Problem 21 - Calculus for Biology and Medicine

Question

Let $$A=\left[\begin{array}{rr} -1 & 0 \\ 1 & 2 \end{array}\right], \quad B=\left[\begin{array}{rr} 2 & 0 \\ -1 & -1 \end{array}\right], \quad C=\left[\begin{array}{rr} 1 & 2 \\ 0 & -1 \end{array}\right]$$ Compute the following: (a) \(A B\) (b) \(B A\)

Step-by-Step Solution

Answer

(a) \( AB = \left[ \begin{array}{rr} -2 & 0 \\ 0 & -2 \end{array} \right] \); (b) \( BA = \left[ \begin{array}{rr} -2 & 0 \\ 0 & -2 \end{array} \right] \).

Step 1: Define Matrix Multiplication

Matrix multiplication involves taking each element of the rows of the first matrix and multiplying them by the corresponding element in the columns of the second matrix. The result is the sum of these products.

Step 2: Compute AB

To find the product \( AB \), calculate each entry of the resulting matrix by multiplying rows of \( A \) with columns of \( B \).For the first row, first column entry:\[ (-1 \times 2) + (0 \times -1) = -2 \]For the first row, second column entry:\[ (-1 \times 0) + (0 \times -1) = 0 \]For the second row, first column entry:\[ (1 \times 2) + (2 \times -1) = 2 - 2 = 0 \]For the second row, second column entry:\[ (1 \times 0) + (2 \times -1) = 0 - 2 = -2 \]So, \( AB = \left[ \begin{array}{rr} -2 & 0 \ 0 & -2 \end{array} \right] \).

Step 3: Compute BA

To find the product \( BA \), calculate each entry of the resulting matrix by multiplying the rows of \( B \) with the columns of \( A \).For the first row, first column entry:\[ (2 \times -1) + (0 \times 1) = -2 \]For the first row, second column entry:\[ (2 \times 0) + (0 \times 2) = 0 \]For the second row, first column entry:\[ (-1 \times -1) + (-1 \times 1) = 1 - 1 = 0 \]For the second row, second column entry:\[ (-1 \times 0) + (-1 \times 2) = 0 - 2 = -2 \]So, \( BA = \left[ \begin{array}{rr} -2 & 0 \ 0 & -2 \end{array} \right] \).

Step 4: Conclusion

Both products \( AB \) and \( BA \) result in the same matrix, which is \( \left[ \begin{array}{rr} -2 & 0 \ 0 & -2 \end{array} \right] \). This is an example of a special case where matrix multiplication is commutative, which is not generally true for most matrices.

Key Concepts

Linear Algebra

Linear algebra is a branch of mathematics that deals with vectors, matrices, and linear transformations. It provides a way to represent and solve systems of linear equations and is widely used in geometry and other fields. In linear algebra, matrices are used to represent complex data and operations. They can transform vector spaces and solve systems of equations, making them fundamental to the subject.
An understanding of linear algebra is essential for studying more advanced mathematical concepts and applications, such as those found in physics, computer science, and engineering. By learning how to manipulate matrices and perform operations like matrix multiplication, students can tackle a wide range of problems in these fields.

Matrix Operations

Matrix operations involve processes like addition, subtraction, scalar multiplication, and matrix multiplication. Of these, matrix multiplication is one of the more complex but powerful operations.
During matrix multiplication, the number of columns in the first matrix must match the number of rows in the second matrix. In our exercise, the matrices involved are given as two-dimensional arrays. Here's a quick recap of how to perform matrix multiplication:

• Take a row from the first matrix and a column from the second matrix.
• Multiply corresponding elements from the row and column.
• Sum up these products to get the resulting element in the new matrix.

Matrix multiplication is not commutative, which means the order of multiplication affects the result. However, while working through some exercises, you might encounter special cases where the result remains the same regardless of the order of multiplication, but these are exceptions, not the rule.

Commutative Property

In most mathematical operations, like addition and multiplication of numbers, the commutative property holds true. This property suggests that the order in which you perform the operation doesn't change the result. For instance, with real numbers, 3 + 5 is the same as 5 + 3.
However, when it comes to matrix multiplication, the commutative property does not generally apply. This means that multiplying matrix A by matrix B does not usually yield the same result as multiplying matrix B by matrix A. This is evident in most calculations involving matrices. That being said, the exercise provided is a special case where matrix multiplication is commutative, as both products, \( AB \) and \( BA \), resulted in the same matrix: \[\left[ \begin{array}{rr} -2 & 0 \ 0 & -2 \end{array} \right] \]When such equality occurs, it is worth examining the matrices involved to understand why this happened. This could be due to the symmetry or specific construction of the matrices.