We first identify our coefficient matrix \(A\) and constant vector \(\mathbf{b}\text{:}\)
\begin{equation*}
A = \begin{bmatrix} 2 \amp -1 \\ 3 \amp -5 \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 5 \\ 11 \end{bmatrix}
\end{equation*}
Step 1: Compute \(\det(A)\)
The determinant of the coefficient matrix is:
\begin{equation*}
\det(A) = (2)(-5) - (-1)(3) = -10 + 3 = -7
\end{equation*}
Since \(\det(A) \neq 0\text{,}\) the matrix \(A\) is invertible and Cramer’s Rule applies.
Step 2: Form \(A_1(\mathbf{b})\) and compute \(x_1\)
To find \(x_1\text{,}\) we replace the first column of \(A\) with \(\mathbf{b}\text{:}\)
\begin{equation*}
A_1(\mathbf{b}) = \begin{bmatrix} 5 \amp -1 \\ 11 \amp -5 \end{bmatrix}
\end{equation*}
The determinant is:
\begin{equation*}
\det(A_1(\mathbf{b})) = (5)(-5) - (-1)(11) = -25 + 11 = -14
\end{equation*}
Therefore, by Cramer’s Rule:
\begin{equation*}
x_1 = \frac{\det(A_1(\mathbf{b}))}{\det(A)} = \frac{-14}{-7} = 2
\end{equation*}
Step 3: Form \(A_2(\mathbf{b})\) and compute \(x_2\)
To find \(x_2\text{,}\) we replace the second column of \(A\) with \(\mathbf{b}\text{:}\)
\begin{equation*}
A_2(\mathbf{b}) = \begin{bmatrix} 2 \amp 5 \\ 3 \amp 11 \end{bmatrix}
\end{equation*}
The determinant is:
\begin{equation*}
\det(A_2(\mathbf{b})) = (2)(11) - (5)(3) = 22 - 15 = 7
\end{equation*}
Therefore, by Cramer’s Rule:
\begin{equation*}
x_2 = \frac{\det(A_2(\mathbf{b}))}{\det(A)} = \frac{7}{-7} = -1
\end{equation*}
Final Answer: The solution to the system is
\(x_1 = 2\) and
\(x_2 = -1\text{.}\)
We can verify this solution by substituting back into the original equations:
\begin{align*}
2(2) - (-1) \amp = 4 + 1 = 5 \quad \checkmark\\
3(2) - 5(-1) \amp = 6 + 5 = 11 \quad \checkmark
\end{align*}
