91Ó°ÊÓ

A system of equations consists of multiple equations that are solved together, with the same set of variables. A common example is two linear equations with two variables, such as:

• Equation 1:
  \( 3x - 2y = 6 \)
• Equation 2:
  \( -x + 5y = -2 \)

In solving such systems, the goal is to find values for the variables that satisfy both equations simultaneously. This ensures that the solution point is a shared point between both lines when plotted on a graph.These systems can be solved in various ways including substitution, elimination, and using matrix methods, which is the focus here. By using a matrix, these equations can be handled with powerful algebraic tools to find the solutions effectively and efficiently.

Expressing a system of equations in matrix form is the first step in using matrix methods to find a solution. For two equations with two variables, we can represent the system like this:

• Coefficient Matrix \( A \):
  \[A = \begin{bmatrix} 3 & -2 \ -1 & 5 \end{bmatrix}\]
• Variable Matrix \( \mathbf{x} \):
  \[\mathbf{x} = \begin{bmatrix} x \ y \end{bmatrix}\]
• Constant Matrix \( \mathbf{b} \):
  \[\mathbf{b} = \begin{bmatrix} 6 \ -2 \end{bmatrix}\]

The system of equations can therefore be succinctly written as \( A \mathbf{x} = \mathbf{b} \). This compact representation is very efficient in algebraically handling and solving the system using matrices, especially for larger systems with more equations and variables.

The determinant is a special number that can be calculated from a square matrix. It plays a crucial role in determining whether a matrix has an inverse or not. For a \(2 \times 2\) matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is calculated as:
\[\text{det}(A) = ad - bc\]In this example, using the coefficient matrix \( A \) from the previous section:

• \( a = 3, \ b = -2, \ c = -1, \ d = 5 \)
• \( ext{det}(A) = (3)(5) - (-1)(-2) = 15 - 2 = 13\)

A non-zero determinant (13 in this case) indicates that the matrix is invertible, meaning it is possible to find an inverse matrix \( A^{-1} \). If the determinant were zero, the matrix would not have an inverse, and the system could either have no solution or infinitely many solutions depending on other factors.

Once the system is expressed in matrix form and the inverse of the matrix \( A \) is found, the next step is to solve for the variables. To do this, you multiply both sides of the equation \( A \mathbf{x} = \mathbf{b} \) by \( A^{-1} \):\[\mathbf{x} = A^{-1} \mathbf{b}\]For our specific equation:

• Inverse Matrix \( A^{-1} \):
  \[A^{-1} = \frac{1}{13} \begin{bmatrix} 5 & 2 \ 1 & 3 \end{bmatrix}\]
• Constant Matrix \( \mathbf{b} \):
  \[\mathbf{b} = \begin{bmatrix} 6 \ -2 \end{bmatrix}\]

Multiply to find \( \mathbf{x} \):
\[\mathbf{x} = \frac{1}{13} \begin{bmatrix} 5 & 2 \ 1 & 3 \end{bmatrix} \begin{bmatrix} 6 \ -2 \end{bmatrix} = \begin{bmatrix} 2 \ 0 \end{bmatrix}\]This calculation shows that \( x = 2 \) and \( y = 0 \) is the solution to the system. The beauty of this method is its application to systems beyond just two variables, showcasing the power of linear algebra.