Chapter 6: Problem 74
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^{3}+p x+q=0\), then the value of the determinant \(\left|\begin{array}{ccc}1+\alpha & 1 & 1 \\ 1 & 1+\beta & 1 \\ 1 & 1 & 1+\gamma\end{array}\right|\) is (A) \(p^{2}-2 q\) (B) \(3 p q\) (C) \(p-q\) (D) None of these
Short Answer
Expert verified
The determinant value is \(p-q\) (Option C).
Step by step solution
01
Expand the Determinant
The given determinant is \( \begin{vmatrix} 1+\alpha & 1 & 1 \ 1 & 1+\beta & 1 \ 1 & 1 & 1+\gamma \end{vmatrix} \). As per the properties of determinants, its expansion along the first row gives us the expression: \( (1+\alpha)\left| \begin{array}{cc} 1+\beta & 1 \ 1 & 1+\gamma \end{array} \right| - 1 \left| \begin{array}{cc} 1 & 1 \ 1 & 1+\gamma \end{array} \right| + 1 \left| \begin{array}{cc} 1 & 1+\beta \ 1 & 1 \end{array} \right| \).
02
Simplify Each Minor
Calculating the minors: - For \( (1+\alpha) \): \( \begin{vmatrix} 1+\beta & 1 \ 1 & 1+\gamma \end{vmatrix} = (1+\beta)(1+\gamma) - 1 = 1 + \beta + \gamma + \beta\gamma - 1 = \beta + \gamma + \beta \gamma \).- For the second minor: \( \begin{vmatrix} 1 & 1 \ 1 & 1+\gamma \end{vmatrix} = (1)(1+\gamma) - 1 = \gamma \).- For the third minor: \( \begin{vmatrix} 1 & 1+\beta \ 1 & 1 \end{vmatrix} = 1\cdot1 - 1\cdot(1+\beta) = -\beta \).
03
Substitute Back into the Determinant Formula
Plug the simplified minors back: \( (1+\alpha)(\beta + \gamma + \beta \gamma) - \gamma - \beta = \alpha(\beta + \gamma + \beta \gamma) + \beta + \gamma + \beta \gamma - \gamma - \beta \).Thus, it simplifies to \( \alpha \beta + \alpha \gamma + \alpha \beta \gamma + \beta \gamma \).
04
Use Vieta's Formulas for Roots of Polynomials
Recall Vieta's formulas for the roots of a cubic \(x^3 + px + q = 0\):- \( \alpha + \beta + \gamma = 0 \) (because there is no \( x^2 \) term), - \( \alpha \beta + \beta \gamma + \gamma \alpha = p \), - \( \alpha \beta \gamma = -q \).Substitute these into the simplified determinant result from the previous step: \( \alpha \beta + \alpha \gamma + \beta \gamma + \beta \gamma = p + \alpha \beta \gamma = p - q \).
05
Compare with the Given Options
The final expression for the determinant, derived using Vieta's formulas, is \( p - q \). Comparing with the given options, indeed, option (C) \( p-q \) matches the result.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vieta's Formulas
Vieta's Formulas offer a fascinating connection between the coefficients of a polynomial and the roots (solutions) of the polynomial equation.
They are especially handy with cubic equations like the one in your exercise:
They are especially handy with cubic equations like the one in your exercise:
- If you have a polynomial: \(x^3 + px + q = 0\), the sum of the roots \(\alpha + \beta + \gamma\) equals zero. This is because there's no \(x^2\) term in our equation, which would otherwise translate to the sum of the roots.
- The sum of the products of the roots taken two at a time, \(\alpha \beta + \beta \gamma + \gamma \alpha\), is equal to the coefficient \(p\).
- The product of the roots \(\alpha \beta \gamma\), multiplied by \(-1\), gives you the constant term \(q\).
Polynomial Roots
The roots of a polynomial, like those from the equation \(x^3 + px + q = 0\), are the values of \(x\) that make the polynomial evaluate to zero. In our context, they are represented by \(\alpha, \beta,\) and \(\gamma\).
Understanding the roots is crucial because:
Understanding the roots is crucial because:
- Each root corresponds to an \(x\) value where the graph of the polynomial crosses the \(x\)-axis.
- Roots provide insight into the behavior and characteristics of the polynomial. They indicate the turning points and the general shape or "wiggles" of its graph.
- Knowing the roots helps tackle other mathematical problems, like evaluating determinants or simplifying expressions, linking back to Vieta’s Formulas.
Matrix Expansion
Matrix expansion is a method to simplify determinants, particularly in matrices like the 3x3 one in your exercise. A determinant is a scalar value that represents the volume distortion when a basis is transformed by the matrix—and tells us if the matrix is invertible.
Here’s how the expansion process works:
Here’s how the expansion process works:
- Start by choosing a row or a column to expand along; in our exercise, the first row was selected.
- Multiply each element in that row or column by their corresponding cofactors (the signed minors).
- A minor is the determinant of the matrix that remains after removing the element's row and column.