Problem 26 - Calculus for Biology and Medicine

Question

$$ A=\left[\begin{array}{rr} -1 & 0 \\ 1 & 2 \end{array}\right], \quad B=\left[\begin{array}{rr} 2 & 3 \\ -1 & 1 \end{array}\right], \quad C=\left[\begin{array}{rr} 1 & 2 \\ 0 & -1 \end{array}\right] $$ $$ \text { Show that } A(B+C)=A B+A C $$

Step-by-Step Solution

Answer

The equation \( A(B+C) = AB + AC \) is true.

Step 1: Calculate \( B + C \)

First, let's add matrices \( B \) and \( C \). $$ B = \begin{bmatrix} 2 & 3 \ -1 & 1 \end{bmatrix}, \quad C = \begin{bmatrix} 1 & 2 \ 0 & -1 \end{bmatrix} $$To add these matrices, add corresponding elements:$$ B + C = \begin{bmatrix} 2+1 & 3+2 \ -1+0 & 1+(-1) \end{bmatrix} = \begin{bmatrix} 3 & 5 \ -1 & 0 \end{bmatrix} $$

Step 2: Calculate \( A(B+C) \)

Now we multiply matrix \( A \) by the result of \( B + C \).\( A = \begin{bmatrix} -1 & 0 \ 1 & 2 \end{bmatrix} \), and \( B+C = \begin{bmatrix} 3 & 5 \ -1 & 0 \end{bmatrix} \).The matrix multiplication is performed as follows:$$ A(B+C) = \begin{bmatrix} -1 \times 3 + 0 \times (-1) & -1 \times 5 + 0 \times 0 \ 1 \times 3 + 2 \times (-1) & 1 \times 5 + 2 \times 0 \end{bmatrix} $$So, we have:\( A(B+C) = \begin{bmatrix} -3 & -5 \ 1 & 5 \end{bmatrix} \)

Step 3: Calculate \( AB \)

Next, multiply matrices \( A \) and \( B \).\( A = \begin{bmatrix} -1 & 0 \ 1 & 2 \end{bmatrix} \), and \( B = \begin{bmatrix} 2 & 3 \ -1 & 1 \end{bmatrix} \).Matrix multiplication is given by:$$ AB = \begin{bmatrix} -1 \times 2 + 0 \times (-1) & -1 \times 3 + 0 \times 1 \ 1 \times 2 + 2 \times (-1) & 1 \times 3 + 2 \times 1 \end{bmatrix} $$Therefore, we obtain:\( AB = \begin{bmatrix} -2 & -3 \ 0 & 5 \end{bmatrix} \)

Step 4: Calculate \( AC \)

Now, multiply matrices \( A \) and \( C \).\( A = \begin{bmatrix} -1 & 0 \ 1 & 2 \end{bmatrix} \), and \( C = \begin{bmatrix} 1 & 2 \ 0 & -1 \end{bmatrix} \).The multiplication is given by:$$ AC = \begin{bmatrix} -1 \times 1 + 0 \times 0 & -1 \times 2 + 0 \times (-1) \ 1 \times 1 + 2 \times 0 & 1 \times 2 + 2 \times (-1) \end{bmatrix} $$Thus, the result is:\( AC = \begin{bmatrix} -1 & -2 \ 1 & 0 \end{bmatrix} \)

Step 5: Calculate \( AB + AC \)

Add matrices \( AB \) and \( AC \).\( AB = \begin{bmatrix} -2 & -3 \ 0 & 5 \end{bmatrix} \), and \( AC = \begin{bmatrix} -1 & -2 \ 1 & 0 \end{bmatrix} \).Add corresponding elements:$$ AB + AC = \begin{bmatrix} -2+(-1) & -3+(-2) \ 0+1 & 5+0 \end{bmatrix} = \begin{bmatrix} -3 & -5 \ 1 & 5 \end{bmatrix} $$

Step 6: Conclusion

We found that both \( A(B+C) = \begin{bmatrix} -3 & -5 \ 1 & 5 \end{bmatrix} \) and \( AB + AC = \begin{bmatrix} -3 & -5 \ 1 & 5 \end{bmatrix} \).Thus, we have shown that:\( A(B+C) = AB + AC \), confirming the distributive property of matrix multiplication.

Key Concepts

Distributive Property

In mathematics, the distributive property is a fundamental principle that applies to operations like addition and multiplication across various mathematical structures, including matrix multiplication found in linear algebra. When applied to matrices, the distributive property allows us to multiply a matrix by the sum of two others in one step and get the same result as if we distributed the multiplication over each matrix in the sum individually.

For matrices, this property is stated as:

• Let matrices \( A \), \( B \), and \( C \) be compatible for multiplication, then \( A(B + C) = AB + AC \).

Matrix multiplication is distributive over matrix addition, much like numbers. This means we can choose the more convenient way to compute the result, which can be handy in simplifying matrix operations and proofs. When dealing with matrices, always ensure they are of appropriate dimensions to avoid calculation errors.

This property is particularly useful in solving equations that involve matrices, as it reduces complex expressions into simpler components.

Matrix Addition

Matrix addition is a straightforward process that involves adding the corresponding elements of two matrices of the same dimensions. To better understand how it works, consider matrices \( B \) and \( C \) from the problem.

• Matrix \( B = \begin{bmatrix} 2 & 3 \ -1 & 1 \end{bmatrix} \)
• Matrix \( C = \begin{bmatrix} 1 & 2 \ 0 & -1 \end{bmatrix} \)

To add these matrices, we simply perform element-wise addition:

• Top-left element: \( 2 + 1 = 3 \)
• Top-right element: \( 3 + 2 = 5 \)
• Bottom-left element: \( -1 + 0 = -1 \)
• Bottom-right element: \( 1 + (-1) = 0 \)

This gives us the resulting matrix \( B + C = \begin{bmatrix} 3 & 5 \ -1 & 0 \end{bmatrix} \).

Notice that matrix addition is both straightforward and commutative, meaning \( B + C = C + B \). The simplicity of this process facilitates many operations in linear algebra and helps in applying the distributive property during matrix calculations.

Linear Algebra

Linear algebra is a branch of mathematics focusing on vector spaces and linear mappings between these spaces. This field underpins much of modern mathematics, physics, and engineering.

• Linear algebra involves concepts such as matrices, vectors, and systems of linear equations.
• It is central to computer graphics, machine learning, and data science.

A matrix is a rectangular array of numbers representing the coefficients in a system of linear equations or transformations. Understanding how to manipulate these matrices through addition, multiplication, and other operations is crucial for solving problems in linear algebra.

Essential operations in linear algebra include:

• Matrix addition and multiplication that follow certain rules and properties, like associativity, commutativity (only for addition), and distributivity.
• Solving systems of equations, which helps in finding unknown variables within linear systems.
• Determinants and inverses which further lead to understanding more complex structures in algebraic equations.

Linear algebra serves as a foundation for advanced areas like calculus, statistics, and optimization, making it a vital subject for students in mathematics and science-related fields.