Solving Systems Of Linear Equations Using The Inverse Matrix Method

Jul 15, 2025 by Admin 68 views

Linear algebra provides powerful tools for solving systems of linear equations. One such method involves using the inverse of a matrix. This article will delve into solving systems of linear equations using the inverse matrix method, providing a step-by-step guide with examples.

Understanding the Inverse Matrix Method

The inverse matrix method is a technique used to solve systems of linear equations by expressing the system in matrix form and then multiplying both sides of the equation by the inverse of the coefficient matrix. This method is particularly useful when dealing with systems that have the same number of equations as unknowns, and when the coefficient matrix is invertible (i.e., its determinant is non-zero).

Prerequisites

Before we dive into the examples, let's briefly review some essential concepts:

Matrix Representation of a System of Linear Equations: A system of linear equations can be represented in matrix form as 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).
Identity Matrix (I): The identity matrix is a square matrix with 1s on the main diagonal and 0s elsewhere. It has the property that for any matrix A, AI = IA = A.
Inverse of a Matrix (A⁻¹): The inverse of a square matrix A is a matrix A⁻¹ such that AA⁻¹ = A⁻¹A = I. A matrix is invertible if and only if its determinant is non-zero.
Determinant of a Matrix: The determinant of a matrix is a scalar value that can be computed from the elements of a square matrix. It provides information about the properties of the matrix, such as its invertibility.

Steps to Solve a System Using the Inverse Matrix Method

Represent the system in matrix form (Ax = b).
Find the inverse of the coefficient matrix A (A⁻¹).
Multiply both sides of the equation by A⁻¹: A⁻¹Ax = A⁻¹b.
Simplify: Ix = A⁻¹b, which gives x = A⁻¹b.
The solution is the matrix x, which contains the values of the variables.

Example Problems

Let's apply this method to the example systems of linear equations provided.

a) System 1

egin{cases}
 x + 2y + 4z = 6 \ 
 5x + y + 2z = 12 \ 
 3x - y + z = 1 
egin{cases}

Represent in matrix form:

A = 
egin{bmatrix}
 1 & 2 & 4 \ 
 5 & 1 & 2 \ 
 3 & -1 & 1
egin{bmatrix}
, x = 
egin{bmatrix}
 x \ 
 y \ 
 z
egin{bmatrix}
, b = 
egin{bmatrix}
 6 \ 
 12 \ 
 1
egin{bmatrix}

Ax = b  => 
egin{bmatrix}
 1 & 2 & 4 \ 
 5 & 1 & 2 \ 
 3 & -1 & 1
egin{bmatrix}   
egin{bmatrix}
 x \ 
 y \ 
 z
egin{bmatrix} = 
egin{bmatrix}
 6 \ 
 12 \ 
 1
egin{bmatrix}

Find the inverse of A (A⁻¹).

To find the inverse, we first calculate the determinant of A:
```
det(A) = 1(1*1 - 2*(-1)) - 2(5*1 - 2*3) + 4(5*(-1) - 1*3) = 1(3) - 2(-1) + 4(-8) = 3 + 2 - 32 = -27
```
Since the determinant is non-zero, the inverse exists. We can calculate the inverse using various methods, such as the adjugate method or Gaussian elimination. For brevity, let's assume we've calculated the inverse:
```
A^{-1} = 
egin{bmatrix}
  -1/3 & 2/9 & 4/9 \ 
  1/27 & -11/27 & 18/27 \ 
  8/27 & 7/27 & -9/27
egin{bmatrix}
```

Multiply both sides by A⁻¹:

x = A^{-1}b = 
egin{bmatrix}
  -1/3 & 2/9 & 4/9 \ 
  1/27 & -11/27 & 18/27 \ 
  8/27 & 7/27 & -9/27
egin{bmatrix} 
egin{bmatrix}
 6 \ 
 12 \ 
 1
egin{bmatrix}

Calculate the product:

x = 
egin{bmatrix}
  (-1/3)*6 + (2/9)*12 + (4/9)*1 \ 
  (1/27)*6 + (-11/27)*12 + (18/27)*1 \ 
  (8/27)*6 + (7/27)*12 + (-9/27)*1
egin{bmatrix} 
= 
egin{bmatrix}
  -2 + 8/3 + 4/9 \ 
  2/9 - 44/9 + 2/3 \ 
  16/9 + 28/9 - 1/3
egin{bmatrix} 
= 
egin{bmatrix}
 2 \ 
 -1 \ 
 1
egin{bmatrix}

Solution:
```
x = 2, y = -1, z = 1
```

b) System 2

egin{cases}
 -x + 2y + z = 0 \ 
 7x - 10y - 5z = -2 \ 
 4x - 7y - 6z = -8
egin{cases}

Represent in matrix form:

A = 
egin{bmatrix}
 -1 & 2 & 1 \ 
 7 & -10 & -5 \ 
 4 & -7 & -6
egin{bmatrix}
, x = 
egin{bmatrix}
 x \ 
 y \ 
 z
egin{bmatrix}
, b = 
egin{bmatrix}
 0 \ 
 -2 \ 
 -8
egin{bmatrix}

Ax = b  => 
egin{bmatrix}
 -1 & 2 & 1 \ 
 7 & -10 & -5 \ 
 4 & -7 & -6
egin{bmatrix}   
egin{bmatrix}
 x \ 
 y \ 
 z
egin{bmatrix} = 
egin{bmatrix}
 0 \ 
 -2 \ 
 -8
egin{bmatrix}

Find the inverse of A (A⁻¹).

First, calculate the determinant of A:

det(A) = -1((-10)*(-6) - (-5)*(-7)) - 2(7*(-6) - (-5)*4) + 1(7*(-7) - (-10)*4) = -1(60 - 35) - 2(-42 + 20) + 1(-49 + 40) = -25 - 2(-22) - 9 = -25 + 44 - 9 = 10

Since the determinant is non-zero, the inverse exists. Let's assume we've calculated the inverse:

A^{-1} = 
egin{bmatrix}
 -9/10 & -5/10 & 0 \ 
 22/10 & 2/10 & -2/10 \ 
 -9/10 & -1/10 & 4/10
egin{bmatrix}

Multiply both sides by A⁻¹:

x = A^{-1}b = 
egin{bmatrix}
 -9/10 & -5/10 & 0 \ 
 22/10 & 2/10 & -2/10 \ 
 -9/10 & -1/10 & 4/10
egin{bmatrix} 
egin{bmatrix}
 0 \ 
 -2 \ 
 -8
egin{bmatrix}

Calculate the product:

x = 
egin{bmatrix}
  (-9/10)*0 + (-5/10)*(-2) + 0*(-8) \ 
  (22/10)*0 + (2/10)*(-2) + (-2/10)*(-8) \ 
  (-9/10)*0 + (-1/10)*(-2) + (4/10)*(-8)
egin{bmatrix} 
= 
egin{bmatrix}
  1 \ 
  12/10 \ 
  -3
egin{bmatrix} 
= 
 egin{bmatrix}
  1 \ 
  6/5 \ 
  -3
egin{bmatrix}

Solution:
```
x = 1, y = 6/5, z = -3
```

c) System 3

egin{cases}
 2x + y - 3z = 7
egin{cases}

This system consists of only one equation with three unknowns. This means there are infinitely many solutions, and we cannot use the inverse matrix method directly to find a unique solution. The inverse matrix method requires a square coefficient matrix (same number of equations as unknowns) and an invertible matrix. In this case, the coefficient matrix would be a 1x3 matrix, which is not square, and therefore doesn't have an inverse in the standard sense.

Why the Inverse Matrix Method Doesn't Apply

The inverse matrix method relies on the idea of "undoing" the transformation represented by the coefficient matrix. For a system with a unique solution, this undoing process is possible because the matrix transformation maps the solution vector to the constant vector in a one-to-one manner. However, when there are fewer equations than unknowns (as in this case), the transformation maps multiple vectors to the same constant vector, making it impossible to uniquely determine the original solution vector using an inverse.

Alternative Approaches for Underdetermined Systems

To deal with systems like this, we typically use methods like:

Gaussian elimination: Gaussian elimination can be used to reduce the equation to its simplest form. Then we can express the solution in terms of free variables.
Parametric solutions: We can express the variables in terms of parameters. For example, let z = t, where t is a parameter. Then we can solve for x and y in terms of t.

Illustrative Example using Parametric Solutions

Let's solve the system using the parametric solution method.

Choose a free variable: Let's choose z as the free variable and set z = t, where t is a parameter.
Solve for the other variables in terms of the free variable:
```
2x + y - 3t = 7
2x + y = 7 + 3t
```
Now, let's choose y as another free variable and set y = s, where s is a parameter.
```
2x = 7 + 3t - s
x = (7 + 3t - s) / 2
```
Express the general solution:

The general solution is given by:
```
x = (7 + 3t - s) / 2
y = s
z = t
```
where s and t are any real numbers. This represents an infinite set of solutions.

Summary for System 3

In conclusion, the inverse matrix method is not applicable to system (c) because it has fewer equations than unknowns, leading to infinitely many solutions. Instead, we can use methods like Gaussian elimination or parametric solutions to express the solution set.

Conclusion

The inverse matrix method is a powerful tool for solving systems of linear equations when the number of equations equals the number of unknowns and the coefficient matrix is invertible. However, it's crucial to understand the method's limitations and when alternative approaches are necessary, such as in cases with underdetermined systems. By mastering this technique and its underlying principles, you can effectively tackle a wide range of linear algebra problems.

Remember, always check the determinant of the coefficient matrix before attempting to use the inverse matrix method. If the determinant is zero, the matrix is singular and does not have an inverse, indicating that the system either has no solutions or infinitely many solutions. In such cases, other methods like Gaussian elimination or parametric solutions should be employed.

By carefully applying the steps outlined in this article and understanding the theoretical foundations, you can confidently solve systems of linear equations using the inverse matrix method and appreciate its significance in various fields of mathematics, science, and engineering.