{x} A^{-1} mathbf{b} ).
( mathbf{x} begin{pmatrix} 2 -5 1 -1 3 -1 1 -1 1 end{pmatrix} begin{pmatrix} 13 28 32 end{pmatrix} )
Calculating this:
x 2*13 - 5*28 1*32 26 - 140 32 -82
y -1*13 3*28 - 1*32 -13 84 - 32 39
z 1*13 - 1*28 1*32 13 - 28 32 17
Final Solution
Thus, the solution to the system of equations is:
x -82, y 39, z 17
This solution can be verified by substituting the values back into the original equations to ensure that they hold true.