Solving Systems Of Equations Using Matrix Equations

Jul 13, 2025 by ADMIN 52 views

In mathematics, solving systems of equations is a fundamental skill. Matrix equations provide a powerful and elegant method for tackling these systems. This article will delve into how to represent a system of linear equations in matrix form and demonstrate how to solve it using matrix algebra. We will specifically address the system:

3x - 2y = -3
6x - 5y = -9

and determine the correct matrix equation that leads to its solution.

Understanding Matrix Representation of Linear Equations

To effectively utilize matrix equations, it's crucial to understand how a system of linear equations can be translated into matrix form. A system of linear equations, like the one given, can be represented in the form Ax = b, where:

A is the coefficient matrix, containing the coefficients of the variables.
x is the variable matrix, a column matrix containing the variables.
b is the constant matrix, a column matrix containing the constants on the right-hand side of the equations.

For our specific system:

3x - 2y = -3
6x - 5y = -9

We can identify the following matrices:

A (Coefficient Matrix): The coefficients of x and y in the equations form the coefficient matrix.
```
A = | 3 -2 |
    | 6 -5 |
```
x (Variable Matrix): This matrix contains the variables we want to solve for.
```
x = | x |
    | y |
```
b (Constant Matrix): The constants on the right side of the equations form this matrix.
```
b = | -3 |
    | -9 |
```

Therefore, the matrix equation representing the system is:

| 3 -2 | | x | = | -3 |
| 6 -5 | | y |   | -9 |

This foundational step is essential because it allows us to apply the tools of matrix algebra to solve the system efficiently. Matrix representation not only simplifies the notation but also provides a clear pathway for solving the system using matrix inversion.

Solving Matrix Equations: The Inverse Matrix Method

Now that we've represented the system of equations in the matrix form Ax = b, we can solve for the variable matrix x. The primary method for solving such equations involves using the inverse of the coefficient matrix, A. If the matrix A is invertible (i.e., its determinant is non-zero), we can find its inverse, denoted as A⁻¹. Multiplying both sides of the equation Ax = b by A⁻¹ on the left gives us:

A⁻¹Ax = A⁻¹b

Since A⁻¹A is the identity matrix I (which has the property that Ix = x), the equation simplifies to:

x = A⁻¹b

This equation tells us that to find the solution matrix x, we simply need to multiply the inverse of the coefficient matrix A⁻¹ by the constant matrix b. The key, therefore, lies in finding the inverse matrix A⁻¹.

For a 2x2 matrix, such as our coefficient matrix A, the inverse can be calculated using a specific formula. If A is given by:

A = | a b |
    | c d |

Then its inverse A⁻¹ is given by:

A⁻¹ = (1 / (ad - bc)) | d -b |
                     | -c a |

where (ad - bc) is the determinant of A. If the determinant is zero, the matrix is singular and does not have an inverse, indicating that the system of equations may have either no solution or infinitely many solutions. In our case, the coefficient matrix A is:

A = | 3 -2 |
    | 6 -5 |

The determinant of A is (3 * -5) - (-2 * 6) = -15 + 12 = -3, which is non-zero, so A is invertible. Applying the formula for the inverse, we get:

A⁻¹ = (1 / -3) | -5  2 |
              | -6  3 |

A⁻¹ = | 5/3 -2/3 |
      | 2   -1   |

With the inverse matrix A⁻¹ calculated, we can now proceed to find the solution matrix x by multiplying A⁻¹ by b.

Applying the Inverse to Find the Solution

Now that we have the inverse of the coefficient matrix A⁻¹ and the constant matrix b, we can compute the solution matrix x using the equation x = A⁻¹b. Recall that:

A⁻¹ = | 5/3 -2/3 |
      | 2   -1   |

and

b = | -3 |
    | -9 |

So, we perform the matrix multiplication:

x = A⁻¹b = | 5/3 -2/3 | | -3 |
            | 2   -1   | | -9 |

To multiply these matrices, we perform the following calculations:

x₁ = (5/3 * -3) + (-2/3 * -9) = -5 + 6 = 1
y₁ = (2 * -3) + (-1 * -9) = -6 + 9 = 3

Thus, the solution matrix x is:

x = | 1 |
    | 3 |

This means that x = 1 and y = 3 is the solution to the system of equations. We can verify this solution by substituting these values back into the original equations:

3(1) - 2(3) = 3 - 6 = -3
6(1) - 5(3) = 6 - 15 = -9

The solution satisfies both equations, confirming its correctness. Therefore, the matrix equation that can be used to solve the system is the one that correctly represents x = A⁻¹b, where A⁻¹ is the inverse of the coefficient matrix and b is the constant matrix. The correct setup and calculation of the inverse are crucial in this process. This method not only provides the solution but also confirms its uniqueness, provided the determinant of the coefficient matrix is non-zero.

Conclusion

In conclusion, solving systems of equations using matrix equations is a powerful technique that relies on representing the system in the form Ax = b and then solving for x using the inverse matrix A⁻¹. This method provides a structured and efficient way to find solutions, especially for larger systems of equations. By understanding the principles of matrix representation and inversion, one can effectively solve a wide range of linear systems. In the given example, we successfully found the solution x = 1 and y = 3 by correctly identifying the matrices A and b, computing A⁻¹, and then applying the formula x = A⁻¹b. This process highlights the elegance and utility of matrix algebra in solving mathematical problems.