Problem 31 - College Algebra

Question

In Exercises \(27-36,\) find (if possible): \(\begin{array}{llll}\text { a. } A B & \text { and } & \text { b. } B A\end{array}\) $$ A=\left[\begin{array}{rrr} 1 & -1 & 4 \\ 4 & -1 & 3 \\ 2 & 0 & -2 \end{array}\right], \quad B=\left[\begin{array}{rrr} 1 & 1 & 0 \\ 1 & 2 & 4 \\ 1 & -1 & 3 \end{array}\right] $$

Step-by-Step Solution

Answer

The matrix AB is \[ \[ {(1, 1, 0),\n (4, 2, 4),\n (2, -1, 3)} \]\] and the matrix BA is \[ \[ {(5, -2, 7),\n (1, -3, 2),\n (1, 1, -4)} \]\].

Step 1: Matrix Multiplication AB

Matrix multiplication involves taking the dot product of each row of the first matrix with each column of the second. Let's denote the first matrix as A and the second matrix as B.\n To find AB, for each i-th row of A and j-th column of B, we multiply corresponding elements from the row and column, and then add them up:\n AB(i,j) = sum_k (A[i,k]*B[k,j])\n By applying this formula, we find the matrix AB to be:\n AB = \[ \[ {(1, 1, 0),\n (4, 2, 4),\n (2, -1, 3)} \]\]

Step 2: Matrix Multiplication BA

For BA, we take each i-th row of B and j-th column of A. Same as before, we multiply corresponding elements from the row and column, and then add them up:\n BA(i,j) = sum_k (B[i,k]*A[k,j])\n By applying this formula, we find the matrix BA to be:\n BA = \[ \[ {(5, -2, 7),\n (1, -3, 2),\n (1, 1, -4)} \]\]

Step 3: Conclusion

We have successfully computed the products AB and BA. Note, as expected AB ≠ BA which illustrates non-commutativity of matrix multiplication. That means the order of matrices in multiplication matters.

Key Concepts

Dot Product

The dot product is a fundamental operation used in matrix multiplication. It involves multiplying corresponding elements of two sequences of numbers and then summing the results. In the context of matrix multiplication, it is calculated between rows of the first matrix and columns of the second matrix. For example, if you have the row vector from matrix A, \( [a_1, a_2, a_3] \) and the column vector from matrix B, \( [b_1, b_2, b_3]^T \), the dot product will be \( a_1 \times b_1 + a_2 \times b_2 + a_3 \times b_3 \).

Each element of the resulting matrix from multiplying matrices A and B is the dot product of a row from A with a column from B. So for matrix A with dimensions \( m \times n \) and matrix B with dimensions \( n \times p \) (note that the inner dimensions must match), the resulting matrix AB will be of dimension \( m \times p \) and each element \( AB_{ij} \) will be the dot product of the i-th row of A and the j-th column of B.

Non-Commutativity in Matrices

An intriguing aspect of matrix multiplication is non-commutativity, which stands in contrast to multiplication of real numbers we're accustomed to. This means that for two matrices A and B, the product AB is generally not equal to the product BA. This characteristic should always be taken into account when working with matrices.

To illustrate this, let's consider our given matrices A and B from the exercise. When we compute AB, we're effectively combining the rows of A with columns of B. However, when we calculate BA, we're combining the rows of B with the columns of A. Since the rows and columns contain different values and are combinated in different ways, the results AB and BA will not be the same in most cases, as seen in the provided solutions of our exercise.

Order of Matrices

The 'order of matrices,' or the size of the matrices, determines if multiplication is possible and what the size of the resulting matrix will be. The order is given by the number of rows and columns in the matrix, typically denoted as \( m \times n \) for an m-row by n-column matrix. For matrix multiplication to be valid, the number of columns in the first matrix must be equal to the number of rows in the second matrix. This is often referred to as 'conformable for multiplication'.

For example, in our exercise, matrix A is of order \( 3 \times 3 \) and matrix B is also of order \( 3 \times 3 \)—thus, their inner dimensions match (3 and 3), allowing us to multiply them. It's also important to note that even if two matrices are conformable for multiplication, switching their order might render them non-conformable, further demonstrating non-commutativity and the importance of matrix order in matrix multiplication.