Problem 9 - Calculus for Biology and Medicine

Question

$$ \begin{array}{r} A=\left[\begin{array}{rrr} 1 & 3 & -1 \\ 2 & 4 & 1 \\ 0 & -2 & 2 \end{array}\right], \quad B=\left[\begin{array}{rrr} 5 & -1 & 4 \\ 2 & 0 & 1 \\ 1 & -3 & -3 \end{array}\right], \\ \quad C=\left[\begin{array}{rrr} -2 & 0 & 4 \\ 1 & -3 & 1 \\ 0 & 0 & 2 \end{array}\right] \end{array} $$ $$ \text { Determine } D \text { so that } A+B+C+D=\mathbf{0} \text { . } $$

Step-by-Step Solution

Answer

Matrix \( D \) is \( \begin{bmatrix} -4 & -2 & -7 \\ -5 & -1 & -3 \\ -1 & 5 & -1 \end{bmatrix} \).

Step 1: Setup the Equation

We need to find a matrix \( D \) such that \( A + B + C + D = \mathbf{0} \). Start by writing the equation: \( D = - (A + B + C) \). This step involves understanding what the problem is asking and rearranging to focus on finding \( D \).

Step 2: Calculate A + B

Add matrices \( A \) and \( B \). The sum will be:\[A + B = \begin{bmatrix}1+5 & 3+(-1) & -1+4 \2+2 & 4+0 & 1+1 \0+1 & -2+(-3) & 2+(-3) \\end{bmatrix} = \begin{bmatrix}6 & 2 & 3 \4 & 4 & 2 \1 & -5 & -1 \\end{bmatrix}\]

Step 3: Calculate (A + B) + C

Now, add the result from Step 2 to matrix \( C \). The sum will be:\[(A + B) + C = \begin{bmatrix}6+(-2) & 2+0 & 3+4 \4+1 & 4+(-3) & 2+1 \1+0 & -5+0 & -1+2 \\end{bmatrix} = \begin{bmatrix}4 & 2 & 7 \5 & 1 & 3 \1 & -5 & 1 \\end{bmatrix}\]

Step 4: Find -((A + B) + C)

Compute the negation of the matrix from Step 3 to get \( D \):\[D = -\begin{bmatrix}4 & 2 & 7 \5 & 1 & 3 \1 & -5 & 1 \\end{bmatrix} = \begin{bmatrix}-4 & -2 & -7 \-5 & -1 & -3 \-1 & 5 & -1 \\end{bmatrix}\]

Step 5: Verify the Solution

Add \( A + B + C + D \) to ensure the result is \( \mathbf{0} \), confirming that we have correctly found \( D \). Use the matrices calculated:\[\begin{bmatrix}4 & 2 & 7 \5 & 1 & 3 \1 & -5 & 1 \\end{bmatrix} + \begin{bmatrix}-4 & -2 & -7 \-5 & -1 & -3 \-1 & 5 & -1 \\end{bmatrix} = \mathbf{0}\]Each element of the sum is zero, confirming that our solution is correct.

Key Concepts

Matrix Addition

Matrix addition is one of the fundamental operations performed on matrices. To add two matrices, each of the matrices must be of the same dimension. This means they need to have the same number of rows and columns.

For matrix addition, you simply add the corresponding elements of the matrices. For example, if you have matrices \( A \) and \( B \) both of size \( 3 \times 3 \), the element at the first row and first column of the result will be the sum of the element at the same position in each matrix.

If \( A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{bmatrix} \) and \( B = \begin{bmatrix} b_{11} & b_{12} & b_{13} \ b_{21} & b_{22} & b_{23} \ b_{31} & b_{32} & b_{33} \end{bmatrix} \), then their sum \( A + B = \begin{bmatrix} a_{11}+b_{11} & a_{12}+b_{12} & a_{13}+b_{13} \ a_{21}+b_{21} & a_{22}+b_{22} & a_{23}+b_{23} \ a_{31}+b_{31} & a_{32}+b_{32} & a_{33}+b_{33} \end{bmatrix} \).

This operation is both commutative and associative, meaning the order in which you add the matrices doesn't change their sum.

Matrix Negation

Matrix negation involves changing the sign of each element within the matrix. If you have a matrix, say matrix \( A \), negating it will give you \( -A \).

For example, if \( A = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix} \), then negating \( A \) gives \( -A = \begin{bmatrix} -1 & -2 \ -3 & -4 \end{bmatrix} \).

Matrix negation is used in our exercise to find the matrix \( D \) by equating \( D \) to the negation of the sum of other matrices \( A + B + C \). This is a handy operation when solving matrix equations or simplifying linear algebra problems.

Matrix Equation

A matrix equation involves two or more matrices connected through an equation. In linear algebra, these equations are used to represent systems of linear equations in a compact form.

In our example, we used the matrix equation \( A + B + C + D = \mathbf{0} \). This equation represents the sum of matrices \( A \), \( B \), \( C \), and \( D \) being equal to the zero matrix, \( \mathbf{0} \).

Solving this equation provided the necessary steps to calculate the missing matrix \( D \). Rearranging the equation as \( D = - (A + B + C) \) allows us to calculate \( D \) by finding the negation of the sum of the matrices. This step highlights the importance of understanding how to manipulate matrix equations effectively.

Problem Solving in Linear Algebra

Solving problems in linear algebra often involves a sequence of operations like matrix addition, subtraction, or multiplication. Each concept builds on the previous to address a specific problem.

In problem-solving scenarios, such as for the original exercise, cleanly setting up the equation is crucial. Here, by identifying and defining what each matrix represents and what needs to be solved, you ensure an efficient path to the solution.

Analyzing each step, such as calculating \( A + B \), adding \( C \), and eventually applying matrix negation, helps simplify and verify the problem.

Linear algebra often requires verifying the solution by plugging it back into the equation, as demonstrated in the final step of ensuring that the sum \( A + B + C + D \) results in the zero matrix. This verification step is vital in confirming that all calculations have been correctly executed.