Matrix Inverse Method
The matrix inverse method is a powerful approach for solving a system of linear equations using matrices. When we have a system of equations, we can represent it in matrix form and, if the coefficient matrix is invertible, solve it by multiplying both sides of the equation by the inverse of the matrix.
1. Matrix Representation of a System of Linear Equations
Consider a system of n linear equations with n unknowns:
[ a_1x + b_1y + c_1z = d_1 \ a_2x + b_2y + c_2z = d_2 \ a_3x + b_3y + c_3z = d_3 ]
This can be written in matrix form as:
[ AX = B ]
Where:
-
Ais the coefficient matrix, -
Xis the column matrix of unknowns, -
Bis the column matrix of constants.
[ A = \begin{bmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3 \end{bmatrix}, \quad X = \begin{bmatrix} x \ y \ z \end{bmatrix}, \quad B = \begin{bmatrix} d_1 \ d_2 \ d_3 \end{bmatrix} ]
2. Matrix Inverse Method
If the matrix A is invertible (i.e., has an inverse, denoted as A^{-1}), we can solve for X by multiplying both sides of the matrix equation by A^{-1}:
[ AX = B \implies A^{-1}AX = A^{-1}B \implies X = A^{-1}B ]
Thus, the solution is obtained by multiplying the inverse of the coefficient matrix A by the constant matrix B.
Steps:
- Write the system of equations in matrix form.
- Find the inverse of matrix
A. - Multiply the inverse of
Aby the matrixBto find the solutionX.
Example 1: Solving a 2x2 System Using the Matrix Inverse Method
Consider the following system of linear equations:
[ 2x + 3y = 5 \ 4x + y = 6 ]
Step 1: Write the system in matrix form
[ \begin{bmatrix} 2 & 3 \ 4 & 1 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix}
\begin{bmatrix} 5 \ 6 \end{bmatrix} ]
Here: [ A = \begin{bmatrix} 2 & 3 \ 4 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 5 \ 6 \end{bmatrix} ]
Step 2: Find the inverse of matrix A
For a 2x2 matrix A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, the inverse is given by:
[ A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} ]
For the matrix:
[ A = \begin{bmatrix} 2 & 3 \ 4 & 1 \end{bmatrix}, \quad ad - bc = (2 \times 1) - (4 \times 3) = 2 - 12 = -10 ]
Thus, the inverse of A is:
[ A^{-1} = \frac{1}{-10} \begin{bmatrix} 1 & -3 \ -4 & 2 \end{bmatrix} = \begin{bmatrix} -0.1 & 0.3 \ 0.4 & -0.2 \end{bmatrix} ]
Step 3: Multiply the inverse of A by B
[ X = A^{-1}B = \begin{bmatrix} -0.1 & 0.3 \ 0.4 & -0.2 \end{bmatrix} \begin{bmatrix} 5 \ 6 \end{bmatrix} ]
Perform the matrix multiplication:
[ X = \begin{bmatrix} (-0.1 \times 5) + (0.3 \times 6) \ (0.4 \times 5) + (-0.2 \times 6) \end{bmatrix} = \begin{bmatrix} -0.5 + 1.8 \ 2.0 - 1.2 \end{bmatrix} = \begin{bmatrix} 1.3 \ 0.8 \end{bmatrix} ]
Thus, the solution is ( x = 1.3 ) and ( y = 0.8 ).
Example 2: Solving a 3x3 System Using the Matrix Inverse Method
Consider the following system of linear equations:
[ x + 2y + 3z = 9 \ 2x + 3y + 5z = 19 \ 3x + 2y + z = 10 ]
Step 1: Write the system in matrix form
[ \begin{bmatrix} 1 & 2 & 3 \ 2 & 3 & 5 \ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix}
\begin{bmatrix} 9 \ 19 \ 10 \end{bmatrix} ]
Here: [ A = \begin{bmatrix} 1 & 2 & 3 \ 2 & 3 & 5 \ 3 & 2 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 9 \ 19 \ 10 \end{bmatrix} ]
Step 2: Find the inverse of matrix A
Finding the inverse of a 3x3 matrix is more involved than for a 2x2 matrix. For a 3x3 matrix, we can use methods like the adjugate matrix and the determinant of the matrix. For this example, assume that the inverse of matrix A (calculated by hand or using software) is:
[ A^{-1} = \begin{bmatrix} -3 & 1 & 5 \ 5 & -2 & -7 \ 2 & 0 & -3 \end{bmatrix} ]
Step 3: Multiply the inverse of A by B
[ X = A^{-1}B = \begin{bmatrix} -3 & 1 & 5 \ 5 & -2 & -7 \ 2 & 0 & -3 \end{bmatrix} \begin{bmatrix} 9 \ 19 \ 10 \end{bmatrix} ]
Perform the matrix multiplication:
[ X = \begin{bmatrix} (-3 \times 9) + (1 \times 19) + (5 \times 10) \ (5 \times 9) + (-2 \times 19) + (-7 \times 10) \ (2 \times 9) + (0 \times 19) + (-3 \times 10) \end{bmatrix} ]
[ X = \begin{bmatrix} -27 + 19 + 50 \ 45 - 38 - 70 \ 18 + 0 - 30 \end{bmatrix} = \begin{bmatrix} 42 \ -63 \ -12 \end{bmatrix} ]
Thus, the solution is ( x = 42 ), ( y = -63 ), and ( z = -12 ).
Conclusion
The matrix inverse method is a highly efficient technique for solving systems of linear equations, especially when the system is well-structured and the coefficient matrix is invertible. This method relies on matrix operations, providing a concise and systematic way to solve linear equations for any number of variables.
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