Problem 18 - Precalculus: Functions and Graphs

Question

Find, if possible, \(A B\) and \(B A\). $$A=\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{array}\right], \quad B=\left[\begin{array}{lll} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right]$$

Step-by-Step Solution

Answer

\(AB = BA = \begin{bmatrix} 2 & 4 & 6 \\ 4 & 6 & 2 \\ 6 & 2 & 4 \end{bmatrix}\)

Step 1: Check Dimensions for Multiplication A and B

Matrix multiplication is possible only if the number of columns in the first matrix equals the number of rows in the second matrix. Matrix \(A\) is a \(3 \times 3\) matrix and matrix \(B\) is also a \(3 \times 3\) matrix. Since both have the same number of columns as the number of rows, both \(AB\) and \(BA\) multiplications are possible.

Step 2: Multiply A with B to find AB

Perform the matrix multiplication \(AB\) by calculating each element in the resulting matrix. \[AB = \begin{bmatrix}1 \cdot 2 + 2 \cdot 0 + 3 \cdot 0 & 1 \cdot 0 + 2 \cdot 2 + 3 \cdot 0 & 1 \cdot 0 + 2 \cdot 0 + 3 \cdot 2 \2 \cdot 2 + 3 \cdot 0 + 1 \cdot 0 & 2 \cdot 0 + 3 \cdot 2 + 1 \cdot 0 & 2 \cdot 0 + 3 \cdot 0 + 1 \cdot 2 \3 \cdot 2 + 1 \cdot 0 + 2 \cdot 0 & 3 \cdot 0 + 1 \cdot 2 + 2 \cdot 0 & 3 \cdot 0 + 1 \cdot 0 + 2 \cdot 2\end{bmatrix} = \begin{bmatrix}2 & 4 & 6 \4 & 6 & 2 \6 & 2 & 4\end{bmatrix}\]

Step 3: Multiply B with A to find BA

Perform the matrix multiplication \(BA\) by calculating each element in the resulting matrix. \[BA = \begin{bmatrix}2 \cdot 1 + 0 \cdot 2 + 0 \cdot 3 & 2 \cdot 2 + 0 \cdot 3 + 0 \cdot 1 & 2 \cdot 3 + 0 \cdot 1 + 0 \cdot 2 \0 \cdot 1 + 2 \cdot 2 + 0 \cdot 3 & 0 \cdot 2 + 2 \cdot 3 + 0 \cdot 1 & 0 \cdot 3 + 2 \cdot 1 + 0 \cdot 2 \0 \cdot 1 + 0 \cdot 2 + 2 \cdot 3 & 0 \cdot 2 + 0 \cdot 3 + 2 \cdot 1 & 0 \cdot 3 + 0 \cdot 1 + 2 \cdot 2\end{bmatrix} = \begin{bmatrix}2 & 4 & 6 \4 & 6 & 2 \6 & 2 & 4\end{bmatrix}\]

Step 4: Compare Results of AB and BA

Both matrices \(AB\) and \(BA\) result in the same matrix: \[\begin{bmatrix}2 & 4 & 6 \4 & 6 & 2 \6 & 2 & 4\end{bmatrix}\]

Key Concepts

Matrix Dimensions

Understanding the dimensions of a matrix is crucial when discussing matrix multiplication. A matrix's dimensions tell you its size in terms of rows and columns. It is denoted as "rows × columns".
Here’s a simple way to grasp it:

• Each entry in a matrix is identified by two indices: the row position and the column position.
• For a matrix to be represented as 3 × 3, like matrix A and B in the exercise, it needs to have 3 rows and 3 columns.
• This specific structure allows both matrices to interact through multiplication, as the inner dimensions (3 columns of A and 3 rows of B) match.

When multiplying two matrices, the number of columns in the first matrix must equal the number of rows in the second matrix. This ensures each element has a corresponding match.
This understanding forms the basis for multiplying matrices correctly, allowing us to form a new matrix from two source matrices.

Matrix Product

The matrix product refers to the result you get when you multiply two matrices. This process involves each element in the rows of the first matrix being multiplied by the corresponding element in the columns of the second matrix.
Here's a breakdown of how to conduct matrix multiplication:

• You take each row of the first matrix and multiply it element-wise with each column of the second matrix, summing up the products to form a new element in the resultant matrix.
• This element-wise multiplication and summation is repeated for all row-column combinations meeting the required dimensions.

For example, in our step-by-step solution, to find the product matrix AB, you take Row 1 from A and multiply it by each column of B. The same is repeated for Rows 2 and 3, resulting in a new matrix.
The final matrix, as derived in both products AB and BA in the exercise, showcases the product of these meticulous calculations at every element position. This serves as a clear representation of how matrix multiplication operates in practice.

Commutative Property of Matrices

One unique aspect of matrices that often surprises students is their lack of commutative property under multiplication. In math, the commutative property typically means you can change the order of numbers in an operation and get the same result (like with addition or multiplication of regular numbers). However, when it comes to matrices, this often isn't the case.
For two matrices A and B, matrix multiplication isn't generally commutative, which means that typically \(AB eq BA\).
In the given example though, both AB and BA resulted in the same product matrix:

• AB = Matrix C
• BA = Matrix C

This is a special case where the two matrices (A and B) happen to multiply to give the same product when the order is reversed. It is the exception, not the rule.
Understanding this lack of commutative nature is essential when learning about transformations and operations involving matrices as it influences how equations and expressions are solved and interpreted.