RD Sharma Solutions for Class 12 Maths Exercise 6.5 Chapter 6 Determinants is provided here. The RD Sharma textbook contains examples before each exercise to improve problem-solving speed among students. These concepts are explained in simple language, which makes it possible even for a student not proficient in Maths to understand the subject better.
RD Sharma Solutions are prepared by a set of experts at BYJU’S with the aim of helping students boost their exam preparation. The solution of a homogeneous system of linear equations is the main concept covered in this exercise. In order to excel in the board exams, students can access RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Exercise 6.5.
RD Sharma Solutions For Class 12 Chapter 6 – Determinants Exercise 6.5:
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Exercise 6.5 Page No: 6.89
Solve each of the following system of homogeneous linear equations:
1. x + y – 2z = 0
2x + y – 3z =0
5x + 4y – 9z = 0
Solution:
Given x + y – 2z = 0
2x + y – 3z =0
5x + 4y – 9z = 0
Any system of equations can be written in matrix form as AX = B
Now finding the Determinant of this set of equations,
= 1(1 × (– 9) – 4 × (– 3)) – 1(2 × (– 9) – 5 × (– 3)) – 2(4 × 2 – 5 × 1)
= 1(– 9 + 12) – 1(– 18 + 15) – 2(8 – 5)
= 1 × 3 –1 × (– 3) – 2 × 3
= 3 + 3 – 6
= 0
Since D = 0, so the system of equations has an infinite solution.
Now let z = k
⇒ x + y = 2k
And 2x + y = 3k
Now using Cramer’s rule
2. 2x + 3y + 4z = 0
x + y + z = 0
2x + 5y – 2z = 0
Solution:
Given
2x + 3y + 4z = 0
x + y + z = 0
2x + 5y – 2z = 0
Any system of equations can be written in matrix form as AX = B
Now finding the Determinant of this set of equations,
= 2(1 × (– 2) – 1 × 5) – 3(1 × (– 2) – 2 × 1) + 4(1 × 5 – 2 × 1)
= 2(– 2 – 5) – 3(– 2 – 2) + 4(5 – 2)
= 1 × (– 7) – 3 × (– 4) + 4 × 3
= – 7 + 12 + 12
= 17
Since D ≠ 0, so the system of equations has an infinite solution.
Therefore the system of equations has only a solution as x = y = z = 0.
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