Problem 31 - College Algebra with Modeling and Visualization

Question

If possible, find \(A B\) and \(B A\). $$A=\left[\begin{array}{rr}1 & -1 \\\2 & 0\end{array}\right]$$' $$ B=\left[\begin{array}{rr}-2 & 3 \\\1 & 2\end{array}\right]$$

Step-by-Step Solution

Answer

Yes, AB and BA can be computed and are different.

Step 1: Matrix Multiplication Basics

To multiply two matrices, the number of columns in the first matrix must match the number of rows in the second matrix. Both matrices A and B are 2x2, so we can calculate both AB and BA.

Step 2: Calculate AB

To find the product \(AB\), multiply corresponding elements and sum them for each element in the resulting matrix. The elements of \(AB\) are:- First row, first column: \((1)(-2) + (-1)(1) = -2 - 1 = -3\)- First row, second column: \((1)(3) + (-1)(2) = 3 - 2 = 1\)- Second row, first column: \((2)(-2) + (0)(1) = -4 + 0 = -4\)- Second row, second column: \((2)(3) + (0)(2) = 6 + 0 = 6\)Thus, \(AB = \begin{bmatrix} -3 & 1 \ -4 & 6 \end{bmatrix}\).

Step 3: Calculate BA

To find the product \(BA\), again, multiply corresponding elements and sum them for each element in the resulting matrix. The elements of \(BA\) are:- First row, first column: \((-2)(1) + (3)(2) = -2 + 6 = 4\)- First row, second column: \((-2)(-1) + (3)(0) = 2 + 0 = 2\)- Second row, first column: \((1)(1) + (2)(2) = 1 + 4 = 5\)- Second row, second column: \((1)(-1) + (2)(0) = -1 + 0 = -1\)Thus, \(BA = \begin{bmatrix} 4 & 2 \ 5 & -1 \end{bmatrix}\).

Step 4: Verify Results

Both calculations are completed, and the resulting matrices are as follows:- \(AB = \begin{bmatrix} -3 & 1 \ -4 & 6 \end{bmatrix}\)- \(BA = \begin{bmatrix} 4 & 2 \ 5 & -1 \end{bmatrix}\).

Key Concepts

Matrix Algebra

Matrix Algebra is a branch of mathematics focusing on operations involving matrices. Matrices are rectangular arrays of numbers, symbols, or expressions, arranged in rows and columns. This concept is essential because it enables us to perform operations like addition, subtraction, and most importantly, multiplication on these arrays.

It's crucial to understand the basics of Matrix Algebra if you want to work with matrices effectively. The vital aspect to remember is that matrix multiplication is not commutative. This means that the order in which you multiply matrices matters: if you have two matrices, say A and B, then generally, \(AB eq BA\). This is distinct from regular arithmetic where the order of multiplication doesn't affect the result. Take time to practice and become familiar with this unique property of matrices.

2x2 matrices

A 2x2 matrix is a matrix that has two rows and two columns. It's among the simplest forms of matrices yet plays a crucial role in matrix algebra. Understanding the structure of 2x2 matrices allows you to perform various operations common in linear algebra.

• Each element is referred to by its position within the matrix. For example, element \(a_{11}\) is in the first row, first column, while \(a_{12}\) is in the first row, second column.
• These matrices are often used in transformations and in solving systems of equations.

In our example, matrices \(A\) and \(B\) are both 2x2 matrices. They contain four elements each, and we can easily perform operations such as addition or multiplication with them, provided they meet certain conditions in the case of multiplication.

Matrix Product Calculation

When calculating the product of two matrices, you must ensure that the number of columns in the first matrix matches the number of rows in the second matrix. In the case of 2x2 matrices, this condition is naturally satisfied.

For our matrices, \(A = \begin{bmatrix} 1 & -1 \ 2 & 0 \end{bmatrix}\) and \(B = \begin{bmatrix} -2 & 3 \ 1 & 2 \end{bmatrix}\), to perform the multiplication, follow these steps:

• Take each element from a row of the first matrix and multiply it by the corresponding element of a column in the second matrix.
• Sum the products to get the new element in the resulting matrix.

Here’s how it goes for \(AB\): start with the first row of \(A\) and first column of \(B\): \((1)(-2) + (-1)(1) = -3\). Repeat the operation for each corresponding position. For \(BA\), swap the roles of matrix A and B and perform the same calculations. This method yields two distinct matrices:\
\[ AB = \begin{bmatrix} -3 & 1 \ -4 & 6 \end{bmatrix}, \quad BA = \begin{bmatrix} 4 & 2 \ 5 & -1 \end{bmatrix} \]