Q. 25.0( 1 Vote )

# Solve the followi

The system can be written as

A X = 0

Now, |A| = 2(– 15 + 5) + 1(– 25 + 1) + 2(25 – 3)

|A| = – 20 – 24 + 44

|A| = 0

Hence, the system has infinite solutions

Let z = k

2x – y = – 2k

5x + 3y = k

A X = B

|A| = 6 + 5 = 11≠0 So, A – 1 exist

X = A – 1 B =

X =

Hence, x = , y = and z = k

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :

If <span lang="ENMathematics - Board Papers

Using matrices soMathematics - Board Papers

<span lang="EN-USMathematics - Board Papers

Using properties Mathematics - Board Papers

Using matrices, sMathematics - Board Papers

If A is square maMathematics - Exemplar

If <span lang="ENMathematics - Exemplar

The management coMathematics - Board Papers

If A is an invertMathematics - Board Papers

If <iMathematics - Board Papers