To solve this problem, let’s go through each part step-by-step.

Step 1: Set Up the Coefficient Matrix

For both systems, the coefficient matrix $A$ is: $A = \begin{bmatrix} -6 & -4 \\ -2 & 1 \end{bmatrix}$

Step 2: Find the Inverse of $A$

The inverse of a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by: $A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

For our matrix:

• $a = -6$
• $b = -4$
• $c = -2$
• $d = 1$

Now, calculate the determinant: $ad - bc = (-6)(1) - (-4)(-2) = -6 - 8 = -14$

Then, the inverse matrix $A^{-1}$ is: $A^{-1} = \frac{1}{-14} \begin{bmatrix} 1 & 4 \\ 2 & -6 \end{bmatrix} = \begin{bmatrix} -\frac{1}{14} & -\frac{4}{14} \\ -\frac{2}{14} & \frac{6}{14} \end{bmatrix} = \begin{bmatrix} -\frac{1}{14} & -\frac{2}{7} \\ -\frac{1}{7} & \frac{3}{7} \end{bmatrix}$

Step 3: Solve for Each System Using $A^{-1}$

We now use $A^{-1}$ to solve for the solutions by multiplying $A^{-1}$ with $\mathbf{b}$, where $\mathbf{b}$ represents the right-hand side of each system.

System (a):

For system (a), $\mathbf{b} = \begin{bmatrix} 3 \\ -1 \end{bmatrix}$.

Calculate $A^{-1} \mathbf{b}$: $A^{-1} \mathbf{b} = \begin{bmatrix} -\frac{1}{14} & -\frac{2}{7} \\ -\frac{1}{7} & \frac{3}{7} \end{bmatrix} \begin{bmatrix} 3 \\ -1 \end{bmatrix}$

Perform the matrix multiplication: $x = -\frac{1}{14} \cdot 3 + -\frac{2}{7} \cdot (-1) = -\frac{3}{14} + \frac{2}{7} = -\frac{3}{14} + \frac{4}{14} = \frac{1}{14}$ $y = -\frac{1}{7} \cdot 3 + \frac{3}{7} \cdot (-1) = -\frac{3}{7} - \frac{3}{7} = -\frac{6}{7}$

So, for system (a), $x = \frac{1}{14}$ and $y = -\frac{6}{7}$.

System (b):

For system (b), $\mathbf{b} = \begin{bmatrix} 2 \\ 3 \end{bmatrix}$.

Calculate $A^{-1} \mathbf{b}$: $A^{-1} \mathbf{b} = \begin{bmatrix} -\frac{1}{14} & -\frac{2}{7} \\ -\frac{1}{7} & \frac{3}{7} \end{bmatrix} \begin{bmatrix} 2 \\ 3 \end{bmatrix}$

Perform the matrix multiplication: $x = -\frac{1}{14} \cdot 2 + -\frac{2}{7} \cdot 3 = -\frac{2}{14} - \frac{6}{7} = -\frac{2}{14} - \frac{12}{14} = -\frac{14}{14} = -1$ $y = -\frac{1}{7} \cdot 2 + \frac{3}{7} \cdot 3 = -\frac{2}{7} + \frac{9}{7} = \frac{7}{7} = 1$

So, for system (b), $x = -1$ and $y = 1$.

Summary of Solutions

• Inverse Matrix $A^{-1}$: $A^{-1} = \begin{bmatrix} -\frac{1}{14} & -\frac{2}{7} \\ -\frac{1}{7} & \frac{3}{7} \end{bmatrix}$

• Solutions:

  • For system (a): $x = \frac{1}{14}$, $y = -\frac{6}{7}$
  • For system (b): $x = -1$, $y = 1$

Would you like a further breakdown or have any questions on these steps?

Here are 5 related questions for further exploration:

1. How would you verify the solution by substituting $x$ and $y$ back into the original equations?
2. What would happen if the determinant of $A$ were zero?
3. How would you approach finding the inverse of a $3 \times 3$ matrix?
4. Can a matrix with fractional entries have an integer solution?
5. How could Gaussian elimination be used to solve this system instead of using the matrix inverse?

Tip: Always check the determinant of a matrix before trying to find its inverse; if the determinant is zero, the matrix is singular and has no inverse.