RD Sharma Solutions for Class 12 Maths Exercise 8.2 Chapter 8 Solution of Simultaneous Linear Equations consist of problems on finding the homogeneous system of linear equations. These problems are solved by BYJU’S subject experts in a simple way to help with the exam preparation of students. They mainly help during revision and improve confidence among students before appearing for the board exam.
To gain better knowledge about the concepts explained in this exercise, they can refer to the RD Sharma Solutions Class 12 Maths Chapter 8 Solution of Simultaneous Linear Equations Exercise 8.2 free PDF, which are given below.
RD Sharma Solutions for Class 12 Chapter 8 Solution of Simultaneous Linear Equations Exercise 8.2
Access another exercise of RD Sharma Solutions For Class 12 Chapter 8 – Solution of Simultaneous Linear Equations
Access answers to Maths RD Sharma Solutions For Class 12 Chapter 8 – Solution of Simultaneous Linear Equations Exercise 8.2
Exercise 8.2 Page No: 8.20
Solve the following systems of homogeneous linear equations by matrix method:
1. 2x – y + z = 0
3x + 2y – z = 0
x + 4y + 3z = 0
Solution:
Given
2x – y + z = 0
3x + 2y – z = 0
X + 4y + 3z = 0
The system can be written as
A X = 0
Now, |A| = 2(6 + 4) + 1(9 + 1) + 1(12 – 2)
|A| = 2(10) + 10 + 10
|A| = 40 ≠0
Since, |A|≠0, hence x = y = z = 0 is the only solution of this homogeneous equation.
2. 2x – y + 2z = 0
5x + 3y – z = 0
X + 5y – 5z = 0
Solution:
Given 2x – y + 2z = 0
5x + 3y – z = 0
X + 5y – 5z = 0
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
3. 3x – y + 2z = 0
4x + 3y + 3z = 0
5x + 7y + 4z = 0
Solution:
Given 3x – y + 2z = 0
4x + 3y + 3z = 0
5x + 7y + 4z = 0
A X = 0
Now, |A| = 3(12 – 21) + 1(16 – 15) + 2(28 – 15)
|A| = – 27 + 1 + 26
|A| = 0
Hence, the system has infinite solutions
Let z = k
3x – y = – 2k
4x + 3y = – 3k
4. x + y – 6z = 0
x – y + 2z = 0
– 3x + y + 2z = 0
Solution:
Given x + y – 6z = 0
x – y + 2z = 0
– 3x + y + 2z = 0
A X = 0
Now, |A| = 1(– 2 – 2) – 1(2 + 6) – 6(1 – 3)
|A| = – 4 – 8 + 12
|A| = 0
Hence, the system has infinite solutions
Let z = k
x + y = 6k
x – y = – 2k
Comments