Matrix Operations: Finding P⁻¹ And Computing P⁻¹AP

by Alex Johnson 51 views

In this article, we will explore fundamental matrix operations, including finding the inverse of a matrix and computing the product of matrices. Specifically, we will focus on a given matrix AA and PP, and our goal is to determine P−1P^{-1} and then calculate P−1APP^{-1}AP. Matrix operations are crucial in various fields such as linear algebra, computer graphics, and data analysis. Understanding how to perform these operations is essential for solving complex problems in these areas. Let's dive in and explore the steps involved in finding the inverse of a matrix and computing matrix products.

(a) Finding P−1P^{-1}

To find the inverse of matrix PP, denoted as P−1P^{-1}, we will use the formula for the inverse of a 2x2 matrix. Finding the inverse of a matrix is a fundamental operation in linear algebra, with applications ranging from solving systems of linear equations to performing transformations in computer graphics. For a 2x2 matrix P=[abcd]P = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, the inverse P−1P^{-1} is given by:

P−1=1ad−bc[d−b−ca]P^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}

Where (ad−bc)(ad - bc) is the determinant of PP. If the determinant is zero, the matrix does not have an inverse. In our case, the matrix PP is given by:

P=[1221]P = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}

Here, a=1a = 1, b=2b = 2, c=2c = 2, and d=1d = 1. Let's calculate the determinant of PP:

det(P)=ad−bc=(1)(1)−(2)(2)=1−4=−3\text{det}(P) = ad - bc = (1)(1) - (2)(2) = 1 - 4 = -3

Since the determinant is non-zero (-3), the inverse of PP exists. Now, we can find P−1P^{-1} using the formula:

P−1=1−3[1−2−21]=[−132323−13]P^{-1} = \frac{1}{-3} \begin{bmatrix} 1 & -2 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} -\frac{1}{3} & \frac{2}{3} \\ \frac{2}{3} & -\frac{1}{3} \end{bmatrix}

Thus, the inverse of matrix PP is:

P−1=[−132323−13]P^{-1} = \begin{bmatrix} -\frac{1}{3} & \frac{2}{3} \\ \frac{2}{3} & -\frac{1}{3} \end{bmatrix}

This result is crucial for the next part of our task, where we need to compute P−1APP^{-1}AP. The inverse matrix allows us to "undo" the transformation represented by the original matrix, and it plays a key role in solving linear systems and eigenvalue problems. Understanding how to calculate the inverse is therefore a cornerstone of linear algebra.

(b) Find P−1APP^{-1} A P

Now, let's compute the matrix product P−1APP^{-1}AP. This operation is significant in linear algebra as it represents a similarity transformation of matrix AA. Similarity transformations are essential for diagonalizing matrices and simplifying eigenvalue problems. The matrices AA and PP are given as:

A=[02−25]A = \begin{bmatrix} 0 & 2 \\ -2 & 5 \end{bmatrix}

P=[1221]P = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}

We have already found P−1P^{-1} in the previous part:

P−1=[−132323−13]P^{-1} = \begin{bmatrix} -\frac{1}{3} & \frac{2}{3} \\ \frac{2}{3} & -\frac{1}{3} \end{bmatrix}

To compute P−1APP^{-1}AP, we first multiply P−1P^{-1} by AA and then multiply the result by PP. Let's start by calculating P−1AP^{-1}A:

P−1A=[−132323−13][02−25]P^{-1}A = \begin{bmatrix} -\frac{1}{3} & \frac{2}{3} \\ \frac{2}{3} & -\frac{1}{3} \end{bmatrix} \begin{bmatrix} 0 & 2 \\ -2 & 5 \end{bmatrix}

To perform matrix multiplication, we multiply the rows of the first matrix by the columns of the second matrix:

P−1A=[(−13)(0)+(23)(−2)(−13)(2)+(23)(5)(23)(0)+(−13)(−2)(23)(2)+(−13)(5)]P^{-1}A = \begin{bmatrix} (-\frac{1}{3})(0) + (\frac{2}{3})(-2) & (-\frac{1}{3})(2) + (\frac{2}{3})(5) \\ (\frac{2}{3})(0) + (-\frac{1}{3})(-2) & (\frac{2}{3})(2) + (-\frac{1}{3})(5) \end{bmatrix}

P−1A=[0−43−23+1030+2343−53]=[−438323−13]P^{-1}A = \begin{bmatrix} 0 - \frac{4}{3} & -\frac{2}{3} + \frac{10}{3} \\ 0 + \frac{2}{3} & \frac{4}{3} - \frac{5}{3} \end{bmatrix} = \begin{bmatrix} -\frac{4}{3} & \frac{8}{3} \\ \frac{2}{3} & -\frac{1}{3} \end{bmatrix}

Now, we multiply the resulting matrix P−1AP^{-1}A by PP:

(P−1A)P=[−438323−13][1221](P^{-1}A)P = \begin{bmatrix} -\frac{4}{3} & \frac{8}{3} \\ \frac{2}{3} & -\frac{1}{3} \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}

Again, we multiply the rows of P−1AP^{-1}A by the columns of PP:

P−1AP=[(−43)(1)+(83)(2)(−43)(2)+(83)(1)(23)(1)+(−13)(2)(23)(2)+(−13)(1)]P^{-1}AP = \begin{bmatrix} (-\frac{4}{3})(1) + (\frac{8}{3})(2) & (-\frac{4}{3})(2) + (\frac{8}{3})(1) \\ (\frac{2}{3})(1) + (-\frac{1}{3})(2) & (\frac{2}{3})(2) + (-\frac{1}{3})(1) \end{bmatrix}

P−1AP=[−43+163−83+8323−2343−13]=[1230033]P^{-1}AP = \begin{bmatrix} -\frac{4}{3} + \frac{16}{3} & -\frac{8}{3} + \frac{8}{3} \\ \frac{2}{3} - \frac{2}{3} & \frac{4}{3} - \frac{1}{3} \end{bmatrix} = \begin{bmatrix} \frac{12}{3} & 0 \\ 0 & \frac{3}{3} \end{bmatrix}

Simplifying the matrix, we get:

P−1AP=[4001]P^{-1}AP = \begin{bmatrix} 4 & 0 \\ 0 & 1 \end{bmatrix}

So, the final result of the matrix product P−1APP^{-1}AP is a diagonal matrix, which is a significant outcome. This diagonalization process is a key application of similarity transformations and is used extensively in eigenvalue analysis and solving differential equations. Understanding how to perform these calculations is vital for advanced topics in linear algebra and its applications.

Conclusion

In summary, we have successfully found the inverse of matrix PP and computed the matrix product P−1APP^{-1}AP. These operations are fundamental in linear algebra and have wide-ranging applications in various fields. Matrix operations like finding inverses and similarity transformations are essential tools for solving complex problems in mathematics, engineering, and computer science.

For further reading on matrix operations and linear algebra, you might find helpful resources on websites like Khan Academy's Linear Algebra section.