Problem 6 - 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 \(2 A+3 B-C\).

Step-by-Step Solution

Answer

The result is the matrix \( \begin{bmatrix} 4 & -3 & 10 \\ -2 & 5 & -6 \\ 1 & 5 & 7 \end{bmatrix} \).

Step 1: Matrix Multiplication by Scalar 2

Begin by multiplying matrix \( A \) by scalar 2: \[2A = 2 \times \begin{bmatrix}1 & 0 & 1 \2 & 3 & -1 \0 & -2 & 0\end{bmatrix} = \begin{bmatrix}2 \cdot 1 & 2 \cdot 0 & 2 \cdot 1 \2 \cdot 2 & 2 \cdot 3 & 2 \cdot (-1) \2 \cdot 0 & 2 \cdot (-2) & 2 \cdot 0\end{bmatrix} = \begin{bmatrix}2 & 0 & 2 \4 & 6 & -2 \0 & -4 & 0\end{bmatrix}.\]

Step 2: Matrix Multiplication by Scalar 3

Next, multiply matrix \( B \) by scalar 3:\[3B = 3 \times \begin{bmatrix}1 & -1 & 4 \-2 & 0 & -1 \1 & 3 & 3\end{bmatrix} = \begin{bmatrix}3 \cdot 1 & 3 \cdot (-1) & 3 \cdot 4 \3 \cdot (-2) & 3 \cdot 0 & 3 \cdot (-1) \3 \cdot 1 & 3 \cdot 3 & 3 \cdot 3\end{bmatrix} = \begin{bmatrix}3 & -3 & 12 \-6 & 0 & -3 \3 & 9 & 9\end{bmatrix}.\]

Step 3: Matrix Subtraction

Subtract matrix \( C \) from the result of the addition of \( 2A \) and \( 3B \):Calculate \( 2A + 3B \):\[\begin{bmatrix}2 & 0 & 2 \4 & 6 & -2 \0 & -4 & 0\end{bmatrix} + \begin{bmatrix}3 & -3 & 12 \-6 & 0 & -3 \3 & 9 & 9\end{bmatrix} = \begin{bmatrix}2+3 & 0+(-3) & 2+12 \4+(-6) & 6+0 & -2+(-3) \0+3 & -4+9 & 0+9\end{bmatrix} = \begin{bmatrix}5 & -3 & 14 \-2 & 6 & -5 \3 & 5 & 9\end{bmatrix}.\]Now subtract matrix \( C \):\[\begin{bmatrix}5 & -3 & 14 \-2 & 6 & -5 \3 & 5 & 9\end{bmatrix} - \begin{bmatrix}1 & 0 & 4 \0 & 1 & 1 \2 & 0 & 2\end{bmatrix} = \begin{bmatrix}5-1 & -3-0 & 14-4 \-2-0 & 6-1 & -5-1 \3-2 & 5-0 & 9-2\end{bmatrix} = \begin{bmatrix}4 & -3 & 10 \-2 & 5 & -6 \1 & 5 & 7\end{bmatrix}.\]

Step 4: Final Result

The resulting matrix from the operation \( 2A + 3B - C \) is:\[\begin{bmatrix}4 & -3 & 10 \-2 & 5 & -6 \1 & 5 & 7\end{bmatrix}.\]

Key Concepts

Scalar Multiplication

When you perform scalar multiplication on a matrix, you essentially multiply every element in the matrix by a fixed number, known as the scalar. This operation is quite straightforward. For example, if you have a matrix \( A \) and a scalar 2, then each element of \( A \) is multiplied by 2 to get the new scaled matrix.
This is how it works:

• Take the matrix \( A = \begin{bmatrix} 1 & 0 & 1 \ 2 & 3 & -1 \ 0 & -2 & 0 \end{bmatrix} \).
• Multiply everything in \( A \) by 2: \( 2 \times A = \begin{bmatrix} 2\times 1 & 2\times 0 & 2\times 1 \ 2\times 2 & 2\times 3 & 2\times (-1) \ 2\times 0 & 2\times (-2) & 2\times 0 \end{bmatrix} = \begin{bmatrix} 2 & 0 & 2 \ 4 & 6 & -2 \ 0 & -4 & 0 \end{bmatrix} \).

Scalar multiplication is useful because it changes the magnitude of the matrix's elements without altering its overall structure.

Matrix Addition

Matrix addition is the process of adding two matrices by adding the corresponding elements together. It's important to note that you can only add matrices of the same dimensions. This consistency is crucial because each element at the same position in both matrices is combined.
To add matrices:

• Use matrices \( 2A = \begin{bmatrix} 2 & 0 & 2 \ 4 & 6 & -2 \ 0 & -4 & 0 \end{bmatrix} \) and \( 3B = \begin{bmatrix} 3 & -3 & 12 \ -6 & 0 & -3 \ 3 & 9 & 9 \end{bmatrix} \).
• Add corresponding elements: \( \begin{bmatrix} 2+3 & 0+(-3) & 2+12 \ 4 +(-6) & 6 + 0 & -2 + (-3) \ 0 + 3 & -4 + 9 & 0 + 9 \end{bmatrix} \).
• This results in: \( \begin{bmatrix} 5 & -3 & 14 \ -2 & 6 & -5 \ 3 & 5 & 9 \end{bmatrix} \).

Matrix addition helps to compile results of operations into an aggregate form for further exploration or application.

Matrix Subtraction

Matrix subtraction is similar to matrix addition, but you subtract the elements of one matrix from the corresponding elements of another. As with addition, both matrices must share the same dimensions for subtraction to take place. Subtraction can be visualized as the removal of quantities from their respective places.
Consider our new matrix and matrix \( C \):

• You have \( 2A + 3B = \begin{bmatrix} 5 & -3 & 14 \ -2 & 6 & -5 \ 3 & 5 & 9 \end{bmatrix} \).
• Subtract \( C = \begin{bmatrix} 1 & 0 & 4 \ 0 & 1 & 1 \ 2 & 0 & 2 \end{bmatrix} \) from it.
• The result of the subtraction is \( \begin{bmatrix} 5-1 & -3-0 & 14-4 \ -2-0 & 6-1 & -5-1 \ 3-2 & 5-0 & 9-2 \end{bmatrix} = \begin{bmatrix} 4 & -3 & 10 \ -2 & 5 & -6 \ 1 & 5 & 7 \end{bmatrix} \).

Matrix subtraction serves to highlight the differences between data sets or operations, providing insight into what changes or remains consistent when data sets are juxtaposed.