If (a - b) (b - c) (c - a) = 2 and abc = 6, then what is the value of \(\begin{vmatrix}a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix}?\)
12
Evaluating a Determinant using the Vandermonde Property
This problem requires us to calculate the determinant of a specific matrix, given some conditions about the variables involved. The structure of the determinant points towards using properties related to Vandermonde determinants.
Understanding the Problem and Given Information
We are provided with the following information:
- The relationship \((a - b) (b - c) (c - a) = 2\).
- The product \(abc = 6\).
- The matrix whose determinant we need to find: \(A = \begin{pmatrix} a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{pmatrix}\).
Our objective is to determine the numerical value of \(\det(A)\), which is the determinant \(\begin{vmatrix}a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix}\).
Calculating the Determinant Value
Let's compute the determinant of the given matrix:
\[ \begin{vmatrix} a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix} \]We can observe that each column has a common factor. We can factor out \(a\) from the first column, \(b\) from the second column, and \(c\) from the third column. This gives us:
\[ = a \cdot b \cdot c \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} \]The remaining determinant, \(\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix}\), is a standard Vandermonde determinant with elements \(a, b,\) and \(c\). The general formula for a 3x3 Vandermonde determinant is \(\begin{vmatrix} 1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \end{vmatrix} = (y-x)(z-x)(z-y)\).
Applying this formula with \(x=a, y=b, z=c\), the Vandermonde determinant part evaluates to \((b-a)(c-a)(c-b)\).
So, the determinant of the original matrix is:
\[ \det(A) = abc \times (b-a)(c-a)(c-b) \]Connecting with the Given Condition
We are given the condition \((a - b) (b - c) (c - a) = 2\). Let's compare the terms in the Vandermonde expansion \((b-a)(c-a)(c-b)\) with the terms in the given condition.
- \((b-a)\) is the negative of \((a-b)\), i.e., \((b-a) = -(a-b)\).
- \((c-b)\) is the negative of \((b-c)\), i.e., \((c-b) = -(b-c)\).
- \((c-a)\) is one of the terms in the given expression.
Now let's rewrite \((b-a)(c-a)(c-b)\) using these relationships:
\[ (b-a)(c-a)(c-b) = (-(a-b)) \times (c-a) \times (-(b-c)) \] \[ = (-1) \times (-1) \times (a-b) \times (c-a) \times (b-c) \] \[ = (a-b)(c-a)(b-c) \]Since multiplication is commutative, the expression \((a-b)(c-a)(b-c)\) is the same as \((a-b)(b-c)(c-a)\).
We are given that \((a - b) (b - c) (c - a) = 2\).
Therefore, \((b-a)(c-a)(c-b) = 2\).
Performing the Final Calculation
Now we substitute the known values into the determinant expression we derived:
\[ \det(A) = abc \times (b-a)(c-a)(c-b) \]We are given \(abc = 6\) and we found that \((b-a)(c-a)(c-b) = 2\).
\[ \det(A) = 6 \times 2 = 12 \]Conclusion
The value of the determinant is 12.
Revision Table: Essential Determinant Concepts
| Concept | Explanation |
|---|---|
| Determinant Properties | Factoring out a constant from a column (or row) multiplies the determinant by that constant. The determinant is a scalar value derived from a square matrix. |
| Vandermonde Matrix | A square matrix with terms in a geometric progression in each row (or column). A standard form has columns \(1, x, x^2, \dots\). |
| Vandermonde Determinant Formula | For distinct \(x_1, x_2, \dots, x_n\), the determinant of a Vandermonde matrix with columns \(1, x_i, x_i^2, \dots, x_i^{n-1}\) is the product of all differences \((x_j - x_i)\) for \(j > i\). For 3x3 with \(a, b, c\): \(\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} = (b-a)(c-a)(c-b)\). |
Additional Information: Applications of Determinants
Determinants are fundamental in linear algebra and have various applications:
- Solving Linear Equations: Used in Cramer's rule to find unique solutions for systems of linear equations.
- Eigenvalues: Determinants are used in finding eigenvalues of a matrix by solving the characteristic equation \(\det(A - \lambda I) = 0\).
- Area and Volume: In geometry, the absolute value of a determinant represents the scale factor of volume/area transformation by the matrix, or the volume/area of a shape defined by the matrix's column vectors.
- Linear Dependence: The column vectors (or row vectors) of a square matrix are linearly dependent if and only if its determinant is zero.
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The determinant \(\left| {\begin{array}{*{20}{c}} {xp + y}&x&y\\ {yp + z}&y&z\\ 0&{xp + y}&{yp + z} \end{array}} \right| = 0,\) if
