Class 12 RD Sharma Solutions - Chapter 6 Determinants - Exercise 6.5

Question 1. Solve each of the following system of homogeneous linear equations:

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

This system of equations can be expressed in the form of a matrix AX = B

Now find the determinant,

= 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

So, D = 0, that means this system of equations has infinite solution.

Now, 

Let z = k

⇒ x + y = 2k

And 2x + y = 3k

Now using the Cramer’s rule

x = 

x = 

x = 

x = k

Similarly,

y = 

y = 

y = 

y = k

Therefore,

x = y = z = k.

Question 2. Solve each of the following system of homogeneous linear equations:

2x + 3y + 4z = 0

x + y + z = 0

2x + 5y - 2z = 0

Solution:

2x + 3y + 4z = 0

x + y + z = 0

2x + 5y - 2z = 0

This system of equations can be expressed in the form of a matrix AX = B

Find the determinant

= 2(1 × (-2) - 1 × 5) - 3(1 × (-2) - 2 × 1) + 4(1 × 5 - 2 × 1)

= 2(-2 - 5) - 3(-2 - 2) + 4(5 - 2)

= 2 × (-7) - 3 × (-4) + 4 × 3

= -14 + 12 + 12

= -10

Hence, D ≠ 0, so the system of equation has trivial solution.

Therefore, the system of equation has only solution as x = y = z = 0.

Question 3. Solve each of the following system of homogeneous linear equations:

3x + y + z = 0

x - 4y + 3z = 0

2x +5y - 2z = 0

Solution:

Given:

3x + y + z = 0

x - 4y + 3z = 0

2x +5y - 2z = 0

This system of equations can be expressed in the form of a matrix AX = B

Find the determinant

= 3(8 - 15) - 1(-2 - 6) + 1(13)

= -21 + 8 + 13

= 0

So, the system has infinite solutions:

Let z = k,

So, 

3x + y = -k

x - 4y = -3k

Now,

x = 

y = 

x = 

y = 

z = k

and there values satisfy equation 3

Hence, x = -7k, y = 8k, z = 13k

Question 4. Find the real values of λ for which the following system of linear equations has non-trivial solutions

2λx – 2y + 3z = 0

x + λy + 2z = 0

2x + λz = 0

Solution:

Finding the determinant

= 3λ³ + 2λ - 8 - 6λ 

= 2λ³ - 4λ - 8

Which is satisfied by λ = 2 {for non-trivial solutions λ =2}

Now let z = k

4x - 2y = -3k

x + 2y = -3k

x = 

y =

Hence, the solution is x = -k, y = , z = k

Question 5. If a, b, c are non-zero real numbers and if the system of equations

(a - 1)x = y + z

(b - 1)y = z + x

(c - 1)z = x + y

has a non-trivial solution, then prove that ab + bc + ca = abc

Solution:

Finding the determinant

Now for non-trivial solution, D = 0

0 = (a - 1)[(b - 1)(c - 1) - 1]+1[-c + 1 - 1] + [-c + 1 - 1] - [ 1 + b - 1]

0 = (a - 1)[bc - b - c + 1 - 1] - c - b

0 = abc - ab -ac + b + c - c - b

ab + bc + ac = abc

Hence proved

Summary

Chapter 6, Exercise 6.5 typically focuses on advanced applications and properties of determinants. The main topics covered usually include:

• Complex determinant problems and their solutions
• Applications of determinants in various mathematical fields
• Determinants in coordinate geometry
• Special determinants and their properties
• Proofs involving determinants
• Solving equations using determinants
• Determinants in matrix algebra

This exercise emphasizes problem-solving skills, analytical thinking, and the ability to apply determinant concepts in diverse and challenging scenarios.