1. **Problem Statement:** Solve each system of linear equations using the matrix method, i.e., express as $AX = B$ and find $X = A^{-1}B$ if $A$ is invertible. 2. **Matrix Method Formula:** For system $AX = B$, where $A$ is coefficient matrix, $X$ is variable vector, and $B$ is constants vector, solution is $X = A^{-1}B$ if $\det(A) \neq 0$. 3. **Important Rule:** If $\det(A) = 0$, the system may have no unique solution or infinitely many solutions. --- ### a. System: $$\begin{cases} x + 2y = 4 \\ 3x - 5y = 1 \end{cases}$$ Matrix form: $$A = \begin{bmatrix}1 & 2 \\ 3 & -5\end{bmatrix},\quad X = \begin{bmatrix}x \\ y\end{bmatrix},\quad B = \begin{bmatrix}4 \\ 1\end{bmatrix}$$ Calculate $\det(A) = 1 \times (-5) - 3 \times 2 = -5 - 6 = -11 \neq 0$. Find $A^{-1} = \frac{1}{-11} \begin{bmatrix} -5 & -2 \\ -3 & 1 \end{bmatrix}$. Multiply: $$X = A^{-1}B = \frac{1}{-11} \begin{bmatrix} -5 & -2 \\ -3 & 1 \end{bmatrix} \begin{bmatrix}4 \\ 1\end{bmatrix} = \frac{1}{-11} \begin{bmatrix} -5 \times 4 + (-2) \times 1 \\ -3 \times 4 + 1 \times 1 \end{bmatrix} = \frac{1}{-11} \begin{bmatrix} -20 - 2 \\ -12 + 1 \end{bmatrix} = \frac{1}{-11} \begin{bmatrix} -22 \\ -11 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$$ Answer: $x=2$, $y=1$. --- ### b. System: $$\begin{cases} 5x + y = 13 \\ 3x + 2y = 5 \end{cases}$$ Matrix form: $$A = \begin{bmatrix}5 & 1 \\ 3 & 2\end{bmatrix},\quad B = \begin{bmatrix}13 \\ 5\end{bmatrix}$$ Calculate $\det(A) = 5 \times 2 - 3 \times 1 = 10 - 3 = 7 \neq 0$. Find $A^{-1} = \frac{1}{7} \begin{bmatrix} 2 & -1 \\ -3 & 5 \end{bmatrix}$. Multiply: $$X = A^{-1}B = \frac{1}{7} \begin{bmatrix} 2 & -1 \\ -3 & 5 \end{bmatrix} \begin{bmatrix}13 \\ 5\end{bmatrix} = \frac{1}{7} \begin{bmatrix} 2 \times 13 - 1 \times 5 \\ -3 \times 13 + 5 \times 5 \end{bmatrix} = \frac{1}{7} \begin{bmatrix} 26 - 5 \\ -39 + 25 \end{bmatrix} = \frac{1}{7} \begin{bmatrix} 21 \\ -14 \end{bmatrix} = \begin{bmatrix} 3 \\ -2 \end{bmatrix}$$ Answer: $x=3$, $y=-2$. --- ### c. System: $$\begin{cases} 3x + 2y = -2 \\ x + 4y = 6 \end{cases}$$ Matrix form: $$A = \begin{bmatrix}3 & 2 \\ 1 & 4\end{bmatrix},\quad B = \begin{bmatrix}-2 \\ 6\end{bmatrix}$$ Calculate $\det(A) = 3 \times 4 - 1 \times 2 = 12 - 2 = 10 \neq 0$. Find $A^{-1} = \frac{1}{10} \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix}$. Multiply: $$X = A^{-1}B = \frac{1}{10} \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix} \begin{bmatrix}-2 \\ 6\end{bmatrix} = \frac{1}{10} \begin{bmatrix} 4 \times (-2) - 2 \times 6 \\ -1 \times (-2) + 3 \times 6 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} -8 - 12 \\ 2 + 18 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} -20 \\ 20 \end{bmatrix} = \begin{bmatrix} -2 \\ 2 \end{bmatrix}$$ Answer: $x=-2$, $y=2$. --- ### d. System: $$\begin{cases} 2x + 3y = 6 \\ x - y = \frac{1}{2} \end{cases}$$ Matrix form: $$A = \begin{bmatrix}2 & 3 \\ 1 & -1\end{bmatrix},\quad B = \begin{bmatrix}6 \\ \frac{1}{2}\end{bmatrix}$$ Calculate $\det(A) = 2 \times (-1) - 1 \times 3 = -2 - 3 = -5 \neq 0$. Find $A^{-1} = \frac{1}{-5} \begin{bmatrix} -1 & -3 \\ -1 & 2 \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 1 & 3 \\ 1 & -2 \end{bmatrix}$. Multiply: $$X = A^{-1}B = \frac{1}{5} \begin{bmatrix} 1 & 3 \\ 1 & -2 \end{bmatrix} \begin{bmatrix}6 \\ \frac{1}{2}\end{bmatrix} = \frac{1}{5} \begin{bmatrix} 1 \times 6 + 3 \times \frac{1}{2} \\ 1 \times 6 - 2 \times \frac{1}{2} \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 6 + 1.5 \\ 6 - 1 \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 7.5 \\ 5 \end{bmatrix} = \begin{bmatrix} \frac{7.5}{5} \\ 1 \end{bmatrix} = \begin{bmatrix} \frac{3}{2} \\ 1 \end{bmatrix}$$ Note: The problem states $x=\frac{5}{7}$ and $y=\frac{1}{7}$, so let's re-check calculations. Recalculate $A^{-1}$ carefully: $$A^{-1} = \frac{1}{-5} \begin{bmatrix} -1 & -3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} \frac{1}{5} & \frac{3}{5} \\ \frac{1}{5} & -\frac{2}{5} \end{bmatrix}$$ Multiply: $$X = A^{-1}B = \begin{bmatrix} \frac{1}{5} & \frac{3}{5} \\ \frac{1}{5} & -\frac{2}{5} \end{bmatrix} \begin{bmatrix}6 \\ \frac{1}{2}\end{bmatrix} = \begin{bmatrix} \frac{6}{5} + \frac{3}{5} \times \frac{1}{2} \\ \frac{6}{5} - \frac{2}{5} \times \frac{1}{2} \end{bmatrix} = \begin{bmatrix} \frac{6}{5} + \frac{3}{10} \\ \frac{6}{5} - \frac{1}{5} \end{bmatrix} = \begin{bmatrix} \frac{12}{10} + \frac{3}{10} \\ \frac{6}{5} - \frac{1}{5} \end{bmatrix} = \begin{bmatrix} \frac{15}{10} \\ \frac{5}{5} \end{bmatrix} = \begin{bmatrix} 1.5 \\ 1 \end{bmatrix}$$ This is $x=1.5$, $y=1$, not matching given answer. Check original system: $$2x + 3y = 6$$ $$x - y = \frac{1}{2}$$ Try substitution: $$x = y + \frac{1}{2}$$ Substitute into first: $$2(y + \frac{1}{2}) + 3y = 6 \Rightarrow 2y + 1 + 3y = 6 \Rightarrow 5y = 5 \Rightarrow y = 1$$ Then: $$x = 1 + \frac{1}{2} = \frac{3}{2}$$ So correct solution is $x=\frac{3}{2}$, $y=1$, not $\frac{5}{7}$ and $\frac{1}{7}$. Answer: $x=\frac{3}{2}$, $y=1$. --- ### e. System: $$\begin{cases} 4x + 3y = 11 \\ x - 3y = -1 \end{cases}$$ Matrix form: $$A = \begin{bmatrix}4 & 3 \\ 1 & -3\end{bmatrix},\quad B = \begin{bmatrix}11 \\ -1\end{bmatrix}$$ Calculate $\det(A) = 4 \times (-3) - 1 \times 3 = -12 - 3 = -15 \neq 0$. Find $A^{-1} = \frac{1}{-15} \begin{bmatrix} -3 & -3 \\ -1 & 4 \end{bmatrix} = \begin{bmatrix} \frac{1}{5} & \frac{1}{5} \\ \frac{1}{15} & -\frac{4}{15} \end{bmatrix}$. Multiply: $$X = A^{-1}B = \begin{bmatrix} \frac{1}{5} & \frac{1}{5} \\ \frac{1}{15} & -\frac{4}{15} \end{bmatrix} \begin{bmatrix}11 \\ -1\end{bmatrix} = \begin{bmatrix} \frac{11}{5} - \frac{1}{5} \\ \frac{11}{15} + \frac{4}{15} \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$$ Answer: $x=2$, $y=1$. --- ### f. System: $$\begin{cases} 2x + 3y = 8 \\ 4x + 6y = 20 \end{cases}$$ Matrix form: $$A = \begin{bmatrix}2 & 3 \\ 4 & 6\end{bmatrix},\quad B = \begin{bmatrix}8 \\ 20\end{bmatrix}$$ Calculate $\det(A) = 2 \times 6 - 4 \times 3 = 12 - 12 = 0$. Since $\det(A) = 0$, matrix $A$ is singular, so no unique solution. Check if system is consistent: Multiply first equation by 2: $$2(2x + 3y) = 2 \times 8 \Rightarrow 4x + 6y = 16$$ But second equation is: $$4x + 6y = 20$$ Contradiction means no solution. Answer: No solution (inconsistent system). --- **Summary:** - a: $x=2$, $y=1$ - b: $x=3$, $y=-2$ - c: $x=-2$, $y=2$ - d: $x=\frac{3}{2}$, $y=1$ - e: $x=2$, $y=1$ - f: No solution