Problem 27 - Algebra for College Students

Question

For Problems \(27-30\), use the following matrices. \(\begin{aligned} &A=\left[\begin{array}{rr} -2 & 3 \\ 5 & 4 \end{array}\right] \quad B=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\\\ &C=\left[\begin{array}{ll} 1 & 0 \\ 1 & 0 \end{array}\right] \quad D=\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]\\\ &I=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \end{aligned}\) $$ \text { Compute } A B \text { and } B A \text {. } $$

Step-by-Step Solution

Answer

AB = \(\begin{bmatrix} 3 & -2 \\ 4 & 5 \end{bmatrix}\), BA = \(\begin{bmatrix} 5 & 4 \\ -2 & 3 \end{bmatrix}\).

Step 1: Determine the size of matrices A and B

Matrix \( A \) is a \( 2 \times 2 \) matrix, and matrix \( B \) is also a \( 2 \times 2 \) matrix. Therefore, the multiplication \( AB \) and \( BA \) is defined, and both results will be \( 2 \times 2 \) matrices.

Step 2: Calculate AB

To find the product \( AB \), take each row of \( A \) and multiply it with each column of \( B \).- For the first element of \( AB \), multiply the first row of \( A \) with the first column of \( B \): \((-2)\cdot 0 + 3\cdot 1 = 3\).- For the second element, multiply the first row of \( A \) with the second column of \( B \): \((-2)\cdot 1 + 3\cdot 0 = -2\).- For the third element, multiply the second row of \( A \) with the first column of \( B \): \(5\cdot 0 + 4\cdot 1 = 4\).- For the fourth element, multiply the second row of \( A \) with the second column of \( B \): \(5\cdot 1 + 4\cdot 0 = 5\).Thus, \( AB = \begin{bmatrix} 3 & -2 \ 4 & 5 \end{bmatrix} \).

Step 3: Calculate BA

To find the product \( BA \), take each row of \( B \) and multiply it with each column of \( A \).- For the first element of \( BA \), multiply the first row of \( B \) with the first column of \( A \): \(0\cdot(-2) + 1\cdot 5 = 5\).- For the second element, multiply the first row of \( B \) with the second column of \( A \): \(0\cdot 3 + 1\cdot 4 = 4\).- For the third element, multiply the second row of \( B \) with the first column of \( A \): \(1\cdot(-2) + 0\cdot 5 = -2\).- For the fourth element, multiply the second row of \( B \) with the second column of \( A \): \(1\cdot 3 + 0\cdot 4 = 3\).Thus, \( BA = \begin{bmatrix} 5 & 4 \ -2 & 3 \end{bmatrix} \).

Step 4: Compare AB and BA

Note that \( AB \) and \( BA \) are not equal. \( AB = \begin{bmatrix} 3 & -2 \ 4 & 5 \end{bmatrix} \) while \( BA = \begin{bmatrix} 5 & 4 \ -2 & 3 \end{bmatrix} \). In general, matrix multiplication is not commutative, which means \( AB eq BA \).

Key Concepts

2x2 Matrices

In the world of matrix operations, a 2x2 matrix is one of the simplest forms, consisting of four elements. These matrices are arranged in two rows and two columns. A matrix like this can be expressed in the form:

• A = \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \)

This means there are four specific positions: the top left, top right, bottom left, and bottom right. Understanding the layout of a 2x2 matrix is crucial since you'll be dealing a lot with these structures in linear algebra.
In our exercise, matrix A with elements \([-2, 3, 5, 4]\) and matrix B with elements \([0, 1, 1, 0]\) are both examples of 2x2 matrices.
When you multiply a 2x2 matrix by another 2x2 matrix, the resulting product will also be a 2x2 matrix. This is a standard property of matrices that makes them predictable and thus easier to handle in mathematical computations.

Non-Commutative Property

One of the critical differences between matrix multiplication and regular multiplication we perform in arithmetic is that matrix multiplication is non-commutative. This means that the order in which you multiply two matrices matters.
In other words, given two matrices A and B, the product AB is not necessarily equal to BA. Our earlier exercise serves as a practical example of this concept.
We calculated AB and obtained \( \begin{bmatrix} 3 & -2 \ 4 & 5 \end{bmatrix} \), while for BA we found \( \begin{bmatrix} 5 & 4 \ -2 & 3 \end{bmatrix} \).
Clearly, the resulting matrices from these operations are not the same, highlighting the non-commutative nature of matrix multiplication.

• Takeaway: With matrices, swapping the order of multiplication usually changes the result, unlike with numbers.

The non-commutative property is foundational in understanding matrix operations and has significant implications in fields like physics and computer science.

Matrix Operations

Matrix operations include addition, subtraction, and multiplication, each with its own rules and properties.
The focus here is matrix multiplication, where we multiply rows by columns. Let me simplify it for you:

• Take a row from the first matrix.
• Take a column from the second matrix.
• Multiply corresponding elements and sum them up for each element in the new matrix.

In our exercise, this methodology was used to compute AB and BA. For example, to find the first element of AB, we took the first row of A and multiplied it by the first column of B, yielding \((-2) \cdot 0 + 3 \cdot 1 = 3\).
It's important to note:

• Matrix multiplication is only possible when the number of columns in the first matrix matches the number of rows in the second.
• The size of the resulting matrix will be determined by the number of rows in the first matrix and the number of columns in the second.

Through understanding these simple matrix operations, you can solve problems involving more complex matrices, leading to applications in varied disciplines like engineering, computer graphics, and more.