To solve the system of linear equations using Cramer's rule, follow these steps:
Step 1: Define the Coefficient Matrix and Its Determinant
The given system is:
[
\begin{cases}
ax + by = e \
cx + dy = f
\end{cases}
]
The coefficient matrix (D) is:
[
D = \begin{vmatrix} a & b \ c & d \end{vmatrix} = ad - bc
]
Condition: For a unique solution, (D \neq 0).
Step 2: Compute Determinants for (x) and (y)
-
Determinant (D_x) (replace the first column with constants ([e, f])):
[ D_x = \begin{vmatrix} e & b \ f & d \end{vmatrix} = ed - bf ] -
Determinant (D_y) (replace the second column with constants ([e, f])):
[ D_y = \begin{vmatrix} a & e \ c & f \end{vmatrix} = af - ce ]
Step 3: Solve for (x) and (y)
[ x = \frac{D_x}{D} = \frac{ed - bf}{ad - bc} \quad \text{and} \quad y = \frac{D_y}{D} = \frac{af - ce}{ad - bc} ]
Final Answer:
[
\boxed{x = \frac{ed - bf}{ad - bc}, \quad y = \frac{af - ce}{ad - bc}}
] (if (ad - bc \neq 0))
Or, in boxed form for the values:
If we need to present as separate boxes (though usually combined):
(\boxed{\frac{ed - bf}{ad - bc}}) for (x) and (\boxed{\frac{af - ce}{ad - bc}}) for (y).
But per standard notation, the combined solution is:
[
\boxed{\left( \frac{ed - bf}{ad - bc}, \frac{af - ce}{ad - bc} \right)}
]
However, if the problem expects individual values, then:
(x = \boxed{\frac{ed - bf}{ad - bc}}) and (y = \boxed{\frac{af - ce}{ad - bc}}).
Assuming the latter is needed, the answer is:
(\boxed{\frac{ed - bf}{ad - bc}}) (for (x)) and (\boxed{\frac{af - ce}{ad - bc}}) (for (y)).
But since the user might want both in one line, the combined box is acceptable. Let's go with the combined form as it's a solution pair:
(\boxed{\left( \frac{ed - bf}{ad - bc}, \frac{af - ce}{ad - bc} \right)})
Alternatively, if separate boxes are required:
First box: (\boxed{\frac{ed - bf}{ad - bc}})
Second box: (\boxed{\frac{af - ce}{ad - bc}})
But given the problem statement, probably the pair is expected. So the final answer is:
(\boxed{\left( \frac{ed - bf}{ad - bc}, \frac{af - ce}{ad - bc} \right)})
(Note: If the system has no solution or infinitely many solutions, (D = 0), but the question likely assumes a unique solution.)
Final Answer
(\boxed{\left( \dfrac{ed - bf}{ad - bc}, \dfrac{af - ce}{ad - bc} \right)})
作者声明:本文包含人工智能生成内容。
