Problem 8 - Calculus for Biology and Medicine

Question

Let $$A=\left[\begin{array}{rrr}1 & 0 & 1 \\ 2 & 3 & -1 \\ 0 & -2 & 0\end{array}\right] ,\boldsymbol{B}=\left[\begin{array}{rrr}\mathbf{1} & \mathbf{- 1} & \mathbf{4} \\\ \mathbf{- 2} & \mathbf{0} & \mathbf{- 1} \\ \mathbf{1} & \mathbf{3} & \mathbf{3}\end{array}\right] ,\boldsymbol{C}=\left[\begin{array}{lll}\mathbf{1} & \mathbf{0} & \mathbf{4} \\\ \mathbf{0} & \mathbf{1} & \mathbf{1} \\ \mathbf{2} & \mathbf{0} & \mathbf{2}\end{array}\right]$$ Find \(3 C-B+\frac{1}{2} A\).

Step-by-Step Solution

Answer

The result is \(\begin{bmatrix} 2.5 & 1 & 8.5 \\ 3 & 4.5 & 3.5 \\ 5 & -4 & 3 \end{bmatrix}\).

Step 1: Multiply Matrix C by 3

To find \(3C\), multiply each element of matrix \(C\) by 3. \[3C = \begin{bmatrix} 3 \times 1 & 3 \times 0 & 3 \times 4 \ 3 \times 0 & 3 \times 1 & 3 \times 1 \ 3 \times 2 & 3 \times 0 & 3 \times 2 \end{bmatrix} = \begin{bmatrix} 3 & 0 & 12 \ 0 & 3 & 3 \ 6 & 0 & 6 \end{bmatrix}\]

Step 2: Subtract Matrix B from 3C

To find \(3C - B\), subtract each element of \(B\) from \(3C\). \[3C - B = \begin{bmatrix} 3 & 0 & 12 \ 0 & 3 & 3 \ 6 & 0 & 6 \end{bmatrix} - \begin{bmatrix} 1 & -1 & 4 \ -2 & 0 & -1 \ 1 & 3 & 3 \end{bmatrix} = \begin{bmatrix} 2 & 1 & 8 \ 2 & 3 & 4 \ 5 & -3 & 3 \end{bmatrix}\]

Step 3: Multiply Matrix A by 0.5

To find \(\frac{1}{2}A\), multiply each element of matrix \(A\) by 0.5. \[\frac{1}{2}A = \begin{bmatrix} 0.5 \times 1 & 0.5 \times 0 & 0.5 \times 1 \ 0.5 \times 2 & 0.5 \times 3 & 0.5 \times -1 \ 0.5 \times 0 & 0.5 \times -2 & 0.5 \times 0 \end{bmatrix} = \begin{bmatrix} 0.5 & 0 & 0.5 \ 1 & 1.5 & -0.5 \ 0 & -1 & 0 \end{bmatrix}\]

Step 4: Add the Resulting Matrices

To find \(3C - B + \frac{1}{2}A\), add the matrices \(3C - B\) and \(\frac{1}{2}A\). \[3C - B + \frac{1}{2}A = \begin{bmatrix} 2 & 1 & 8 \ 2 & 3 & 4 \ 5 & -3 & 3 \end{bmatrix} + \begin{bmatrix} 0.5 & 0 & 0.5 \ 1 & 1.5 & -0.5 \ 0 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 2.5 & 1 & 8.5 \ 3 & 4.5 & 3.5 \ 5 & -4 & 3 \end{bmatrix}\]

Key Concepts

Matrix Multiplication

Matrix multiplication is a vital concept in linear algebra where two matrices are combined to produce a new matrix. Unlike simple arithmetic, matrix multiplication is not as straightforward. The product of two matrices is found by taking the dot product of rows from the first matrix with columns of the second matrix.
To multiply two matrices, say matrix \( A \) of size \( m \times n \) and matrix \( B \) of size \( n \times p \), the resulting matrix \( C \) will be of size \( m \times p \). Each element of \( C \) is calculated as follows:

• Select a row from matrix \( A \).
• Select a column from matrix \( B \).
• Multiply each element of the row by the corresponding element of the column and sum the products.

A key point to note is that matrix multiplication is not commutative. This means \( AB eq BA \) in most cases. Also, both matrices involved need to be compatible in size, meaning the number of columns in the first matrix should equal the number of rows in the second matrix.
It’s important to keep in mind that matrix multiplication can represent transformations in multiple dimensions, making it crucial in various fields including computer graphics and statistics.

Matrix Addition

Matrix addition is a simple operation where two matrices of the same dimensions are added together by adding corresponding elements. If two matrices \( A \) and \( B \) have the same dimensions, say \( m \times n \), then their sum \( C = A + B \) will also have dimensions \( m \times n \). Each element \( c_{ij} \) of the resulting matrix \( C \) is obtained by adding elements \( a_{ij} \) and \( b_{ij} \):

• Identify the corresponding elements in \( A \) and \( B \).
• Add each pair of corresponding elements.

Matrix addition is both commutative and associative, which means \( A + B = B + A \) and \( (A + B) + C = A + (B + C) \), respectively. These properties make addition a versatile and powerful tool in algebra, especially when combined with other operations like scalar multiplication.

Matrix Subtraction

Matrix subtraction involves subtracting corresponding elements of two matrices of the same size. Like matrix addition, both matrices involved must have identical dimensions. If matrix \( A \) and matrix \( B \) each have dimensions \( m \times n \), their difference \( C = A - B \) also possesses dimensions \( m \times n \). The specific elements are calculated as follows:

• Take the element \( a_{ij} \) from matrix \( A \).
• Subtract the corresponding element \( b_{ij} \) from matrix \( B \).

The result will be a new matrix \( C \) where each element is the difference \( c_{ij} = a_{ij} - b_{ij} \). This operation is more direct than multiplication, simply owing to the straightforward nature of subtraction between numbers. It also maintains linearity similar to addition.

Scalar Multiplication

Scalar multiplication refers to multiplying each element of a matrix by a single number called a scalar. This operation is straightforward and is used to scale matrices up or down. For matrix \( A \) with dimensions \( m \times n \), and a scalar \( k \), the result of scalar multiplication is denoted as \( kA \).
Here's how it's done:

• Take each element \( a_{ij} \) in the matrix \( A \).
• Multiply each element by the scalar \( k \).

The resulting matrix maintains the same size as the original, with each element \( ka_{ij} \) scaled according to the scalar. Scalar multiplication is distributive over matrix addition, meaning \( k(A + B) = kA + kB \), and it plays a crucial role in various computations involving matrices, particularly in simplifying linear equations and transformations.