We will solve the system of equations using both the Inverse Method and Cramer's Rule.

The system of equations:

1. $2x + 5y = 1$
2. $x - 3y = 4$

We can rewrite the system in matrix form as:

$\begin{pmatrix} 2 & 5 \\ 1 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}$

The matrix equation can be expressed as $A \mathbf{x} = \mathbf{b}$, where:

• $A = \begin{pmatrix} 2 & 5 \\ 1 & -3 \end{pmatrix}$
• $\mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}$
• $\mathbf{b} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}$

1. Inverse Method:

We first find the inverse of the matrix $A$.

The inverse of a 2x2 matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is given by: $A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ where $\text{det}(A) = ad - bc$.

For our matrix $A = \begin{pmatrix} 2 & 5 \\ 1 & -3 \end{pmatrix}$:

• $a = 2$, $b = 5$, $c = 1$, $d = -3$
• $\text{det}(A) = (2)(-3) - (1)(5) = -6 - 5 = -11$

Thus, the inverse of $A$ is:

= \begin{pmatrix} \frac{3}{11} & \frac{5}{11} \\ \frac{1}{11} & -\frac{2}{11} \end{pmatrix}$$ Now, to find $$\mathbf{x}$$, we multiply the inverse of $$A$$ by $$\mathbf{b}$$: $$\mathbf{x} = A^{-1} \mathbf{b} = \begin{pmatrix} \frac{3}{11} & \frac{5}{11} \\ \frac{1}{11} & -\frac{2}{11} \end{pmatrix} \begin{pmatrix} 1 \\ 4 \end{pmatrix}$$ Performing the matrix multiplication: $$x = \frac{3}{11} \times 1 + \frac{5}{11} \times 4 = \frac{3}{11} + \frac{20}{11} = \frac{23}{11}$$ $$y = \frac{1}{11} \times 1 + \left(-\frac{2}{11}\right) \times 4 = \frac{1}{11} - \frac{8}{11} = -\frac{7}{11}$$ So the solution is: $$x = \frac{23}{11}, \quad y = -\frac{7}{11}$$ #### 2. **Cramer's Rule:** Cramer's Rule states that for a system of equations $$A \mathbf{x} = \mathbf{b}$$, the solution is given by: $$x = \frac{\text{det}(A_x)}{\text{det}(A)}, \quad y = \frac{\text{det}(A_y)}{\text{det}(A)}$$ where $$A_x$$ and $$A_y$$ are the matrices formed by replacing the corresponding columns of $$A$$ with $$\mathbf{b}$$. The determinant of $$A$$ is $$\text{det}(A) = -11$$ (as calculated earlier). Now, we compute $$A_x$$ and $$A_y$$: - $$A_x = \begin{pmatrix} 1 & 5 \\ 4 & -3 \end{pmatrix}$$ $$\text{det}(A_x) = (1)(-3) - (4)(5) = -3 - 20 = -23$$ - $$A_y = \begin{pmatrix} 2 & 1 \\ 1 & 4 \end{pmatrix}$$ $$\text{det}(A_y) = (2)(4) - (1)(1) = 8 - 1 = 7$$ Now, using Cramer's rule: $$x = \frac{\text{det}(A_x)}{\text{det}(A)} = \frac{-23}{-11} = \frac{23}{11}$$ $$y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{7}{-11} = -\frac{7}{11}$$ Thus, the solution is: $$x = \frac{23}{11}, \quad y = -\frac{7}{11}$$ ### Final Answer: $$x = \frac{23}{11}, \quad y = -\frac{7}{11}$$ Would you like more details or explanations on any part? Here are 5 related questions: 1. How does Cramer's Rule extend to larger systems (3x3 or 4x4)? 2. What are the conditions for the existence of an inverse matrix? 3. How is the determinant used to test if a system has a unique solution? 4. Can a matrix with a determinant of zero be solved by inverse methods? 5. How does Cramer's Rule compare to Gaussian elimination? **Tip**: If the determinant of the matrix is zero, the system has no unique solution, meaning it is either inconsistent or has infinitely many solutions.