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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. 53 (page 741)

In Problems 51鈥�56, solve for x,
$|\begin{array}{r} x & 1 & 1 \\ 4 & 3 & 2 \\ - 1 & 2 & 5 \end{array}| = 2$

Short Answer

Expert verified

The value of xis $\frac{13}{11}$ .

Step by step solution

01

Step 1. Given Information

We are given,

$|\begin{array}{r} x & 1 & 1 \\ 4 & 3 & 2 \\ - 1 & 2 & 5 \end{array}| = 2$

We have to find the value of x.

02

Step 2. Solving the determinant

Solving the determinant, we get

$x |\begin{array}{l} 3 & 2 \\ 2 & 5 \end{array}| - 1 |\begin{matrix} 4 & 2 \\ - 1 & 5 \end{matrix}| + 1 |\begin{matrix} 4 & 3 \\ - 1 & 2 \end{matrix}| = 2 x (3 路 5 - 2 路 2) - 1 (4 路 5 - 2 (- 1)) + 1 (4 路 2 - 3 (- 1)) = 2 x (15 - 4) - 1 (20 + 2) + 1 (8 + 3) = 2 11 x - 11 = 2 11 x = 13$

Dividing by $11$ , we get

$x = \frac{13}{11}$

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Most popular questions from this chapter

Graph the equation: $y^{2} = x^{2} - 1$

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