/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q 225 Solve each system of equations u... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve each system of equations using a matrix.

4x-3y+2z=0-2x+3y-7z=12x-2y+3z=6

Short Answer

Expert verified

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

Step by step solution

01

Step 1. Given information.

Consider the given system of equations,

4x-3y+2z=0-2x+3y-7z=12x-2y+3z=6

02

Step 2. Write in augmented form.

The augmented matrix for the given system of equations is

4-320-23-712-236

03

Step 3. Apply row operations.

Apply R14→R1and R2+2×R1→R2,

1-34120032-612-236

Apply R3-2×R1→R3and R2→R2×23,

1-3412001-4230-1226

Apply R3+12×R2→R3,

1-3412001-423000193

Apply R3×319→R3,

1-3412001-4230001

Now, the matrix is in row-echelon form.

04

Step 4. Write in system of equations.

Writing the corresponding system of equations,

x-34y+12z=0.......(i)y-4z=23.......(ii)0=1.......(iii)

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.