Matrix Multiplication Find Products AB And BA For Diagonal Matrices

Find the product of matrices AB and BA for the diagonal matrices A and B.

In linear algebra, matrix multiplication is a fundamental operation. When dealing with diagonal matrices, the process simplifies significantly due to their unique structure. This article will walk you through finding the products AB and BA for the given diagonal matrices:

$A=\begin{bmatrix} 4 & 0 \ 0 & -2 \end{bmatrix}, \quad B=\begin{bmatrix} -3 & 0 \ 0 & 5 \end{bmatrix}$

Understanding Diagonal Matrices

Before diving into the calculations, let's briefly define what diagonal matrices are. A diagonal matrix is a square matrix in which all the elements outside the main diagonal (from the top-left corner to the bottom-right corner) are zero. The matrices A and B provided above are examples of 2x2 diagonal matrices. Diagonal matrices have several useful properties, one of which is that their multiplication is relatively straightforward.

Properties of Diagonal Matrices

Diagonal matrices possess unique characteristics that simplify many matrix operations. These properties not only make calculations easier but also provide valuable insights into the behavior of linear transformations. One key property is that the product of two diagonal matrices is also a diagonal matrix. This significantly reduces the computational complexity when multiplying such matrices, as we only need to focus on the diagonal elements. Another important property is that the determinant of a diagonal matrix is simply the product of its diagonal entries. This makes it very easy to check if a diagonal matrix is invertible, as a diagonal matrix is invertible if and only if all its diagonal elements are non-zero. Furthermore, the eigenvalues of a diagonal matrix are precisely its diagonal entries, which simplifies eigenvalue-related calculations considerably. These properties make diagonal matrices a fundamental concept in various applications, including solving linear systems, analyzing dynamical systems, and more. Understanding these properties can greatly aid in simplifying complex mathematical problems and enhancing problem-solving skills in linear algebra.

Calculating AB

To find the product AB, we multiply matrix A by matrix B. The process involves taking the dot product of each row of A with each column of B. For 2x2 matrices, this is a manageable task.

$AB = \begin{bmatrix} 4 & 0 \ 0 & -2 \end{bmatrix} \begin{bmatrix} -3 & 0 \ 0 & 5 \end{bmatrix}$

To compute the elements of the resulting matrix, we perform the following calculations:

• Element (1,1): (4 * -3) + (0 * 0) = -12
• Element (1,2): (4 * 0) + (0 * 5) = 0
• Element (2,1): (0 * -3) + (-2 * 0) = 0
• Element (2,2): (0 * 0) + (-2 * 5) = -10

Thus, the product AB is:

$AB = \begin{bmatrix} -12 & 0 \ 0 & -10 \end{bmatrix}$

Step-by-Step Calculation of AB

The multiplication of matrices A and B involves a methodical process of taking the dot product of rows from the first matrix with columns from the second matrix. Understanding this process is crucial for handling more complex matrix multiplications. Let's break down the calculation of AB step by step. First, consider the first row of A [4, 0] and the first column of B [-3, 0]. The dot product is calculated as (4 * -3) + (0 * 0) = -12, which becomes the element in the first row and first column of AB. Next, we take the first row of A [4, 0] and the second column of B [0, 5]. The dot product here is (4 * 0) + (0 * 5) = 0, which fills the first row and second column of AB. Moving on, we consider the second row of A [0, -2] and the first column of B [-3, 0]. The dot product is (0 * -3) + (-2 * 0) = 0, giving us the element for the second row and first column of AB. Finally, we compute the dot product of the second row of A [0, -2] and the second column of B [0, 5], which is (0 * 0) + (-2 * 5) = -10. This value goes into the second row and second column of AB. By following these steps carefully, we arrive at the product AB, which is a new diagonal matrix. This detailed walkthrough not only clarifies the multiplication process but also highlights the importance of methodical calculation in linear algebra.

Calculating BA

Now, let's find the product BA by multiplying matrix B by matrix A. This is another straightforward calculation due to the matrices being diagonal.

$BA = \begin{bmatrix} -3 & 0 \ 0 & 5 \end{bmatrix} \begin{bmatrix} 4 & 0 \ 0 & -2 \end{bmatrix}$

We compute the elements of the resulting matrix as follows:

• Element (1,1): (-3 * 4) + (0 * 0) = -12
• Element (1,2): (-3 * 0) + (0 * -2) = 0
• Element (2,1): (0 * 4) + (5 * 0) = 0
• Element (2,2): (0 * 0) + (5 * -2) = -10

Thus, the product BA is:

$BA = \begin{bmatrix} -12 & 0 \ 0 & -10 \end{bmatrix}$

Step-by-Step Calculation of BA

Calculating BA follows the same principles as calculating AB, but it's essential to maintain the correct order of multiplication. Here, we are multiplying matrix B by matrix A, so we take the rows of B and the columns of A to compute the dot products. First, we consider the first row of B [-3, 0] and the first column of A [4, 0]. The dot product is (-3 * 4) + (0 * 0) = -12, which is the element in the first row and first column of BA. Next, we take the first row of B [-3, 0] and the second column of A [0, -2]. The dot product is (-3 * 0) + (0 * -2) = 0, which fills the first row and second column of BA. Moving on, we consider the second row of B [0, 5] and the first column of A [4, 0]. The dot product is (0 * 4) + (5 * 0) = 0, giving us the element for the second row and first column of BA. Finally, we compute the dot product of the second row of B [0, 5] and the second column of A [0, -2], which is (0 * 0) + (5 * -2) = -10. This value goes into the second row and second column of BA. Just as with AB, the meticulous execution of each step ensures the accuracy of the final result. This process highlights that while matrix multiplication is not commutative in general, in this specific case with diagonal matrices, the result happens to be the same.

Observation

In this particular case, AB and BA are equal. This is not always the case for matrix multiplication, which is generally not commutative. However, for diagonal matrices, the order of multiplication does not affect the result.

$AB = BA = \begin{bmatrix} -12 & 0 \ 0 & -10 \end{bmatrix}$

Why AB = BA for These Diagonal Matrices?

The equality of AB and BA in this scenario is a specific property arising from the nature of diagonal matrices. While matrix multiplication is generally not commutative, diagonal matrices form an exception under certain conditions. The key reason for this commutativity lies in how diagonal matrices interact during multiplication. When you multiply two diagonal matrices, the resulting matrix's elements are simply the product of the corresponding diagonal elements from the original matrices. In other words, the element in the i-th row and i-th column of the product is the product of the i-th diagonal elements of the two matrices being multiplied. This can be expressed mathematically as follows: if A and B are diagonal matrices, then (AB)ᵢᵢ = Aᵢᵢ * Bᵢᵢ and (BA)ᵢᵢ = Bᵢᵢ * Aᵢᵢ. Since scalar multiplication is commutative (Aᵢᵢ * Bᵢᵢ = Bᵢᵢ * Aᵢᵢ), it follows that (AB)ᵢᵢ = (BA)ᵢᵢ. Furthermore, all off-diagonal elements remain zero because of the zeros in the original matrices. Thus, when we calculate AB and BA for diagonal matrices, the results are identical because the multiplication of the diagonal elements is commutative. This observation is not just a mathematical curiosity but has practical implications, such as simplifying calculations in linear transformations and eigenvalue problems where diagonal matrices frequently appear. Understanding this commutative property can save computational effort and provide deeper insights into the structure and behavior of linear systems.

Conclusion

Finding the products AB and BA for diagonal matrices is a straightforward process. The key takeaway is that while matrix multiplication is generally not commutative, diagonal matrices provide an exception in some cases. This understanding is crucial for various applications in linear algebra and beyond.

Importance of Understanding Matrix Multiplication with Diagonal Matrices

Understanding matrix multiplication, particularly with diagonal matrices, is crucial for several reasons in mathematics, engineering, and computer science. Diagonal matrices appear frequently in various applications, including solving systems of linear equations, eigenvalue problems, and principal component analysis. The simplicity of multiplying diagonal matrices—only the diagonal elements are involved—makes them computationally efficient in large-scale calculations. For instance, in computer graphics, scaling transformations can be represented using diagonal matrices, and the efficiency of multiplying these matrices directly impacts the performance of graphics rendering. In control systems, diagonal matrices can represent decoupled systems, simplifying analysis and design. Moreover, the eigenvectors of a diagonalizable matrix form a basis that transforms the matrix into a diagonal matrix, a process used extensively in solving differential equations and analyzing stability in dynamical systems. Additionally, the eigenvalues, which are the diagonal elements of a diagonal matrix, provide critical information about the matrix's behavior, such as its invertibility and the stability of linear transformations it represents. In machine learning, diagonal matrices are used in covariance matrices for feature scaling and in regularization techniques. Therefore, mastering the properties and operations of diagonal matrices enhances problem-solving capabilities across various fields and lays a solid foundation for advanced topics in linear algebra and its applications. This thorough understanding allows for more efficient and insightful handling of complex mathematical and computational problems.