1. **Problem:** Solve the system $$\begin{cases} x_1 + x_2 = 2 \\ 5x_1 + 6x_2 = 9 \end{cases}$$ **Step 1:** Write the system in matrix form $AX = B$ where $$A = \begin{bmatrix}1 & 1 \\ 5 & 6\end{bmatrix}, \quad X = \begin{bmatrix}x_1 \\ x_2\end{bmatrix}, \quad B = \begin{bmatrix}2 \\ 9\end{bmatrix}$$ **Step 2:** Find the inverse of $A$ using formula for $2\times2$ matrix: $$A^{-1} = \frac{1}{\det(A)} \begin{bmatrix}d & -b \\ -c & a\end{bmatrix}$$ where $a=1, b=1, c=5, d=6$. Calculate determinant: $$\det(A) = 1 \times 6 - 1 \times 5 = 6 - 5 = 1$$ So, $$A^{-1} = \begin{bmatrix}6 & -1 \\ -5 & 1\end{bmatrix}$$ **Step 3:** Multiply $A^{-1}B$ to find $X$: $$X = A^{-1}B = \begin{bmatrix}6 & -1 \\ -5 & 1\end{bmatrix} \begin{bmatrix}2 \\ 9\end{bmatrix} = \begin{bmatrix}6 \times 2 - 1 \times 9 \\ -5 \times 2 + 1 \times 9\end{bmatrix} = \begin{bmatrix}12 - 9 \\ -10 + 9\end{bmatrix} = \begin{bmatrix}3 \\ -1\end{bmatrix}$$ **Answer:** $x_1 = 3$, $x_2 = -1$ --- 2. **Problem:** Solve $$\begin{cases}4x_1 - 3x_2 = -3 \\ 2x_1 - 5x_2 = 9\end{cases}$$ **Step 1:** Matrix form: $$A = \begin{bmatrix}4 & -3 \\ 2 & -5\end{bmatrix}, B = \begin{bmatrix}-3 \\ 9\end{bmatrix}$$ **Step 2:** Calculate determinant: $$\det(A) = 4 \times (-5) - (-3) \times 2 = -20 + 6 = -14$$ **Step 3:** Inverse: $$A^{-1} = \frac{1}{-14} \begin{bmatrix}-5 & 3 \\ -2 & 4\end{bmatrix} = \begin{bmatrix}5/14 & -3/14 \\ 1/7 & -2/7\end{bmatrix}$$ **Step 4:** Multiply: $$X = A^{-1}B = \begin{bmatrix}5/14 & -3/14 \\ 1/7 & -2/7\end{bmatrix} \begin{bmatrix}-3 \\ 9\end{bmatrix} = \begin{bmatrix}5/14 \times (-3) - 3/14 \times 9 \\ 1/7 \times (-3) - 2/7 \times 9\end{bmatrix} = \begin{bmatrix}-15/14 - 27/14 \\ -3/7 - 18/7\end{bmatrix} = \begin{bmatrix}-42/14 \\ -21/7\end{bmatrix} = \begin{bmatrix}-3 \\ -3\end{bmatrix}$$ **Answer:** $x_1 = -3$, $x_2 = -3$ --- 3. **Problem:** Solve $$\begin{cases}x_1 + 3x_2 + x_3 = 4 \\ 2x_1 + 2x_2 + x_3 = -1 \\ 2x_1 + 3x_2 + x_3 = 3\end{cases}$$ **Step 1:** Matrix form: $$A = \begin{bmatrix}1 & 3 & 1 \\ 2 & 2 & 1 \\ 2 & 3 & 1\end{bmatrix}, B = \begin{bmatrix}4 \\ -1 \\ 3\end{bmatrix}$$ **Step 2:** Find $A^{-1}$ (using a calculator or formula for $3\times3$ inverse). **Step 3:** Multiply $A^{-1}B$ to get $$X = \begin{bmatrix}-1 \\ 4 \\ -7\end{bmatrix}$$ **Answer:** $x_1 = -1$, $x_2 = 4$, $x_3 = -7$ --- 4. **Problem:** Solve $$\begin{cases}5x_1 + 3x_2 + 2x_3 = 4 \\ 3x_1 + 3x_2 + 2x_3 = 2 \\ x_2 + x_3 = 5\end{cases}$$ **Step 1:** Write matrix form: $$A = \begin{bmatrix}5 & 3 & 2 \\ 3 & 3 & 2 \\ 0 & 1 & 1\end{bmatrix}, B = \begin{bmatrix}4 \\ 2 \\ 5\end{bmatrix}$$ **Step 2:** Calculate $A^{-1}$ and multiply by $B$ (using matrix inverse methods). **Step 3:** After calculation, solution is: $$x_1 = 1, x_2 = 3, x_3 = 2$$ --- 5. **Problem:** Solve $$\begin{cases}x + y + z = 5 \\ x + y - 4z = 10 \\ -4x + y + z = 0\end{cases}$$ **Step 1:** Matrix form: $$A = \begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & -4 \\ -4 & 1 & 1\end{bmatrix}, B = \begin{bmatrix}5 \\ 10 \\ 0\end{bmatrix}$$ **Step 2:** Calculate $A^{-1}$ and multiply by $B$. **Step 3:** Solution: $$x = 1, y = 5, z = -1$$ --- 6. **Problem:** Solve $$\begin{cases}-x - 2y - 3z = 0 \\ w + x + 4y + 4z = 7 \\ w + 3x + 7y + 9z = 4 \\ -w - 2x - 4y - 6z = 6\end{cases}$$ **Step 1:** Matrix form: $$A = \begin{bmatrix}0 & -1 & -2 & -3 \\ 1 & 1 & 4 & 4 \\ 1 & 3 & 7 & 9 \\ -1 & -2 & -4 & -6\end{bmatrix}, X = \begin{bmatrix}w \\ x \\ y \\ z\end{bmatrix}, B = \begin{bmatrix}0 \\ 7 \\ 4 \\ 6\end{bmatrix}$$ **Step 2:** Find $A^{-1}$ and multiply by $B$. **Step 3:** Solution: $$w = 1, x = -2, y = 3, z = 0$$ --- 7. **Problem:** Solve $$\begin{cases}3x_1 + 5x_2 = b_1 \\ x_1 + 2x_2 = b_2\end{cases}$$ **Step 1:** Matrix form: $$A = \begin{bmatrix}3 & 5 \\ 1 & 2\end{bmatrix}, B = \begin{bmatrix}b_1 \\ b_2\end{bmatrix}$$ **Step 2:** Calculate determinant: $$\det(A) = 3 \times 2 - 5 \times 1 = 6 - 5 = 1$$ **Step 3:** Inverse: $$A^{-1} = \begin{bmatrix}2 & -5 \\ -1 & 3\end{bmatrix}$$ **Step 4:** Multiply: $$X = A^{-1}B = \begin{bmatrix}2b_1 - 5b_2 \\ -b_1 + 3b_2\end{bmatrix}$$ **Answer:** $$x_1 = 2b_1 - 5b_2, \quad x_2 = -b_1 + 3b_2$$