Problem 13 - Calculus for Biology and Medicine

Question

$$ \begin{array}{r} A=\left[\begin{array}{rrr} 1 & 3 & -1 \\ 2 & 4 & 1 \\ 0 & -2 & 2 \end{array}\right], \quad B=\left[\begin{array}{rrr} 5 & -1 & 4 \\ 2 & 0 & 1 \\ 1 & -3 & -3 \end{array}\right], \\ \quad C=\left[\begin{array}{rrr} -2 & 0 & 4 \\ 1 & -3 & 1 \\ 0 & 0 & 2 \end{array}\right] \end{array} $$ $$ \text { Show that if } A+B=C, \text { then } A=C-B $$

Step-by-Step Solution

Answer

The matrices \( A + B \neq C \), hence \( A \neq C - B \) based on the given condition.

Step 1: Understand the Problem

We need to show that if the matrix equation \( A + B = C \) holds true, then \( A = C - B \). We will do this by performing the operations given on the matrices \( A \, \), \( B \, \), and \( C \).

Step 2: Set Up the Equation

Start with the equation given in the problem: \( A + B = C \). Substitute the matrices to get: \[\begin{bmatrix} 1 & 3 & -1 \2 & 4 & 1 \0 & -2 & 2 \end{bmatrix}+ \begin{bmatrix} 5 & -1 & 4 \2 & 0 & 1 \1 & -3 & -3 \end{bmatrix}= \begin{bmatrix} -2 & 0 & 4 \1 & -3 & 1 \0 & 0 & 2 \end{bmatrix}\]

Step 3: Verify A + B = C

Add the matrices \( A \) and \( B \). Calculate each element of the resulting matrix:\[\begin{align*}1+5 & = 6, & 3 - 1 & = 2, & -1 + 4 & = 3,\2+2 & = 4, & 4 + 0 & = 4, & 1 + 1 & = 2,\0+1 & = 1, & -2 - 3 & = -5, & 2 - 3 & = -1.\end{align*}\]This results in \( A+B = \begin{bmatrix} 6 & 2 & 3 \ 4 & 4 & 2 \ 1 & -5 & -1 \end{bmatrix} \), which is not equal to \( C \). The initial condition given by the problem \( A + B = C \) actually does not hold in this case.

Step 4: Rearrange the Equation

Despite the incorrect initial condition, we can still set up the equation \( A = C - B \) for theoretical purposes. This means we rewrite the condition as:\[A = C - B\]. Substitute matrices for \( C \) and \( B \) and then subtract:\[C - B = \begin{bmatrix} -2 & 0 & 4 \1 & -3 & 1 \0 & 0 & 2 \end{bmatrix}- \begin{bmatrix} 5 & -1 & 4 \2 & 0 & 1 \1 & -3 & -3 \end{bmatrix}\]

Step 5: Perform the Matrix Subtraction

Calculate each element of the resulting matrix by subtracting the corresponding elements of matrices \( C \) and \( B \):\[\begin{align*}-2 - 5 & = -7, & 0 - (-1) & = 1, & 4 - 4 & = 0,\1 - 2 & = -1, & -3 - 0 & = -3, & 1 - 1 & = 0,\0 - 1 & = -1, & 0 - (-3) & = 3, & 2 - (-3) & = 5.\end{align*}\]This results in \( C-B = \begin{bmatrix} -7 & 1 & 0 \ -1 & -3 & 0 \ -1 & 3 & 5 \end{bmatrix} \) which does not equal \( A \).

Step 6: Conclusion

The conclusion shows that due to incorrect initial condition \( A + B eq C \), rearranging the equation confirms the inconsistency. Therefore, that exercise's verification unintentionally shows that \( A = C - B \) is not valid when the given initial condition \( A + B = C \) is incorrect.

Key Concepts

Matrix Addition

Matrix addition is an essential operation in linear algebra. This concept involves adding two matrices, which must have the same dimensions. For two matrices to be added, each corresponding pair of elements is summed.

• If you have matrix \( A \) and matrix \( B \), both of size \( m \times n \), then the sum \( A + B \) is also a matrix of size \( m \times n \).
• The process involves taking each element \( a_{ij} \) from matrix \( A \) and adding it to the corresponding element \( b_{ij} \) from matrix \( B \).

Let's consider an example:For matrices \( A = \begin{bmatrix} 1 & 3 \ 2 & 4 \end{bmatrix} \) and \( B = \begin{bmatrix} 5 & -1 \ 2 & 0 \end{bmatrix} \), the resulting matrix from their addition is calculated as follows:

• \( 1 + 5 = 6 \)
• \( 3 + (-1) = 2 \)
• \( 2 + 2 = 4 \)
• \( 4 + 0 = 4 \)

Thus, \( A + B = \begin{bmatrix} 6 & 2 \ 4 & 4 \end{bmatrix} \). Note that the initial exercise setup incorrectly stated \( A + B = C \), but here, understanding matrix addition's fundamental method can pinpoint successful addition even when theoretical assumptions fall short.

Matrix Subtraction

Matrix subtraction functions similarly to matrix addition but involves subtracting corresponding elements of two matrices. The two matrices must still be the same dimension, \( m \times n \).

• For matrices \( C \) and \( B \), the subtraction \( C - B \) yields another matrix of size \( m \times n \).
• Subtracting each element \( b_{ij} \) in matrix \( B \) from the corresponding \( c_{ij} \) in matrix \( C \), results in the matrix difference.

For instance, consider matrices \( C = \begin{bmatrix} -2 & 0 \ 1 & -3 \end{bmatrix} \) and \( B = \begin{bmatrix} 5 & -1 \ 2 & 0 \end{bmatrix} \):

• To calculate, \( -2 - 5 = -7 \)
• \( 0 - (-1) = 1 \)
• \( 1 - 2 = -1 \)
• \( -3 - 0 = -3 \)

This results in \( C - B = \begin{bmatrix} -7 & 1 \ -1 & -3 \end{bmatrix} \). Though matrices \( A \) and the resulting matrix \( C - B \) were not equal in the original exercise, it's a crucial practice for comprehending subtraction's role in verifying equations in linear algebra.

Linear Algebra

Linear algebra is a branch of mathematics concerning linear equations, matrices, and vector spaces. It establishes the foundation for solving systems of linear equations through various operations, particularly matrix manipulations.

• Understanding matrix operations like addition and subtraction is pivotal since they are frequently used for solving linear equations and transforming vector spaces.
• Matrix operations also support sophisticated concepts such as matrix transformations, determinants, and eigenvalues.

In linear algebra, matrices serve as representations of linear transformations, allowing you to simplify and solve complex problems involving multi-dimensional spaces. The exercise clearly illustrates the use of matrix operations to explore theoretical conditions even when a given mathematical setup may be inaccurate or not hold true, fostering a deeper comprehension of the underlying structures and relationships in linear systems. This exploration echoes linear algebra's core principle of expressing and manipulating linear equations in clear, structured ways.