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Chapter 3: Problem 74

Evaluate the determinants to verify the equation. $$\left|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)$$

Short Answer

Expert verified
The calculation validates that the given equation is correct. The determinant is indeed \((a - b)(b - c)(c - a)(a+b+c)\).

Step by step solution

01

Setup the matrix

First and foremost, start by setting up the given matrix as demonstrated below. \[ \begin{bmatrix} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{bmatrix} \]
02

Matrix simplification

To simplify the calculation of the determinant, subtract the second row from the third, leading to: \[ \begin{bmatrix} 1 & 1 & 1 \\ a & b & c \\ a^{3}-a & b^{3}-b & c^{3}-c \end{bmatrix} \] Now make use of the formula \(a^3 - a = a(a^2 - 1) = a(a - 1)(a + 1)\), we have: \[ \begin{bmatrix} 1 & 1 & 1 \\ a & b & c \\ a(a^{2} - 1) & b(b^{2} - 1) & c(c^{2} - 1) \end{bmatrix} = (a - b)(b - c)(c - a) * \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ a+b+c & a+b+c & a+b+c \end{bmatrix} \]
03

Calculating the determinant

The determinant of the given matrix can now be calculated. Applying the properties of determinants, namely that the determinant of a matrix with two equal rows is zero, we have the final expression. The determinant of the matrix equals to zero, therefore the expression equals to \((a - b)(b - c)(c - a)(a+b+c)\).

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Most popular questions from this chapter

Use a graphing utility or a computer software program with matrix capabilities and Cramer's Rule to solve for \(x_{1}\) if possible. $$\begin{aligned} 5 x_{1}-3 x_{2}+2 x_{3} &=2 \\ 2 x_{1}+2 x_{2}-3 x_{3} &=3 \\ x_{1}-7 x_{2}+8 x_{3} &=-4 \end{aligned}$$

Use Cramer's Rule to solve the system of linear equations, if possible. $$\begin{array}{l} 13 x_{1}-6 x_{2}=17 \\ 26 x_{1}-12 x_{2}=8 \end{array}$$

If \(A\) is an idempotent matrix \(\left(A^{2}=A\right),\) then prove that the determinant of \(A\) is either 0 or 1

Find an equation of the line passing through the given points. $$(-4,7),(2,4)$$

Determine whether the points are collinear. $$(-1,-3),(-4,7),(2,-13)$$
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