91Ó°ÊÓ

When working with a system of linear equations, a very systematic way to represent them is through a matrix equation. Essentially, this allows us to use the power of matrix operations to find solutions efficiently. In the given problem, we have two equations:

• \(3y = 6\)
• \(x + y = 7\)

To express this as a matrix equation, we define a coefficient matrix \(A\), a variable matrix \(\mathbf{x}\), and a constant matrix \(\mathbf{b}\). The coefficient matrix contains the coefficients of the variables \( x \) and \( y \), while the constant matrix comes from the right side of the equations.For this exercise, we have:\[A = \begin{bmatrix} 0 & 3 \ 1 & 1 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \ y \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 6 \ 7 \end{bmatrix}\]Thus, the matrix equation is \( A\mathbf{x} = \mathbf{b} \). This equation forms the basis for utilizing matrix methods, like finding the inverse to solve the system.

The determinant of a matrix tells us a lot about its properties, especially when dealing with square matrices like our coefficient matrix \(A\). It is a single number that can hint whether a matrix is invertible or not. In our case, the determinant of \( A \) is calculated as follows:\[\det(A) = (0)(1) - (3)(1) = -3\]Because \( \det(A) eq 0 \), we know that matrix \(A\) is invertible, meaning we can technically "divide" by \(A\) in our matrix equation to solve for the variable matrix \(\mathbf{x}\). An invertible matrix is essential for solving systems of equations using matrices. Anytime the determinant equals zero, a matrix lacks an inverse, and different methods might be required to solve the system.

Finding the inverse of a matrix is akin to finding a multiplicative counterpart that, when multiplied by the original matrix, results in the identity matrix. This concept is crucial in solving matrix equations like \( A\mathbf{x} = \mathbf{b} \). Once you have determined that a matrix is invertible, calculating its inverse follows.The inverse \(A^{-1}\) of a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\) is given by:\[A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]For our problem:\[A^{-1} = \frac{1}{-3} \begin{bmatrix} 1 & -3 \ -1 & 0 \end{bmatrix} = \begin{bmatrix} -\frac{1}{3} & 1 \ \frac{1}{3} & 0 \end{bmatrix}\]Having \(A^{-1}\) allows us to transform the original matrix equation to directly solve for \(\mathbf{x}\) by multiplying both sides of \(A\mathbf{x} = \mathbf{b}\) by \(A^{-1}\).

The solution of linear systems using matrices involves a few straightforward steps. Once you represent the equations as a matrix equation and find the inverse matrix (if the determinant is not zero), you approach the solution straightforwardly.In our exercise, after obtaining \(A^{-1}\), substitute it into the equation \(\mathbf{x} = A^{-1}\mathbf{b}\) to find \(\mathbf{x}\):\[\mathbf{x} = \begin{bmatrix} -\frac{1}{3} & 1 \ \frac{1}{3} & 0 \end{bmatrix} \begin{bmatrix} 6 \ 7 \end{bmatrix}\]Perform the multiplication:

• For \(x\), calculate: \(-\frac{1}{3} \cdot 6 + 1 \cdot 7 = -2 + 7 = 5\)
• For \(y\), calculate: \((\frac{1}{3} \cdot 6) + (0 \cdot 7) = 2 + 0 = 2 \)

Thus, the solution to the system \(3y = 6\) and \(x + y = 7\) is \(x = 5\) and \(y = 2\). This shows how matrices can simplify solving linear equations, providing a structured and reliable method.