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0.875V, – 0.125V2 – 0.5V3 = 2 → Eq. 1 -0.125V + 0.875V2 – 0.25V3 = 0 → Eq. 2 -0.5V1 – 0.25V2 + 0.875V3 = 0 → Eq. 3 Answer: V1 = 10V V2 = 4. 933V V3 = 12.267V

0.875V, – 0.125V2 – 0.5V3 = 2 → Eq. 1 -0.125V + 0.875V2 – 0.25V3 = 0 → Eq. 2 -0.5V1 – 0.25V2 + 0.875V3 = 0 → Eq. 3 Answer: V1 = 10V V2 = 4. 933V V3 = 12.267V

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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USE CRAMERS RULE (MATRIX) TO SOLVE THE EQUATION 1 , 2 AND 3 TO GET THE CORRECT ANSWER WHICH IS ALREADY GIVEN.

0.875V – 0.125V2 – 0.5V3 = 2 → Eq. 1 -0.125V + 0.875V2 – 0.25V3 = 0 → Eq. 2 -0.5V – 0.25V2 + 0.875V3 = 0 → Eq. 3 %3D Answer: V1 = 10V V2 = 4. 933V V3 = 12.267V
Transcribed Image Text:0.875V – 0.125V2 – 0.5V3 = 2 → Eq. 1 -0.125V + 0.875V2 – 0.25V3 = 0 → Eq. 2 -0.5V – 0.25V2 + 0.875V3 = 0 → Eq. 3 %3D Answer: V1 = 10V V2 = 4. 933V V3 = 12.267V
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