Problem 15 - Algebra for College Students

Question

For Problems \(13-26\), compute \(A B\) and \(B A\). $$ A=\left[\begin{array}{rr} 1 & -3 \\ -4 & 6 \end{array}\right], \quad B=\left[\begin{array}{rr} 7 & -3 \\ 4 & 5 \end{array}\right] $$

Step-by-Step Solution

Answer

\(A \cdot B = \begin{bmatrix} -5 & -18 \\ -4 & 42 \end{bmatrix}\), \(B \cdot A = \begin{bmatrix} 19 & -39 \\ -16 & 18 \end{bmatrix}\).

Step 1: Initialize Matrix Sizes

First, identify the sizes of matrices \(A\) and \(B\). Matrix \(A\) is a \(2 \times 2\) matrix, and matrix \(B\) is also a \(2 \times 2\) matrix. This means that both \(A \cdot B\) and \(B \cdot A\) results will be \(2 \times 2\) matrices.

Step 2: Compute Matrix Product \(A \cdot B\)

To compute \(A \cdot B\), multiply each element of the rows of matrix \(A\) by the corresponding element of the columns of matrix \(B\).\[ A \cdot B = \begin{bmatrix} 1 & -3 \ -4 & 6 \end{bmatrix} \cdot \begin{bmatrix} 7 & -3 \ 4 & 5 \end{bmatrix} \]Calculate each element:- First row, first column: \((1 \cdot 7) + (-3 \cdot 4) = 7 - 12 = -5\)- First row, second column: \((1 \cdot -3) + (-3 \cdot 5) = -3 - 15 = -18\)- Second row, first column: \((-4 \cdot 7) + (6 \cdot 4) = -28 + 24 = -4\)- Second row, second column: \((-4 \cdot -3) + (6 \cdot 5) = 12 + 30 = 42\)\[ A \cdot B = \begin{bmatrix} -5 & -18 \ -4 & 42 \end{bmatrix} \]

Step 3: Compute Matrix Product \(B \cdot A\)

Similarly, compute \(B \cdot A\) by multiplying each element of the rows of matrix \(B\) by the columns of matrix \(A\).\[ B \cdot A = \begin{bmatrix} 7 & -3 \ 4 & 5 \end{bmatrix} \cdot \begin{bmatrix} 1 & -3 \ -4 & 6 \end{bmatrix} \]Calculate each element:- First row, first column: \((7 \cdot 1) + (-3 \cdot -4) = 7 + 12 = 19\)- First row, second column: \((7 \cdot -3) + (-3 \cdot 6) = -21 - 18 = -39\)- Second row, first column: \((4 \cdot 1) + (5 \cdot -4) = 4 - 20 = -16\)- Second row, second column: \((4 \cdot -3) + (5 \cdot 6) = -12 + 30 = 18\)\[ B \cdot A = \begin{bmatrix} 19 & -39 \ -16 & 18 \end{bmatrix} \]

Step 4: Validate Non-Commutativity

The results from Step 2 and Step 3 show that \(A \cdot B eq B \cdot A\), which illustrates that matrix multiplication is not commutative.

Key Concepts

Matrix Dimensions

Understanding the dimensions of a matrix is crucial in matrix arithmetic. Matrices are represented in a row-column format, denoted as "rows by columns". This is often symbolized as \(m \times n\), where \(m\) represents the number of rows and \(n\) represents the number of columns. For example, matrix \(A\) with the elements \([1, -3, -4, 6]\) is a \(2 \times 2\) matrix because it has 2 rows and 2 columns. Similarly, matrix \(B\) is also a \(2 \times 2\) matrix.
It is important to know the dimensions of a matrix before performing any operations. In matrix multiplication, particularly, the inner dimensions (the columns of the first matrix and the rows of the second matrix) must be the same for the multiplication to be possible.

• If matrix \(A\) is \(m \times n\) and matrix \(B\) is \(n \times p\), the resulting matrix \(AB\) will be \(m \times p\).
• In our example, both matrices are \(2 \times 2\), meaning they can indeed be multiplied in both possible orders, but they will not yield the same results due to the non-commutative property of matrices.

Non-Commutative Property

In mathematics, multiplication is usually commutative; however, this is not the case with matrices. The non-commutative property of matrix multiplication means that the order in which you multiply matrices affects the result. That is, \(A \cdot B eq B \cdot A\), unless both \(A\) and \(B\) are specifically designed to commute, such as identity matrices.
This property is intrinsic to how matrix multiplication is structured. Each element of the resulting matrix depends on an entire row from the first matrix and an entire column from the second matrix.

• This dependence makes matrix multiplication sensitive to the arrangement of terms, unlike regular multiplication of numbers.
• The exercise provides a practical demonstration: ensuring both \(A \cdot B\) and \(B \cdot A\) are correct, we find different matrices as a result.

Understanding this property helps in fields such as linear transformations and computer graphics, where matrices are used to perform operations on data in an ordered sequence.

Matrix Product Calculation

Matrix multiplication can seem daunting, but breaking it down makes it manageable. The procedure involves taking rows from the first matrix and columns from the second and calculating their dot products to get the elements of the resulting matrix. This requires the inner dimensions of the matrices to be compatible, meaning the number of columns in the first matrix must equal the number of rows in the second.
Here is a step-by-step approach to calculate the matrix product:\( A \cdot B \):

• Identify row 1 of \(A\) and column 1 of \(B\). Calculate their dot product: \((1 \cdot 7) + (-3 \cdot 4) = -5\).
• Proceed similarly for all row-column combinations to complete the result.
• The final matrix \(A \cdot B\) will be derived from these calculations.

The pattern is repeated for \(B \cdot A\), confirming the need for careful calculation as differing order affects outcome. These calculations do more than give results; they provide insight into how matrices interact with datasets, influencing everything from data arrangement in databases to animations in digital media.