Q. 51 In Problems 41鈥� 60, use the in... [FREE SOLUTION]

91影视

Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 11: Q. 51 (page 756)

In Problems 41鈥� 60, use the inverses found in Problems 31鈥�40 to solve each system of equations
$\{\begin{array}{r} 2 x + y = \frac{7}{a} \\ a x + a y = 5 \end{array} a 鈮� 0$

Short Answer

Expert verified

The solution of the matrix is $x = \frac{2}{a} and y = \frac{3}{a}$ .

Step by step solution

Step 1. Given information

A matrix is given as,

$\{\begin{array}{l} 2 x + y = \frac{7}{a} \\ a x + a y = 5 \end{array} a 鈮� 0$

We need to find the solution of the matrix using inverse of the matrix.

Step 2. Inverse of the matrix.

The inverse of the matrix can be found by :

$[A 鈭� I_{2}] = [\begin{array}{l} 2 & 1 & 1 & 0 \\ a & a & 0 & 1 \end{array}] R_{1} = \frac{1}{2} r_{1} and then R_{2} = r_{2} - a r_{1} [\begin{array}{l} 2 & 1 & 1 & 0 \\ a & a & 0 & 1 \end{array}] 鈫� [\begin{array}{l} 1 & \frac{1}{2} & \frac{1}{2} & 0 \\ a & a & 0 & 1 \end{array}] 鈫� [\begin{matrix} 1 & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & \frac{a}{2} & - \frac{a}{2} & 1 \end{matrix}] R_{2} = \frac{2}{a} r_{2} and R_{1} = r_{1} - \frac{1}{2} r_{2} . [\begin{matrix} 1 & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & \frac{a}{2} & - \frac{a}{2} & 1 \end{matrix}] 鈫� [\begin{matrix} 1 & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 1 & - 1 & \frac{2}{a} \end{matrix}] 鈫� [\begin{matrix} 1 & 0 & 1 & - \frac{1}{a} \\ 0 & 1 & - 1 & \frac{2}{a} \end{matrix}] T h e r e f o r e, A^{- 1} = [\begin{matrix} 1 & - \frac{1}{a} \\ - 1 & \frac{2}{a} \end{matrix}]$

Step 3. Solution of the matrix

We know,

$X = A^{- 1} B$

Therefore,

$[\begin{array}{l} x \\ y \end{array}] = [\begin{matrix} 1 & - \frac{1}{a} \\ - 1 & \frac{2}{a} \end{matrix}] 路 [\begin{array}{l} \frac{7}{a} \\ 5 \end{array}] = [\begin{matrix} 1 路 \frac{7}{a} + - \frac{1}{a} 路 5 \\ - 1 路 \frac{7}{a} + \frac{2}{a} 路 5 \end{matrix}] = [\begin{matrix} \frac{7}{a} - \frac{5}{a} \\ - \frac{7}{a} + \frac{10}{a} \end{matrix}] = [\begin{array}{l} \frac{2}{a} \\ \frac{3}{a} \end{array}] T h e r e f o r e, x = \frac{2}{a} and y = \frac{3}{a}$

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91影视

In Problems 41鈥� 60, use the inverses found in Problems 31鈥�40 to solve each system of equations
$\{\begin{array}{r} 2 x + y = \frac{7}{a} \\ a x + a y = 5 \end{array} a 鈮� 0$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Inverse of the matrix.

Step 3. Solution of the matrix

One App. One Place for Learning.

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91影视

In Problems 41鈥� 60, use the inverses found in Problems 31鈥�40 to solve each system of equations2x+y=7aax+ay=5a鈮�0

Short Answer

Step by step solution

Step 1. Given information

Step 2. Inverse of the matrix.

Step 3. Solution of the matrix

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Theoretical and Mathematical Physics

Applied Mathematics

Decision Maths

Probability and Statistics

Statistics

Study anywhere. Anytime. Across all devices.

In Problems 41鈥� 60, use the inverses found in Problems 31鈥�40 to solve each system of equations
$\{\begin{array}{r} 2 x + y = \frac{7}{a} \\ a x + a y = 5 \end{array} a 鈮� 0$