Problem 13 - Algebra for College Students

Question

For Problems \(9-20\), find \(A B\) and \(B A\), whenever they exist. $$ A=\left[\begin{array}{rrr} 1 & -1 & 2 \\ 0 & 1 & -2 \\ 3 & 1 & 4 \end{array}\right], \quad B=\left[\begin{array}{rrr} 2 & 3 & -1 \\ 4 & 0 & 2 \\ -5 & 1 & -1 \end{array}\right] $$

Step-by-Step Solution

Answer

The product \(AB\) is \(\begin{bmatrix} -12 & 4 & -1 \\ -14 & -1 & 0 \\ 10 & 10 & -1 \end{bmatrix}\) and \(BA\) is \(\begin{bmatrix} 7 & -1 & 10 \\ 2 & -4 & 12 \\ -2 & 4 & -12 \end{bmatrix}\).

Step 1: Verify Matrix Dimensions for Multiplication

To find \(A B\), check that the number of columns in matrix \(A\) matches the number of rows in matrix \(B\). Matrix \(A\) is \(3 \times 3\) and so is matrix \(B\), thus multiplication is possible for both \(A B\) and \(B A\).

Step 2: Compute Matrix Product AB

Calculate each element of \(AB\) using the formula for matrix multiplication. For element in row \(i\), column \(j\) in \(AB\), compute it as: \[(AB)_{ij} = \sum_{k=1}^3 A_{ik}B_{kj}\]For \((AB)_{11}\), \(1 \cdot 2 + (-1) \cdot 4 + 2 \cdot (-5) = 2 - 4 - 10 = -12\).Continue similarly for all elements to get: \(AB = \begin{bmatrix} -12 & 4 & -1 \ -14 & -1 & 0 \ 10 & 10 & -1 \end{bmatrix}\).

Step 3: Compute Matrix Product BA

Use the same process for \(BA\). Calculate each element as:\[(BA)_{ij} = \sum_{k=1}^3 B_{ik}A_{kj}\]For \((BA)_{11}\), \(2 \cdot 1 + 3 \cdot 0 + (-1) \cdot 3 = 2 + 0 - 3 = -1\).Continue similarly for all elements to get: \(BA = \begin{bmatrix} 7 & -1 & 10 \ 2 & -4 & 12 \ -2 & 4 & -12 \end{bmatrix}\).

Step 4: Verify and Conclude

Confirm your results by re-evaluating if necessary. Conclude that \(AB\) and \(BA\) are not equal, which is generally expected for non-commutative matrix multiplication.

Key Concepts

Matrix Dimensions

When working with matrix multiplication, understanding matrix dimensions is crucial.
Dimensions define the structure and size of a matrix: the number of rows and columns it contains.
In matrix notation, a matrix with 'm' rows and 'n' columns is described as an \(m \times n\) matrix. Here, both matrices A and B are \(3 \times 3\), since they contain 3 rows and 3 columns.
This compatibility ensures that these matrices can be multiplied because, for matrix multiplication to be valid, the number of columns in the first matrix must match the number of rows in the second matrix.
This condition allows us to calculate products like \(A B\) and \(B A\), resulting in another \(3 \times 3\) matrix.

Commutative Property

One key aspect of matrices is that they do not generally abide by the commutative property of multiplication.
The commutative property in arithmetic operations states that the order of numbers does not affect the product: for example, \(a \times b = b \times a\).
However, matrix multiplication is not commutative, meaning \(A B\) does not necessarily equal \(B A\).
In the given solution, matrices \(A\) and \(B\) illustrate this as \(A B\) results in a different matrix than \(B A\).
This non-commutativity is a fundamental trait of matrix operations and highlights the importance of order in matrix multiplication.

Matrix Product

The matrix product involves systematic calculations to find each element of the resultant matrix.
This process uses the formula for matrix multiplication, where each element \((C)_{ij}\) in the product matrix \(C\) is derived by summing the products of elements from the corresponding row of the first matrix \(A\) and the column of the second matrix \(B\).
The general formula is

• \((C)_{ij} = \sum_{k=1}^{n} A_{ik} B_{kj}\)

where 'k' iterates over the shared dimension of both matrices.
For example, the element in the first row and first column of \(A B\) is calculated as \((1 \cdot 2) + (-1 \cdot 4) + (2 \cdot -5) = -12\).
This methodology is applied to every element of the resulting matrix, requiring precision and careful calculation.

3x3 Matrices

A \(3 \times 3\) matrix is a square matrix with 3 rows and 3 columns, making it especially convenient for operations like multiplication or determinant calculation.
The structure of \(3 \times 3\) matrices ensures symmetry, simplifying certain operations when compared to non-square matrices.
The given example involving matrices \(A\) and \(B\) uses this symmetry to demonstrate multiplication.
In such matrices, every element plays a role in determining the final product.
While simple in structure, handling \(3 \times 3\) matrices needs meticulous attention especially when performing operations like inversion or finding eigenvalues due to their complexity compared to smaller matrices.