Solving Linear Equations Using Inverse Matrices: A Step-by-Step Guide

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Hey there, math enthusiasts! Ever feel like you're staring at a system of linear equations and just wishing there was an easier way to solve them? Well, guess what? There is! And it involves the magic of inverse matrices. Today, we're diving deep into the world of inverse matrices and how they can be used to conquer those pesky equations. Specifically, we'll walk through examples from Exercises 45 and 46, breaking down each step to make sure you understand the concept inside and out. Get ready to flex those math muscles and unlock a powerful tool for solving linear systems!

Understanding Inverse Matrices: The Basics

Before we jump into the examples, let's get our bearings. What exactly is an inverse matrix, and why should we care? Think of an inverse matrix as the opposite of a regular matrix. If you multiply a matrix by its inverse, you get the identity matrix (a matrix with 1s on the diagonal and 0s everywhere else). This property is super useful because it allows us to isolate variables and solve for them. Basically, the inverse matrix helps us undo what the original matrix did.

So, how do we find an inverse matrix? Well, for a 2x2 matrix like the ones in our examples, there's a straightforward formula. For a matrix A=[abcd]{ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} }, the inverse Aβˆ’1{ A^{-1} } is given by: Aβˆ’1=1adβˆ’bc[dβˆ’bβˆ’ca]{ A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} }. The term adβˆ’bc{ ad - bc } is the determinant of the matrix, and if the determinant is zero, the matrix doesn't have an inverse (it's non-invertible). Keep this in mind, as it's a crucial part of the process! Understanding the concept of inverse matrices will not only help you solve linear equations but will also provide a strong foundation for more advanced topics in linear algebra. It's like building the foundation of a house; without it, you can't go anywhere. Therefore, make sure you understand the basics before you move forward. Now, let's roll up our sleeves and solve some equations!

Example 45: Solving Systems of Equations

Part (a): {x+2y=βˆ’1xβˆ’2y=3{ \begin{cases} x + 2y = -1 \\ x - 2y = 3 \end{cases} }

Alright, let's tackle the first example! We have the system of equations: {x+2y=βˆ’1xβˆ’2y=3{ \begin{cases} x + 2y = -1 \\ x - 2y = 3 \end{cases} }. Our goal is to use an inverse matrix to find the values of x{ x } and y{ y } that satisfy both equations. First, let's represent this system in matrix form. We can write it as AX=B{ AX = B }, where:

  • A=[121βˆ’2]{ A = \begin{bmatrix} 1 & 2 \\ 1 & -2 \end{bmatrix} } (the coefficient matrix)
  • X=[xy]{ X = \begin{bmatrix} x \\ y \end{bmatrix} } (the variable matrix)
  • B=[βˆ’13]{ B = \begin{bmatrix} -1 \\ 3 \end{bmatrix} } (the constant matrix)

Now, the plan is to find the inverse of matrix A Aβˆ’1{ A^{-1} }, and then multiply both sides of the equation AX=B{ AX = B } by Aβˆ’1{ A^{-1} }. This gives us X=Aβˆ’1B{ X = A^{-1}B }, which will directly give us the solution for x{ x } and y{ y }. So, let's find Aβˆ’1{ A^{-1} }. The determinant of A is (1)(βˆ’2)βˆ’(2)(1)=βˆ’2βˆ’2=βˆ’4{ (1)(-2) - (2)(1) = -2 - 2 = -4 }. Since the determinant is not zero, the inverse exists. Using the formula we mentioned earlier:

Aβˆ’1=1βˆ’4[βˆ’2βˆ’2βˆ’11]=[0.50.50.25βˆ’0.25]{ A^{-1} = \frac{1}{-4} \begin{bmatrix} -2 & -2 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 0.5 & 0.5 \\ 0.25 & -0.25 \end{bmatrix} }

Now, multiply Aβˆ’1{ A^{-1} } by B{ B }: X=[0.50.50.25βˆ’0.25][βˆ’13]=[(βˆ’0.5+1.5)(βˆ’0.25βˆ’0.75)]=[1βˆ’1]{ X = \begin{bmatrix} 0.5 & 0.5 \\ 0.25 & -0.25 \end{bmatrix} \begin{bmatrix} -1 \\ 3 \end{bmatrix} = \begin{bmatrix} (-0.5 + 1.5) \\ (-0.25 - 0.75) \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \end{bmatrix} }. Therefore, x=1{ x = 1 } and y=βˆ’1{ y = -1 }. Voila! We've solved the system using an inverse matrix. Always remember to double-check your calculations, especially with the determinant and the matrix multiplication, to avoid silly mistakes. Doing so will help you gain more confidence in your math skills and make you more successful in the long run. Practicing these types of problems will help you understand the concepts better and make it easier to solve more complex problems in the future.

Part (b): {x+2y=10xβˆ’2y=βˆ’6{ \begin{cases} x + 2y = 10 \\ x - 2y = -6 \end{cases} }

Let's keep the ball rolling with part (b). Here, our system of equations is {x+2y=10xβˆ’2y=βˆ’6{ \begin{cases} x + 2y = 10 \\ x - 2y = -6 \end{cases} }. Notice that the coefficient matrix A{ A } is the same as in part (a): A=[121βˆ’2]{ A = \begin{bmatrix} 1 & 2 \\ 1 & -2 \end{bmatrix} }. So, we can reuse the inverse matrix we calculated earlier: Aβˆ’1=[0.50.50.25βˆ’0.25]{ A^{-1} = \begin{bmatrix} 0.5 & 0.5 \\ 0.25 & -0.25 \end{bmatrix} }. The only thing that changes is the constant matrix B{ B }, which is now B=[10βˆ’6]{ B = \begin{bmatrix} 10 \\ -6 \end{bmatrix} }. Now, multiply Aβˆ’1{ A^{-1} } by the new B{ B }: X=[0.50.50.25βˆ’0.25][10βˆ’6]=[(5βˆ’3)(2.5+1.5)]=[24]{ X = \begin{bmatrix} 0.5 & 0.5 \\ 0.25 & -0.25 \end{bmatrix} \begin{bmatrix} 10 \\ -6 \end{bmatrix} = \begin{bmatrix} (5 - 3) \\ (2.5 + 1.5) \end{bmatrix} = \begin{bmatrix} 2 \\ 4 \end{bmatrix} }. Thus, x=2{ x = 2 } and y=4{ y = 4 }. Pretty neat, right? The beauty of using the inverse matrix is that once you have it, you can solve the system for different constant matrices with ease. Remember that the correct setup of the matrices is crucial; otherwise, the answer will not be correct. This is why you need to understand the concept very well before you start solving equations. Practice, practice, and practice are the key to mastering these types of problems.

Example 46: More Inverse Matrix Magic

Part (a): {2xβˆ’y=βˆ’32x+y=7{ \begin{cases} 2x - y = -3 \\ 2x + y = 7 \end{cases} }

Let's get into the last part of our journey. We have the following system of equations: {2xβˆ’y=βˆ’32x+y=7{ \begin{cases} 2x - y = -3 \\ 2x + y = 7 \end{cases} }. First, let's construct our matrices:

  • A=[2βˆ’121]{ A = \begin{bmatrix} 2 & -1 \\ 2 & 1 \end{bmatrix} }
  • X=[xy]{ X = \begin{bmatrix} x \\ y \end{bmatrix} }
  • B=[βˆ’37]{ B = \begin{bmatrix} -3 \\ 7 \end{bmatrix} }

Now, let's find Aβˆ’1{ A^{-1} }. The determinant of A is (2)(1)βˆ’(βˆ’1)(2)=2+2=4{ (2)(1) - (-1)(2) = 2 + 2 = 4 }. The inverse matrix is: Aβˆ’1=14[11βˆ’22]=[0.250.25βˆ’0.50.5]{ A^{-1} = \frac{1}{4} \begin{bmatrix} 1 & 1 \\ -2 & 2 \end{bmatrix} = \begin{bmatrix} 0.25 & 0.25 \\ -0.5 & 0.5 \end{bmatrix} }. Multiply Aβˆ’1{ A^{-1} } by B{ B }: X=[0.250.25βˆ’0.50.5][βˆ’37]=[(βˆ’0.75+1.75)(1.5+3.5)]=[15]{ X = \begin{bmatrix} 0.25 & 0.25 \\ -0.5 & 0.5 \end{bmatrix} \begin{bmatrix} -3 \\ 7 \end{bmatrix} = \begin{bmatrix} (-0.75 + 1.75) \\ (1.5 + 3.5) \end{bmatrix} = \begin{bmatrix} 1 \\ 5 \end{bmatrix} }. Therefore, x=1{ x = 1 } and y=5{ y = 5 }. And that's a wrap! You've successfully used inverse matrices to solve another system of linear equations. It is worth pointing out that understanding the relationship between the original matrix and its inverse is key. The inverse is designed to