Chapter 4: Q. 4.86 (page 441)

Solve each system of equations using a matrix: $\{\begin{cases} 3 x + 4 y - 3 z = - 2 \\ 2 x + 3 y - z = - 12 \\ x + y - 2 z = 6 \end{cases}$

Short Answer

Expert verified

The system of linear equations doesn't have any solution.

Step by step solution

Step 1. Given information.

Consider the given system of equations,

$\{\begin{cases} 3 x + 4 y - 3 z = - 2 \\ 2 x + 3 y - z = - 12 \\ x + y - 2 z = 6 \end{cases}$

Step 2. Write in augmented form.

The augmented matrix for the given system of equations is

$[\begin{array}{cccr} 3 & 4 & - 3 & - 2 \\ 2 & 3 & - 1 & - 12 \\ 1 & 1 & - 2 & 6 \end{array}]$

Step 3. Apply row operations.

Apply $\frac{R_{1}}{3} \to R_{1}$ and $R_{2} - 2 \times R_{1} \to R_{2}$ ,

$[\begin{array}{cccr} 1 & \frac{4}{3} & - 1 & - \frac{2}{3} \\ 0 & \frac{1}{3} & 1 & - \frac{32}{3} \\ 1 & 1 & - 2 & 6 \end{array}]$

Apply $R_{3} - R_{1} \to R_{3}$ and $R_{2} \times 3 \to R_{2}$ ,

localid="1658938105519" $[\begin{array}{cccr} 1 & \frac{4}{3} & - 1 & - \frac{2}{3} \\ 0 & 1 & 3 & - 32 \\ 0 & - \frac{1}{3} & - 1 & \frac{20}{3} \end{array}]$

Apply $R_{3} + \frac{1}{3} \times R_{2} \to R_{3}$ ,

$[\begin{array}{cccr} 1 & \frac{4}{3} & - 1 & - \frac{2}{3} \\ 0 & 1 & 3 & - 32 \\ 0 & 0 & 0 & - 4 \end{array}]$

Apply $\frac{R_{3}}{- 4} \to R_{3}$ ,

$[\begin{array}{cccr} 1 & \frac{4}{3} & - 1 & - \frac{2}{3} \\ 0 & 1 & 3 & - 32 \\ 0 & 0 & 0 & 1 \end{array}]$

Now, the matrix is in row-echelon form.

Step 4. Write in system of equations.

Writing the corresponding system of equations,

$x + \frac{4}{3} y - z = - \frac{2}{3} . . . . . . (i) y + 3 z = - 32 . . . . . . (i i) 0 = 1 . . . . . . (i i i)$

As equation (iii) is a false statement.

Therefore, it is not possible to solve and is an inconsistent system.

Hence, the system of linear equations has no solution.

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Solve each system of equations using a matrix: $\{\begin{cases} 3 x + 4 y - 3 z = - 2 \\ 2 x + 3 y - z = - 12 \\ x + y - 2 z = 6 \end{cases}$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Write in augmented form.

Step 3. Apply row operations.

Step 4. Write in system of equations.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Discrete Mathematics

Geometry

Applied Mathematics

Logic and Functions

Probability and Statistics

Study anywhere. Anytime. Across all devices.

Solve each system of equations using a matrix:3x+4y-3z=-22x+3y-z=-12x+y-2z=6

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Write in augmented form.

Step 3. Apply row operations.

Step 4. Write in system of equations.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Discrete Mathematics

Geometry

Applied Mathematics

Logic and Functions

Probability and Statistics

Study anywhere. Anytime. Across all devices.

Solve each system of equations using a matrix: $\{\begin{cases} 3 x + 4 y - 3 z = - 2 \\ 2 x + 3 y - z = - 12 \\ x + y - 2 z = 6 \end{cases}$