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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 11: Q. 60 (page 741)

Show that $|\begin{array}{l} x^{2} & x & 1 \\ y^{2} & y & 1 \\ z^{2} & z & 1 \end{array}| = (y - z) (x - y) (x - z)$

Short Answer

Expert verified

The given expression is proved.

Step by step solution

01

Step 1. Given Information

We are given a determinant and we have to prove,

$|\begin{array}{l} x^{2} & x & 1 \\ y^{2} & y & 1 \\ z^{2} & z & 1 \end{array}| = (y - z) (x - y) (x - z)$

02

Step 2. Proving the expression

Applying the row transformation, we get

$R_{2} 鈫� R_{2} - R_{1} R_{3} 鈫� R_{3} - R_{1}$

$= |\begin{matrix} x^{2} & x & 1 \\ y^{2} - x^{2} & y - x & 0 \\ z^{2} - x^{2} & z - x & 0 \end{matrix}| = (y - x) (z - x) |\begin{matrix} x^{2} & x & 1 \\ y + x & 1 & 0 \\ z + x & 1 & 0 \end{matrix}|$

Expanding the determinant, we get

localid="1647445639863" $= (y - x) (z - x) 路 ((y + x) 路 1 - (z + x) 路 1) = (x - y) (y - z) (z - x)$

Hence Proved.

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Most popular questions from this chapter

Solve each system of equations. If the system has no solution, say that it is inconsistent.

$\{\begin{matrix} 3 x - y = 7 \\ 9 x - 3 y = 21 \end{matrix}$

In Problems 5鈥�24, graph each equation of the system. Then solve the system to find the points of intersection.

$x^{2} = y; x y = 1$

Solve each system of equations. If the system has no solution, say that it is inconsistent.

$2 x + y - 3 z = 0 - 2 x + 2 y + z = - 7 3 x - 4 y - 3 z = 7$

Verify that the values of the variables listed are solutions of the system of equations.

$\{\begin{matrix} 3 x + 2 y = 2 \\ x - 7 y = - 30 \end{matrix} x = - 2, y = 4; (- 2, 4)$

Graph each inequality by hand.

$x^{2} + y^{2} > 1$

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