Problem 7 - Calculus for Biology and Medicine

Question

Show that the solution of $$ \begin{array}{l} a_{11} x_{1}+a_{12} x_{2}=b_{1} \\ a_{21} x_{1}+a_{22} x_{2}=b_{2} \end{array} $$ where \(a_{11}, a_{12}, a_{21}, a_{22}, b_{1}\) and \(b_{2}\) are all constants, is given by $$ x_{1}=\frac{a_{22} b_{1}-a_{12} b_{2}}{a_{11} a_{22}-a_{21} a_{12}} $$ and $$ x_{2}=\frac{-a_{21} b_{1}+a_{11} b_{2}}{a_{11} a_{22}-a_{21} a_{12}} $$

Step-by-Step Solution

Answer

The solution is derived using matrix inversion, providing \(x_1 = \frac{a_{22}b_1 - a_{12}b_2}{a_{11}a_{22} - a_{21}a_{12}}\) and \(x_2 = \frac{-a_{21}b_1 + a_{11}b_2}{a_{11}a_{22} - a_{21}a_{12}}\).

Step 1: Understanding the System of Equations

We are given a system of linear equations: \( a_{11} x_{1}+a_{12} x_{2}=b_{1} \) and \( a_{21} x_{1}+a_{22} x_{2}=b_{2} \). The goal is to find expressions for \(x_1\) and \(x_2\).

Step 2: Expressing in Matrix Form

Rewrite the system in matrix form as \( AX = B \), where \( A = \begin{pmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{pmatrix} \), \( X = \begin{pmatrix} x_1 \ x_2 \end{pmatrix} \), and \( B = \begin{pmatrix} b_1 \ b_2 \end{pmatrix} \).

Step 3: Using the Inverse Matrix Method

To solve for \( X \), use the formula \( X = A^{-1}B \), assuming that matrix \( A \) is invertible. The inverse of a 2x2 matrix \( A \) is given by \( A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} a_{22} & -a_{12} \ -a_{21} & a_{11} \end{pmatrix} \), where \( \text{det}(A) = a_{11}a_{22} - a_{21}a_{12} \).

Step 4: Calculating the Determinant of A

Calculate the determinant of \( A \): \( \text{det}(A) = a_{11}a_{22} - a_{21}a_{12} \). This value must be non-zero for the inverse to exist.

Step 5: Finding the Inverse of A

Using the determinant, the inverse of \( A \) is \( A^{-1} = \frac{1}{a_{11}a_{22} - a_{21}a_{12}} \begin{pmatrix} a_{22} & -a_{12} \ -a_{21} & a_{11} \end{pmatrix} \).

Step 6: Multiplying A^{-1} with B

Perform the matrix multiplication \( X = A^{-1}B \) to find \( X \). Compute \( X = \frac{1}{a_{11}a_{22} - a_{21}a_{12}} \begin{pmatrix} a_{22} & -a_{12} \ -a_{21} & a_{11} \end{pmatrix} \begin{pmatrix} b_1 \ b_2 \end{pmatrix} \).

Step 7: Deriving x_1

From the matrix multiplication, \( x_1 = \frac{a_{22}b_1 - a_{12}b_2}{a_{11}a_{22} - a_{21}a_{12}} \). This expression is derived by calculating the first row times the column vector.

Step 8: Deriving x_2

Similarly, derive \( x_2 \) using the second row: \( x_2 = \frac{-a_{21}b_1 + a_{11}b_2}{a_{11}a_{22} - a_{21}a_{12}} \).

Step 9: Conclusion: Verifying the Solution

After obtaining the expressions for \( x_1 \) and \( x_2 \), confirm that these satisfy the original system of equations. Both expressions rely on the determinant being non-zero.

Key Concepts

Inverse Matrix

In linear algebra, the concept of the inverse matrix is fundamental when you want to solve systems of linear equations using matrix notation.
To find an inverse of a matrix, the matrix must be square (same number of rows and columns) and the determinant of the matrix must be non-zero.
For a 2x2 matrix \( A \), the inverse \( A^{-1} \) is calculated using:

• \( A = \begin{pmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{pmatrix} \)
• The inverse: \( A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} a_{22} & -a_{12} \ -a_{21} & a_{11} \end{pmatrix} \)

This formula allows you to find an expression for any linear combination set to be solved.
However, remember it can only be used if \( \text{det}(A) eq 0 \), ensuring \( A \) is invertible.

Determinant

The determinant of a matrix is a special number associated with square matrices. It provides essential information about the matrix, primarily whether it is invertible.
For example, a zero determinant indicates a matrix might not be inverted, a situation called singularity.
The determinant for a 2x2 matrix \( A \) is computed as:

• \( \text{det}(A) = a_{11}a_{22} - a_{21}a_{12} \)

This value needs to be non-zero for the inverse to exist.
If \( \text{det}(A) = 0 \), any attempts to find the inverse will lead to undefined calculations.
Therefore, always check the determinant before proceeding with solving systems using matrix inverses.

Matrix Multiplication

Matrix multiplication is a key operation in linear algebra, allowing transformations of vectors and solving systems altogether.
When multiplying two matrices, follow the formula:

• Multiply rows of the first matrix by columns of the second, summing the products.

For example, with a matrix product \( X = A^{-1}B \), you compute each element by taking the dot product of corresponding pairs in the rows and columns.
Matrix multiplication requires that the number of columns in the first matrix equals the number of rows in the second.
It's crucial, particularly when solving linear equations, to maintain proper dimensions and order to achieve valid solutions.

System of Linear Equations

A system of linear equations consists of multiple equations that can all be solved together.
Each equation is a straight line (in two dimensions) or a hyperplane (in higher dimensions) in some form.
For example, in the exercise, you solve:

• \( a_{11}x_{1} + a_{12}x_{2} = b_{1} \)
• \( a_{21}x_{1} + a_{22}x_{2} = b_{2} \)

By arranging terms into matrices, these can be solved using inverse matrix techniques, simplifying complex problems.
Solve them by finding values for unknowns \( x_1 \) and \( x_2 \) that satisfy all equations.
Systems of equations are common in various fields like science, engineering, and economics.