Problem 25 - Algebra for College Students

Question

For Problems \(13-26\), compute \(A B\) and \(B A\). $$ A=\left[\begin{array}{ll} 5 & 6 \\ 2 & 3 \end{array}\right], \quad B=\left[\begin{array}{rr} 1 & -2 \\ -\frac{2}{3} & \frac{5}{3} \end{array}\right] $$

Step-by-Step Solution

Answer

Both \( A \times B \) and \( B \times A \) equal the identity matrix \( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \).

Step 1: Understand the problem

The task is to compute the product of two matrices, namely the products \( A \times B \) and \( B \times A \). Matrix multiplication involves taking the dot product of the rows of the first matrix with the columns of the second matrix.

Step 2: Compute A * B

For \( A \times B \):The matrices are:\[A = \begin{pmatrix} 5 & 6 \ 2 & 3 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & -2 \ -\frac{2}{3} & \frac{5}{3} \end{pmatrix}. \]Multiply the first row of \( A \) with each column of \( B \):- First element: \(5 \times 1 + 6 \times -\frac{2}{3} = 5 - 4 = 1\).- Second element: \(5 \times -2 + 6 \times \frac{5}{3} = -10 + 10 = 0\).Multiply the second row of \( A \) with each column of \( B \):- Third element: \(2 \times 1 + 3 \times -\frac{2}{3} = 2 - 2 = 0\).- Fourth element: \(2 \times -2 + 3 \times \frac{5}{3} = -4 + 5 = 1\).Thus, \( A \times B = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \).

Step 3: Compute B * A

For \( B \times A \):The matrices are:\[B = \begin{pmatrix} 1 & -2 \ -\frac{2}{3} & \frac{5}{3} \end{pmatrix}, \quad A = \begin{pmatrix} 5 & 6 \ 2 & 3 \end{pmatrix}. \]Multiply the first row of \( B \) with each column of \( A \):- First element: \(1 \times 5 + (-2) \times 2 = 5 - 4 = 1\).- Second element: \(1 \times 6 + (-2) \times 3 = 6 - 6 = 0\).Multiply the second row of \( B \) with each column of \( A \):- Third element: \(-\frac{2}{3} \times 5 + \frac{5}{3} \times 2 = -\frac{10}{3} + \frac{10}{3} = 0\).- Fourth element: \(-\frac{2}{3} \times 6 + \frac{5}{3} \times 3 = -4 + 5 = 1\).Thus, \( B \times A = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \).

Step 4: Conclusion

Both products give the identity matrix:\[A \times B = B \times A = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}.\]This shows that both products are the identity matrices in this problem.

Key Concepts

Dot Product

The dot product is a fundamental concept in matrix multiplication. It involves multiplying corresponding elements and summing them up to produce a single number. This process is essential when finding the product of two matrices.
For example, in the exercise, to compute entries for matrix product \(A \times B\), you take the dot product of each row of matrix A with each column of matrix B.

• Take the first row of A: \([5, 6] \).
• Multiply it by the columns in B: \([1, -\frac{2}{3}] \).

In each multiplication, you multiply corresponding components and add up the results. This gives us the first element of the resulting matrix.
The dot product is repeated for each row-column combination to complete the matrix product. Understanding this concept helps you effectively multiply matrices.

Identity Matrix

The identity matrix is a special type of matrix that acts like 1 in matrix multiplication. It is a square matrix with ones on the diagonal and zeros elsewhere.
In this exercise, the result of multiplying matrices \(A \times B\) and \(B \times A\) is the identity matrix:
\[\begin{pmatrix} 1 & 0 \0 & 1 \end{pmatrix}\]
This indicates that the original matrices are inverses of each other within the context of this operation.

• The identity matrix, when multiplied with any compatible matrix, leaves that matrix unchanged.
• Similarly, if any matrix multiplied by the identity matrix retains its original form, this property is key in solving equations and transformations in matrix algebra.

Recognizing the identity matrix can simplify solving matrix equations significantly.

Matrix Algebra

Matrix algebra is a system that allows us to work with matrices using operations similar to basic algebra. Matrices can be added, subtracted, and multiplied, analogous to arithmetic with numbers. However, matrix multiplication is unique and not commutative, meaning that \(A \times B \) is generally not the same as \(B \times A \).
Nonetheless, in the exercise, both products result in the identity matrix, which may indicate special properties of the given matrices.

• Matrix algebra encompasses various rules, such as the distributive, associative, and commutative properties for addition.
• It is also used to solve linear equations, find determinants, and invert matrices.

Understanding matrix algebra forms the foundation for more complex mathematical analysis and applications, ranging from computer graphics to systems engineering.

Matrix Operations

Matrix operations include a variety of transformations that can be performed on matrices. The most common operations are addition, subtraction, and multiplication. Each has specific rules and properties.
In matrix multiplication, as shown in the exercise, the product involves combining rows with columns.

• Matrix addition and subtraction require matrices to be of the same dimensions.
• Multiplication of matrices entails calculating the dot product of rows and columns, as demonstrated in \(A \times B\) and \(B \times A\).

Moreover, specific operations like transposition can switch rows with columns.
Understanding these operations allows us to manipulate and analyze matrices, applying them in various fields like physics, statistics, and computer science. Mastery of matrix operations is essential to unlocking more complex problem-solving abilities in both theoretical and applied mathematics.