Problem 8 - Calculus for Biology and Medicine

Question

Show that if $$ a_{11} a_{22}-a_{21} a_{12} \neq 0 $$ then the system $$ \begin{array}{l} a_{11} x_{1}+a_{12} x_{2}=0 \\ a_{21} x_{1}+a_{22} x_{2}=0 \end{array} $$ has exactly one solution, namely, \(x_{1}=0\) and \(x_{2}=0\).

Step-by-Step Solution

Answer

The system has only the trivial solution \((x_1, x_2) = (0, 0)\) because the determinant is non-zero, making the matrix invertible.

Step 1: Understanding the System of Equations

The lines in the system of equations are given as: \(a_{11} x_{1} + a_{12} x_{2} = 0\) and \(a_{21} x_{1} + a_{22} x_{2} = 0\). These lines form a homogeneous system of linear equations, implying there is always the trivial solution \((x_1, x_2) = (0,0)\). However, you must prove it's the only solution.

Step 2: Using the Non-zero Determinant Condition

The condition given, \(a_{11} a_{22} - a_{21} a_{12} eq 0\), indicates that the matrix \(A\) formed by the coefficients has a non-zero determinant. This means the matrix is invertible, suggesting that the only solution to the homogeneous system \(A X = 0\) is the trivial solution.

Step 3: Writing the Matrix Equation

Form the matrix equation based on the system: \[ A \cdot \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \] where \(A = \begin{pmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{pmatrix}\).

Step 4: Analyzing the Invertibility of the Matrix

Since \( ext{det}(A) = a_{11} a_{22} - a_{21} a_{12} eq 0 \), matrix \(A\) has full rank and is invertible. An invertible matrix ensures that the only solution to the equation \(A \cdot X = 0\) is \(X = 0\).

Step 5: Conclusion about the Solution

Given that \(A\) is invertible, it means the solution set to the system of equations is just the trivial solution: \(x_1 = 0\) and \(x_2 = 0\). Thus, the condition \(a_{11} a_{22} - a_{21} a_{12} eq 0\) ensures the system has a unique solution.

Key Concepts

System of Linear Equations

A system of linear equations consists of two or more equations where each term is either a constant or the product of a constant and a single variable.
In the exercise provided, we have a homogeneous system represented as:

• \( a_{11} x_1 + a_{12} x_2 = 0 \)
• \( a_{21} x_1 + a_{22} x_2 = 0 \)

A homogeneous system is named so because all terms without the variables equal zero.
One solution always exists for homogeneous systems: the trivial solution where all variables are zero.
Therefore, the system of equations simplifies to having at least one solution, \((x_1, x_2) = (0, 0)\).
Identifying solutions beyond this requires evaluating other properties of the system, like invertibility or determinant, to ensure that no other solutions exist.

Matrix Invertibility

Matrix invertibility is a critical concept in linear algebra. It refers to the condition when a square matrix has an inverse.
In this exercise, we represent the coefficients of the system in a 2x2 matrix \( A \).
A matrix is invertible if there exists another matrix that, when multiplied with the original one, yields the identity matrix:

• \( A \cdot A^{-1} = I \)
• Where \( I \) is the identity matrix

If a matrix is invertible, it means the linear system it represents has a unique solution. The exercise indicates this by employing the condition that the determinant is non-zero.
Therefore, if matrix \( A \) is invertible, it guarantees that the only solution to the system \( A \cdot X = 0 \) is the trivial one, \( X = 0 \).
This property is crucial since it determines the uniqueness of solutions from a system of equations.

Determinant

The determinant is a special number that can be calculated from a square matrix.
It gives insight into matrix properties, like invertibility and the volume scale factor of the transformation described by the matrix.
In a 2x2 matrix \( A = \begin{pmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{pmatrix} \), the determinant is calculated as:

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

For our exercise, the determinant condition given as non-zero implies that the matrix is invertible.
When the determinant is zero, it means the matrix approaches singularity, indicating a lack of invertibility and potentially infinite solutions to the system.
However, because the determinant here is non-zero, it confirms that our linear system is linearly independent with only one solution \( (x_1, x_2) = (0, 0) \).
This illustrates the importance of determinants in assessing the nature of systems' solutions.