Lesson 4.4: Solving Systems of Linear Equations

Introduction

Welcome to Lesson 4.4, students! Today, we're diving into a key concept in linear algebra: solving systems of linear equations. 🌟 Our objectives for this lesson are to:

• Write a linear system in matrix form $ Ax = b $.
• Solve systems using both the inverse matrix method and Gaussian elimination.
• Understand consistent, inconsistent, and dependent systems, as well as their geometric interpretations.
• Express and solve a $ 3 \times 3 $ linear system using the inverse matrix.
• Apply row reduction to solve linear systems.

This lesson will help you grasp how matrices and systems of equations work, and how you can apply them in real-world scenarios such as engineering and computer graphics.

Understanding Matrix Form

To solve a system of equations using matrices, we can represent the system in a matrix form. Let's consider a simple example:

Example 1

Suppose we have the following system of equations:

1. $ 2x + 3y = 6 $
2. $ 4x + y = 5 $

We can rewrite this in matrix form as follows:

$$egin{bmatrix} 2 & 3 \ 4 & 1 \end{bmatrix}egin{bmatrix} x \ y \end{bmatrix} = egin{bmatrix} 6 \ 5 \end{bmatrix}$$

Here, the matrix on the left represents the coefficients of $ x $ and $ y $, the vector in the middle contains the variables, and the vector on the right represents the constants from the equations.

Solving by the Inverse Matrix

One powerful method to solve the system is by using the inverse of a matrix. If we have a matrix $ A $ and its inverse $ A^{-1} $, the solution can be found using the formula:

$$ x = A^{-1}b $$

Example 2

Let’s go back to our earlier example. The matrix $ A $ is:

$$ A = egin{bmatrix} 2 & 3 \ 4 & 1 \end{bmatrix} $$

To find the inverse of $ A $, we use the formula for the inverse of a $ 2 \times 2 $ matrix:

$$ A^{-1} = \frac{1}{ad - bc} egin{bmatrix} d & -b \ -c & a \end{bmatrix} $$

For our matrix:

• $ a = 2, b = 3, c = 4, d = 1 $
• The determinant $ ad - bc = 2(1) - 3(4) = 2 - 12 = -10 $

Thus:

$$ A^{-1} = \frac{1}{-10} egin{bmatrix} 1 & -3 \ -4 & 2 \end{bmatrix} = egin{bmatrix} -0.1 & 0.3 \ 0.4 & -0.2 \end{bmatrix} $$

Now we can find $ x $:

$$ x = A^{-1}egin{bmatrix} 6 \ 5 \end{bmatrix} = egin{bmatrix} -0.1 & 0.3 \ 0.4 & -0.2 \end{bmatrix}egin{bmatrix} 6 \ 5 \end{bmatrix} $$

Calculating this gives:

$$ x = egin{bmatrix} -0.1(6) + 0.3(5) \ 0.4(6) - 0.2(5) \end{bmatrix} = egin{bmatrix} -0.6 + 1.5 \ 2.4 - 1.0 \end{bmatrix} = egin{bmatrix} 0.9 \ 1.4 \end{bmatrix} $$

So, $ x = 0.9 $ and $ y = 1.4 $. 🎉 This method can be handy, but it's only applicable when the matrix is invertible (i.e., the determinant is not zero).

Solving by Gaussian Elimination

An alternative method for solving a system of linear equations is Gaussian elimination, which transforms the matrix into row echelon form.

Example 3

Let’s use our previous equations:

1. $ 2x + 3y = 6 $
2. $ 4x + y = 5 $

We form the augmented matrix:

$$ \left[egin{array}{cc|c} 2 & 3 & 6 \ 4 & 1 & 5 \end{array}

ight] $$

Now we apply row operations to arrive at row echelon form. Start by making the first column below the leading 2 into zeros. We can replace $ R_2 $ with $ R_2 - 2R_1 $:

$$ \left[egin{array}{cc|c} 2 & 3 & 6 \ 0 & -5 & -7 \end{array}

ight] $$

Next, we isolate $ y $ from the second row:

$$ -5y = -7 \Rightarrow y = \frac{7}{5} = 1.4 $$

Now substituting $ y = 1.4 $ back into the first equation to find $ x $:

$$ 2x + 3(1.4) = 6 \Rightarrow 2x + 4.2 = 6 \Rightarrow 2x = 6 - 4.2 = 1.8 $$

Thus, $ x = \frac{1.8}{2} = 0.9 $.

So we again have $ x = 0.9 $ and $ y = 1.4 $. 😊

Consistency and Dependency of Systems

When dealing with systems of equations, it's crucial to identify their types:

• Consistent Systems: Have at least one solution. Graphically, the lines intersect at a point or coincide completely.
• Inconsistent Systems: Have no solutions. Graphically, the lines are parallel and never meet.
• Dependent Systems: Have infinitely many solutions. Graphically, the lines overlap completely.

Conclusion

In this lesson, we explored methods for solving systems of linear equations using matrices. We learned how to express a system of equations in matrix form, utilize the inverse matrix and Gaussian elimination, and identify the type of system we are dealing with. Understanding these concepts lays the groundwork for more complex topics in linear algebra and their applications in various fields.

Study Notes

• Represent a system of equations in matrix form: $ Ax = b $.
• Use the inverse matrix to solve: $ x = A^{-1}b $.
• Apply Gaussian elimination to reduce the matrix to row echelon form.
• Distinguish between consistent, inconsistent, and dependent systems with geometric interpretations.
• Explore practical applications of matrices in real-world scenarios, such as graphics and engineering.