Problem 44 - Calculus for Biology and Medicine

Question

Show that the inverse of $$ A=\left[\begin{array}{lll} 2 & 3 & 1 \\ 5 & 2 & 3 \\ 1 & 2 & 0 \end{array}\right] $$ is $$ B=\left[\begin{array}{rrr} -\frac{6}{5} & \frac{2}{5} & \frac{7}{5} \\ \frac{3}{5} & -\frac{1}{5} & -\frac{1}{5} \\ \frac{8}{5} & -\frac{1}{5} & -\frac{11}{5} \end{array}\right] $$

Step-by-Step Solution

Answer

The inverse of matrix \( A \) is matrix \( B \) as \( AB = I \).

Step 1: Understand the Properties of Matrices

To verify that matrix \( B \) is the inverse of matrix \( A \), recall that for two matrices \( A \) and \( B \), if \( AB = I \) and \( BA = I \), where \( I \) is the identity matrix, then \( B \) is the inverse of \( A \). Since \( A \) is a 3x3 matrix, \( I \) must be a 3x3 identity matrix.

Step 2: Prepare for Matrix Multiplication

First, prepare to calculate the product \( AB \). Write out matrices \( A \) and \( B \) for reference: \[ A = \begin{bmatrix} 2 & 3 & 1 \ 5 & 2 & 3 \ 1 & 2 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} -\frac{6}{5} & \frac{2}{5} & \frac{7}{5} \ \frac{3}{5} & -\frac{1}{5} & -\frac{1}{5} \ \frac{8}{5} & -\frac{1}{5} & -\frac{11}{5} \end{bmatrix} \]

Step 3: Compute \( AB \) - First Row

Multiply the first row of \( A \) with each column of \( B \) to get the first row of \( AB \):- First element: \( 2(-\frac{6}{5}) + 3(\frac{3}{5}) + 1(\frac{8}{5}) = -\frac{12}{5} + \frac{9}{5} + \frac{8}{5} = 1 \)- Second element: \( 2(\frac{2}{5}) + 3(-\frac{1}{5}) + 1(-\frac{1}{5}) = \frac{4}{5} - \frac{3}{5} - \frac{1}{5} = 0 \)- Third element: \( 2(\frac{7}{5}) + 3(-\frac{1}{5}) + 1(-\frac{11}{5}) = \frac{14}{5} - \frac{3}{5} - \frac{11}{5} = 0 \)The first row of \( AB \) is \([1, 0, 0]\).

Step 4: Compute \( AB \) - Second Row

Multiply the second row of \( A \) with each column of \( B \) to get the second row of \( AB \):- First element: \( 5(-\frac{6}{5}) + 2(\frac{3}{5}) + 3(\frac{8}{5}) = -6 + \frac{6}{5} + \frac{24}{5} = 0 \)- Second element: \( 5(\frac{2}{5}) + 2(-\frac{1}{5}) + 3(-\frac{1}{5}) = 2 - \frac{2}{5} - \frac{3}{5} = 1 \)- Third element: \( 5(\frac{7}{5}) + 2(-\frac{1}{5}) + 3(-\frac{11}{5}) = 7 - \frac{2}{5} - \frac{33}{5} = 0 \)The second row of \( AB \) is \([0, 1, 0]\).

Step 5: Compute \( AB \) - Third Row

Multiply the third row of \( A \) with each column of \( B \) to get the third row of \( AB \):- First element: \( 1(-\frac{6}{5}) + 2(\frac{3}{5}) + 0(\frac{8}{5}) = -\frac{6}{5} + \frac{6}{5} + 0 = 0 \)- Second element: \( 1(\frac{2}{5}) + 2(-\frac{1}{5}) + 0(-\frac{1}{5}) = \frac{2}{5} - \frac{2}{5} = 0 \)- Third element: \( 1(\frac{7}{5}) + 2(-\frac{1}{5}) + 0(-\frac{11}{5}) = \frac{7}{5} - \frac{2}{5} = 1 \)The third row of \( AB \) is \([0, 0, 1]\).

Step 6: Verify the Result

The product \( AB \) is \[ AB = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \]which is the identity matrix. This confirms that \( B \) is indeed the inverse of \( A \).

Key Concepts

Matrix Multiplication

Matrix multiplication is a fundamental concept in linear algebra. It involves combining two matrices to produce another matrix. To multiply two matrices, say matrices \( A \) and \( B \), the number of columns in \( A \) must match the number of rows in \( B \). If \( A \) is an \( m \times n \) matrix and \( B \) is an \( n \times p \) matrix, then their product is an \( m \times p \) matrix.

Each element of the resulting matrix is computed by taking the dot product of the respective row from the first matrix with the column of the second matrix:

• Multiply corresponding elements of the row and column together.
• Add up all these products.

This method of matrix multiplication is critical in determining inverses and is often used in various applications like computer graphics and solving systems of equations.

Identity Matrix

An identity matrix is a special type of square matrix where all the elements on the main diagonal are 1s and all other elements are 0s. For a 3x3 identity matrix, it looks like this:

\[ I = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \]
The identity matrix is significant because it acts as a multiplicative identity in matrix multiplication. This means if you multiply any matrix by its corresponding identity matrix, the result is the original matrix itself:

• For any matrix \( A \), \( AI = A \)
• Equivalently, \( IA = A \)

In the context of matrix inverses, if \( B \) is the inverse of \( A \), then their product \( AB \) (as well as \( BA \)) will result in the identity matrix, confirming \( B \) is indeed the correct inverse.

3x3 Matrix

A 3x3 matrix is a grid with 3 rows and 3 columns. Each position in the grid contains a number (an entry). Understanding 3x3 matrices is essential because they can represent transformations in three-dimensional space, among other applications.

In practical applications, 3x3 matrices are often used to perform operations such as rotations and scaling in computer graphics. In systems of equations, 3x3 matrices can describe a system with three equations and three unknowns.

When discussing matrix inverses, a 3x3 matrix \( A \) will have an inverse \( B \) if and only if \( AB \) and \( BA \) produce an identity matrix. Calculating the inverse of a 3x3 matrix involves several steps, including finding the determinant and the adjoint of the matrix. If the determinant is non-zero, the matrix is invertible, and the inverse can be computed.