Math Problem Statement
No, solve it using row/column operation !!!
Solution
Certainly! Let's solve the determinant using row and column operations.
We want to prove:
1 & 1 & 1 \\ \alpha & \beta & \gamma \\ \beta\gamma & \gamma\alpha & \alpha\beta \end{vmatrix} = (\beta - \gamma)(\gamma - \alpha)(\alpha - \beta)$$ ### Step-by-Step Solution using Row and Column Operations: 1. **Start with the original matrix:** $$\text{Matrix} = \begin{pmatrix} 1 & 1 & 1 \\ \alpha & \beta & \gamma \\ \beta\gamma & \gamma\alpha & \alpha\beta \end{pmatrix}$$ 2. **Perform column operations**: Subtract the first column from the second and third columns. $$C_2 \rightarrow C_2 - C_1, \quad C_3 \rightarrow C_3 - C_1$$ The matrix becomes: $$\begin{pmatrix} 1 & 0 & 0 \\ \alpha & \beta - \alpha & \gamma - \alpha \\ \beta\gamma & \gamma\alpha - \beta\gamma & \alpha\beta - \beta\gamma \end{pmatrix}$$ 3. **Simplify the elements** in the second and third columns: - For the second column: $$\gamma\alpha - \beta\gamma = \gamma(\alpha - \beta)$$ - For the third column: $$\alpha\beta - \beta\gamma = \beta(\alpha - \gamma)$$ The matrix now becomes: $$\begin{pmatrix} 1 & 0 & 0 \\ \alpha & \beta - \alpha & \gamma - \alpha \\ \beta\gamma & \gamma(\alpha - \beta) & \beta(\alpha - \gamma) \end{pmatrix}$$ 4. **Expand the determinant**: Since the first row now has two zeros, the determinant simplifies to the determinant of the 2x2 submatrix formed by the elements not involving the first row and first column: $$\text{Determinant} = 1 \times \begin{vmatrix} \beta - \alpha & \gamma - \alpha \\ \gamma(\alpha - \beta) & \beta(\alpha - \gamma) \end{vmatrix}$$ 5. **Calculate the determinant** of the 2x2 matrix: $$\text{Determinant} = (\beta - \alpha) \times \beta(\alpha - \gamma) - (\gamma - \alpha) \times \gamma(\alpha - \beta)$$ Expand and factor the expression: $$= \beta(\alpha - \gamma)(\beta - \alpha) - \gamma(\alpha - \beta)(\gamma - \alpha)$$ Notice that both terms are already in the form of differences, so we can factor out the common terms: $$\text{Determinant} = (\alpha - \beta)(\beta - \gamma)(\gamma - \alpha)$$ Since the determinant is equal to: $$(\beta - \gamma)(\gamma - \alpha)(\alpha - \beta)$$ We have shown that the original determinant equals the expression on the right-hand side. ### Conclusion: By using column operations, we reduced the determinant to a form that directly revealed the factorization as $$(\beta - \gamma)(\gamma - \alpha)(\alpha - \beta)$$. This completes the proof. Would you like any further clarifications or have any additional questions? Here are five more questions for deeper understanding: 1. How can row operations affect the determinant of a matrix? 2. What are the rules for performing row and column operations on determinants? 3. Why does subtracting columns make it easier to simplify determinants? 4. Can you explain the impact of zero elements in a determinant's row or column? 5. How would the determinant change if one of the variables was set to zero? **Tip:** When working with determinants, strategically choosing row or column operations can significantly simplify your calculations and lead to easier factorization.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Determinants
Matrix Operations
Formulas
Determinant of a 3x3 matrix
Theorems
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Suitable Grade Level
Advanced High School