Problem 12 - Precalculus: Functions and Graphs

Question

Find, if possible, \(A B\) and \(B A\). $$A=\left[\begin{array}{rr} 4 & -2 \\ -2 & 1 \end{array}\right], \quad B=\left[\begin{array}{ll} 2 & 1 \\ 4 & 2 \end{array}\right]$$

Step-by-Step Solution

Answer

The product \(AB\) is a zero matrix, and \(BA\) is \(\left[\begin{array}{cc} 6 & -3 \\ 12 & -6 \end{array}\right]\)."

Step 1: Check Dimensions

First, check the dimensions of matrices \(A\) and \(B\). Matrix \(A\) is a \(2 \times 2\) matrix, and matrix \(B\) is also a \(2 \times 2\) matrix. Since they are both \(2 \times 2\) matrices, both products \(AB\) and \(BA\) are possible as long as the number of columns in the first matrix matches the number of rows in the second matrix.

Step 2: Calculate \(AB\)

To find \(AB\), multiply each row of \(A\) by each column of \(B\):- First row of \(A\): \(4 \times 2 + (-2) \times 4 = 8 - 8 = 0\)- First row of \(A\): \(4 \times 1 + (-2) \times 2 = 4 - 4 = 0\)- Second row of \(A\): \((-2) \times 2 + 1 \times 4 = -4 + 4 = 0\)- Second row of \(A\): \((-2) \times 1 + 1 \times 2 = -2 + 2 = 0\)Thus, \(AB = \left[\begin{array}{cc} 0 & 0 \ 0 & 0 \end{array}\right]\).

Step 3: Calculate \(BA\)

To find \(BA\), multiply each row of \(B\) by each column of \(A\):- First row of \(B\): \(2 \times 4 + 1 \times (-2) = 8 - 2 = 6\)- First row of \(B\): \(2 \times (-2) + 1 \times 1 = -4 + 1 = -3\)- Second row of \(B\): \(4 \times 4 + 2 \times (-2) = 16 - 4 = 12\)- Second row of \(B\): \(4 \times (-2) + 2 \times 1 = -8 + 2 = -6\)Thus, \(BA = \left[\begin{array}{cc} 6 & -3 \ 12 & -6 \end{array}\right]\).

Key Concepts

Understanding a 2x2 Matrix

A 2x2 matrix is a rectangular array of numbers that consists of two rows and two columns. It is compact and is often used in various mathematical operations such as solving systems of equations and performing transformations. Each element in a 2x2 matrix is identified by its position, ordered first by the row and then by the column it occupies.

For example, the 2x2 matrix: \[ A = \begin{bmatrix} 4 & -2 \ -2 & 1 \end{bmatrix} \] Here, "4" is in the first row, first column (position \((1,1)\)), "-2" on the first row, second column (position \((1,2)\)), "-2" again on the second row, first column (position \((2,1)\)), and "1" is in the second row, second column (position \((2,2)\)).

• Rows run horizontally; in this matrix, there are two rows.
• Columns run vertically; again, there are two columns in this configuration.

Matrix multiplication differs from the usual arithmetic multiplication, and understanding the parts of a 2x2 matrix is crucial in grasping this concept fully.

Exploring Matrix Dimensions

Matrix dimensions are essentially the size identifiers of matrices and are crucial when performing operations such as addition, subtraction, and multiplication. Dimensions are given in the format \(m \times n\), where \(m\) is the number of rows and \(n\) is the number of columns. Knowing the dimensions helps determine whether certain operations can be performed or not.

In the case of our example matrices:

• Matrix \(A\) has dimensions \(2 \times 2\), meaning it has 2 rows and 2 columns.
• Matrix \(B\) also has dimensions \(2 \times 2\), with 2 rows and 2 columns.

For matrix multiplication to be possible, the number of columns in the first matrix must equal the number of rows in the second matrix. With both \(A\) and \(B\) being \(2 \times 2\) matrices, both products \(AB\) and \(BA\) can be calculated. Understanding these basics ensures that we can properly approach problems involving matrix operations with clarity and precision.

The Commutative Property of Matrices

In mathematics, many operations possess a commutative property, meaning the order of operation does not affect the result. However, this is not generally the case with matrix multiplication. It is essential to understand that while you can multiply matrices in any order, the results are not necessarily the same.

For instance, given matrices \(A\) and \(B\) as:\[ A = \begin{bmatrix} 4 & -2 \ -2 & 1 \end{bmatrix} \] \[ B = \begin{bmatrix} 2 & 1 \ 4 & 2 \end{bmatrix} \] Calculating \(AB\):\[ AB = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix} \] Whereas calculating \(BA\): \[ BA = \begin{bmatrix} 6 & -3 \ 12 & -6 \end{bmatrix} \]

Clearly, \(AB eq BA\). This shows that the commutative property does not apply to matrix multiplication, underlining the importance of multiplying matrices in the intended order. Understanding this concept helps prevent mistakes and ensures accurate calculations.