20.9 - Solved Examples

In the previous section, we completed the discussion on properties of determinants. In this section, we will see some solved examples.

Solved example 20.8
Without expanding, prove that

Solution:

 

◼ Remarks:
• 2 (magenta color): Here we apply property VI
• 3(magenta color):
   ♦ Consider R₁ and R₃. Take any two corresponding elements from them.
   ♦ The element from R₁ will be proportional to the element from R₃ in the ratio (x+y+z) : 1.
   ♦ So value of the determinant is zero.

Solved example 20.9
Evaluate

Solution:

◼ Remarks:
• 2 and 3 (magenta color): Here we apply property VI
• 5(magenta color): Here we expand along C₁.

Solved example 20.10
Prove that

Solution:

1. Split the given determinant as |A| + |B| by applying property V.

◼ Remarks:
• 2 (magenta color): Here we apply property V and split Δ into |A| and |B|.
• 3 (magenta color): Here we apply property V and split |A|.
• 5(magenta color): Line 4 has four determinants.
   ♦ First is expanded along R₁.
   ♦ Second is expanded along R₃.
   ♦ Third is expanded along R₂.
   ♦ Fourth is expanded along C₁.

2. Evaluate |B|:

◼ Remarks:

• 2 (magenta color): Here we apply property V and split |B|.
• 3 (magenta color): Line 2 has two determinants.
   ♦ First is expanded along R₁.
   ♦ Second is expanded along R₁.

3. Find the sum:
Δ = |A| + |B|
= 3abc - a²c - ac² + ac² + a²c + abc
= 4abc

Solved example 20.11
If x, y, z are different and

then show that 1 + xyz = 0
Solution:
1. Split the given determinant as |A| + |B| by applying property V.

◼ Remarks:
• 2 (magenta color): Here we apply property V and split Δ into |A| and |B|.
• 3 (magenta color): Here we simplify |A|.
• 6(magenta color): Here we expand the determinant along C₃.

2. Evaluate |B|:

◼ Remarks:

• 2 (magenta color): Here we apply property IV and take out the common factors x, y and z.
• 3 (magenta color): Here we apply property II and interchange C₁ and C₂. So the sign of the determinant will change.
• 4 (magenta color): Here we apply property II and interchange C₂ and C₃. So the sign of the determinant will change again.
• 5 (magenta color):
Line 4 has a determinant on the R.H.S. This determinant is |A|.

3. Find the sum:
Δ = |A| + |B|
= |A| + xyz |A|
= (1+xyz)|A|
= (1+xyz)(x−z)(y−z)(y−x)

4. Given that Δ = 0.
• So we can write: (1+xyz)(x−z)(y−z)(y−x) = 0
• Also given that, x, y, z are different.
• So (x−z) ≠ 0, (y−z) ≠ 0, (y−x) ≠ 0
• Thus we get: 1+xyz = 0

Solved example 20.12
Show that

Solution:
1. Split the given determinant as |A| + |B| by applying property V.

◼ Remarks:

• 2 (magenta color): Here we apply property V and split Δ into |A| and |B|.
• 3 (magenta color): Here we apply property V and split |A|.
• 4 (magenta color): Line 3 has two determinants.
   ♦ The first has two identical columns. So by applying property III, it becomes zero.
• 5 (magenta color): Line 4 has a determinant. It is expanded along the second row.
• 8 (magenta color): |B| is expanded along the first row.

3. Find the sum:

◼ Remarks:
• 4 (magenta color): Here we take out abc as a common factor.

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A few more Solved examples can be seen in the following video:

Exercise 20.2

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In the next section, we will see Area of a triangle.

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