Problem 33 - Calculus for Biology and Medicine

Question

33\. Let $$ A=\left[\begin{array}{rr} 2 & 1 \\ -1 & -3 \end{array}\right] $$ Find \(A^{2}, A^{3}\), and \(A^{4}\).

Step-by-Step Solution

Answer

\(A^2 = \begin{bmatrix} 3 & -1 \\ 1 & 10 \end{bmatrix}, A^3 = \begin{bmatrix} 7 & 6 \\ -8 & -29 \end{bmatrix}, A^4 = \begin{bmatrix} 8 & -11 \\ 13 & 79 \end{bmatrix}\).

Step 1: Multiply A by Itself to Find A²

To find \(A^2\), we need to multiply \(A\) by itself. The matrix \(A\) is given by:\[A = \begin{bmatrix} 2 & 1 \ -1 & -3 \end{bmatrix}\]Calculate the product:\[A^2 = A \times A = \begin{bmatrix} 2 & 1 \ -1 & -3 \end{bmatrix} \times \begin{bmatrix} 2 & 1 \ -1 & -3 \end{bmatrix} = \begin{bmatrix} (2)(2) + (1)(-1) & (2)(1) + (1)(-3) \ (-1)(2) + (-3)(-1) & (-1)(1) + (-3)(-3) \end{bmatrix} = \begin{bmatrix} 3 & -1 \ 1 & 10 \end{bmatrix}\]

Step 2: Multiply A² by A to Find A³

Now that we have \(A^2\), we multiply it with \(A\) to find \(A^3\):\[A^3 = A^2 \times A = \begin{bmatrix} 3 & -1 \ 1 & 10 \end{bmatrix} \times \begin{bmatrix} 2 & 1 \ -1 & -3 \end{bmatrix} = \begin{bmatrix} (3)(2) + (-1)(-1) & (3)(1) + (-1)(-3) \ (1)(2) + (10)(-1) & (1)(1) + (10)(-3) \end{bmatrix} = \begin{bmatrix} 7 & 6 \ -8 & -29 \end{bmatrix}\]

Step 3: Multiply A³ by A to Find A⁴

Finally, we will multiply \(A^3\) by \(A\) to get \(A^4\):\[A^4 = A^3 \times A = \begin{bmatrix} 7 & 6 \ -8 & -29 \end{bmatrix} \times \begin{bmatrix} 2 & 1 \ -1 & -3 \end{bmatrix} = \begin{bmatrix} (7)(2) + (6)(-1) & (7)(1) + (6)(-3) \ (-8)(2) + (-29)(-1) & (-8)(1) + (-29)(-3) \end{bmatrix} = \begin{bmatrix} 8 & -11 \ 13 & 79 \end{bmatrix}\]

Key Concepts

Matrix Powers

Matrix powers involve raising a matrix, like \(A\), to an exponent. To understand this, realize that when you see \(A^n\), it means multiplying matrix \(A\) by itself \(n\) times. Here is a simple sequence:

• To compute \(A^2\), multiply \(A\) by \(A\).
• For \(A^3\), you would multiply \(A^2\) by \(A\) again.
• Repeat this process to find higher powers like \(A^4\) and so on.

The concept of matrix powers is essential because it applies to iterative processes and systems where operations recur, like in solving differential equations or analyzing computer graphics. Importantly, unlike scalar exponentiation, matrix powers require consideration of matrix size and structure, especially since multiplication of matrices is not commutative (e.g., \(A \times B eq B \times A\)). This means that order matters a lot in operations.

Algebraic Operations

Algebraic operations with matrices involve addition, subtraction, and multiplication, among others. To work smoothly with matrices, understanding multiplication is crucial. Here's a refresher:
- **Matrix Multiplication:** To multiply matrices, take the rows from the first matrix (\(A\)) and the columns from the second matrix (\(B\)). Multiply corresponding entries, then sum them up to get each element of the resultant matrix. For example, if multiplying a \(2 \times 2\) matrix:\[\begin{bmatrix} a & b \ c & d \end{bmatrix} \times \begin{bmatrix} e & f \ g & h \end{bmatrix} = \begin{bmatrix} ae + bg & af + bh \ ce + dg & cf + dh \end{bmatrix}\]- **Commutative Property:** Remember that matrix multiplication is not commutative, meaning \(A \times B eq B \times A\) in general. Thus, pay attention to the order of factors when calculating matrix powers or products.
These operations are foundational in various applications such as solving systems of equations, transformations in graphics, and state vector evaluations in control systems.

Determinant Calculation

The determinant of a matrix provides a scalar value that conveys essential properties of the matrix. For a \(2 \times 2\) matrix \(A\), determining the determinant is straightforward:
If \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\), then the determinant, denoted as \(\det(A)\), is given by:\[\det(A) = ad - bc\]Understanding the determinant is crucial because:

• It helps in determining if a matrix is invertible. A non-zero determinant indicates that a matrix has an inverse.
• It offers insights into the matrix's properties, such as volume scaling factors in transformations (in geometry and physics).
• In solving systems of linear equations, determinants play a vital role in methods like Cramer's Rule.

Moreover, if a matrix is non-invertible (determinant equals zero), this impacts how the matrix behaves in algebraic calculations and transformations.