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
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.
We are provided with the following information:
Our objective is to determine the numerical value of \(\det(A)\), which is the determinant \(\begin{vmatrix}a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix}\).
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:
\[ \det(A) = abc \times (b-a)(c-a)(c-b) \]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.
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, \((b-a)(c-a)(c-b) = 2\).
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 \((b-a)(c-a)(c-b) = 2\).
\[ \det(A) = 6 \times 2 = 12 \]The value of the determinant is 12.
| 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 \(a, b, c\): \(\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} = (b-a)(c-a)(c-b)\). |
Determinants are fundamental in linear algebra and have various applications:
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