Problem 7 - Calculus for Biology and Medicine

Question

Show that the solution of $$ \begin{array}{l} a_{11} x_{1}+a_{12} x_{2}=b_{1} \\ a_{21} x_{1}+a_{22} x_{2}=b_{2} \end{array} $$ 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 solutions are \( x_{1} = \frac{a_{22}b_{1} - a_{12}b_{2}}{a_{11}a_{22} - a_{21}a_{12}} \) and \( x_{2} = \frac{-a_{21}b_{1} + a_{11}b_{2}}{a_{11}a_{22} - a_{21}a_{12}} \).

Step 1: Write the System of Equations

The given system of equations is: 1. \( a_{11} x_{1} + a_{12} x_{2} = b_{1} \) 2. \( a_{21} x_{1} + a_{22} x_{2} = b_{2} \)

Step 2: Use the Elimination Method to Solve for x_1

Multiply the first equation by \( a_{22} \) and the second equation by \( a_{12} \), then subtract: \( a_{22}(a_{11}x_{1} + a_{12}x_{2}) - a_{12}(a_{21}x_{1} + a_{22}x_{2}) = a_{22}b_{1} - a_{12}b_{2} \) Simplify to obtain: \( (a_{11}a_{22} - a_{21}a_{12})x_{1} = a_{22}b_{1} - a_{12}b_{2} \)

Step 3: Solve for x_1

To solve for \( x_{1} \), divide both sides of the equation by \( a_{11}a_{22} - a_{21}a_{12} \): \[ x_{1} = \frac{a_{22}b_{1} - a_{12}b_{2}}{a_{11}a_{22} - a_{21}a_{12}} \]

Step 4: Use the Elimination Method to Solve for x_2

Multiply the first equation by \( a_{21} \) and the second equation by \( a_{11} \), then subtract: \( a_{21}(a_{11}x_{1} + a_{12}x_{2}) - a_{11}(a_{21}x_{1} + a_{22}x_{2}) = a_{21}b_{1} - a_{11}b_{2} \) Simplify to obtain: \( (a_{12}a_{21} - a_{22}a_{11})x_{2} = a_{21}b_{1} - a_{11}b_{2} \)

Step 5: Solve for x_2

To solve for \( x_{2} \), divide both sides of the equation by \( -(a_{11}a_{22} - a_{21}a_{12}) \): \[ x_{2} = \frac{-a_{21}b_{1} + a_{11}b_{2}}{a_{11}a_{22} - a_{21}a_{12}} \]

Step 6: Conclusion

Both solutions for \( x_{1} \) and \( x_{2} \) are derived using the elimination method, confirming they satisfy the original system of equations.

Key Concepts

System of Linear Equations

A system of linear equations consists of two or more linear equations that are related by shared variables. The goal is to find values for these variables that satisfy all the equations simultaneously. In mathematical terms, these systems are represented as a set of equations, such as the ones given in the problem:

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

Here, \(x_{1}\) and \(x_{2}\) are the variables we want to solve for and \(a_{ij}\) and \(b_{i}\) are known constants. A solution to this system is any pair \((x_1, x_2)\) that satisfies both equations. Linear systems can be solved using various methods, such as substitution, elimination, or matrix operations. By understanding these systems, you can solve problems in physics, engineering, economics, and more, where relationships can be represented linearly.

Elimination Method

The elimination method is one of the most common techniques for solving a system of linear equations. The idea is to eliminate one of the variables to make it easier to solve for the others. This method involves:

• Multiplying each equation by suitable coefficients so that when added or subtracted, one of the variables gets eliminated.
• Solving the resulting equation for the remaining variable.
• Substituting back to find the other variable.

In our given system, the elimination steps involve:

• Multiplying the first equation by \(a_{22}\) and the second by \(a_{12}\) to eliminate \(x_{2}\), allowing you to solve for \(x_{1}\).
• Similarly, the equations are manipulated to eliminate \(x_1\) and solve for \(x_2\).

Through these steps, the method strategically eliminates variables using basic arithmetic operations, paving the way toward the solution without requiring complex calculations.

Solving Equations

Once you have reduced your system through elimination, solving the resulting simpler equation is the next step. Here, the simplicity lies in the arithmetic:

• Identify the equation containing only one variable.
• Isolate the variable on one side of the equation to solve for it.

For instance, after elimination, we derived:

• For \(x_{1}\), \(x_{1} = \frac{a_{22}b_{1} - a_{12}b_{2}}{a_{11}a_{22} - a_{21}a_{12}}\).
• For \(x_{2}\), \(x_{2} = \frac{-a_{21}b_{1} + a_{11}b_{2}}{a_{11}a_{22} - a_{21}a_{12}}\).

These solutions are contingent upon the determinant \((a_{11}a_{22} - a_{21}a_{12})\) being non-zero, which ensures the system has a unique solution. By understanding how these solutions are derived, you can enhance your problem-solving skills in linear algebra, giving you the tools to tackle similar systems confidently.