Solving Matrix Equations: An In-Depth Guide Using the Given Example

Matrix equations are a fundamental part of linear algebra and are used extensively in various fields such as engineering, physics, and computer science. This tutorial will walk you through solving a specific matrix equation and exploring both the algebraic and analytical methods to find its solution.

Introduction to Matrix Equations

A matrix equation is an equation that involves matrix variables. In this context, we will solve the given matrix equation:

[3/5 1/2] x [-1/3 2/1]

Where the matrix A is defined as A [3/5 1/2] and the vector x [x1, x2]. The right-hand side matrix B is defined as B [-1/3 2/1].

This matrix equation can be expressed in the form:

[3/5 1/2] [x1] [-1/3 2/1]

[x2]

Solution Using Simple Algebra

First, let's use simple algebra to solve the equation. We start by expressing it as:

(3/5)x1 (1/2)x2 -1/3

This is a single linear equation with two variables. To find a solution, we solve for one variable in terms of the other. Let's solve for x2 by rearranging the equation:

(1/2)x2 -1/3 - (3/5)x1

Multiplying both sides by 2 to clear the fraction:

x2 -2/3 - (6/5)x1

From this, we can express x2 in terms of x1:

x2 -2/3 - (6/5)x1

This gives us the general solution as:

[x1, -2/3 - (6/5)x1]

The solution set is parametric, meaning that x1 can be any real number, resulting in infinitely many solutions along the line defined by this equation.

Solution Using Inverse Matrix

Another method to solve matrix equations is by using the concept of an inverse matrix. The given equation can be written as:

A x B

Where A [3/5, 1/2] and B [-1/3, 2/1].

To solve for x, we need to find the inverse of matrix A. A matrix is invertible if its determinant is non-zero. The formula for the inverse of a 2x2 matrix is:

A^{-1} 1 / (ad - bc) [d -b; -c a]

Where a, b, c, and d are the elements of matrix A. For our matrix:

a 3/5 b 1/2 c -1/3 d 2/1

The determinant of A is:

(3/5)(2/1) - (1/2)(-1/3) 6/5 1/6 36/30 5/30 41/30

Therefore, the inverse of A is:

A^{-1} 1 / (41/30) [2/1 -1/2 -3/5 3/5]

After simplification, we get:

A^{-1} [30/41 -30/(41*2) -30/(41*3/5) 30/(41*3/5)] [30/41 -15/41 -50/41 15/13]

Now, we can solve for x by multiplying both sides of the equation with A^{-1}:

x A^{-1}B

Substituting the values of A^{-1} and B we get:

x [30/41 -15/41 -50/41 15/13] [-1/3, 2/1]

Multiplying the matrices on the right-hand side gives:

x [-2/9, 11/9 -4/3, 2]

Conclusion

To summarize, we solved the given matrix equation [3/5 1/2]x [-1/3 2/1] using two methods: simple algebra and inverse matrix. The algebraic solution provided a parametric form of the solution, while the inverse matrix method provided a specific numerical solution.

The techniques discussed here are fundamental in linear algebra and can be applied to solve more complex matrix equations. Understanding these methods is crucial for anyone working in fields that require linear algebra. Whether you are dealing with systems of linear equations or more advanced topics such as eigenvalues and eigenvectors, mastering matrix operations and their applications is vital.