20.5 - Property III

In the previous section, we saw the second property of determinants. In this section, we will see the third property.

Property III
This can be written in 9 steps:
1. Let Δ = $\left |\begin{array}{r}                           
a_1    &{    a_2    }    &{    a_3    }    \\
b_1    &{    b_2    }    &{    b_3    }    \\
a_1    &{    a_2    }    &{    a_3    }    \\
\end{array}\right | $.
• We see that, R₁ and R₃ are identical.
2. Let us evaluate Δ. We will expand along R₁.

 

◼ Remarks:
• In 3(magenta color),
    ♦ First term = -1 × sixth term 
    ♦ Second term = -1 × fourth term 
    ♦ Third term = -1 × fifth term

• So the sum is zero. 

3. In the same way, the reader may check the result when any two columns are identical. It will be zero.

4. Based on the above steps, we can write:
If any two rows (or columns) of a determinant are identical, then the value of that determinant will be zero.

5. The proof can be written in 6 steps:
(i) Suppose that, two rows of a matrix A are identical.
(ii) Calculate det(A)
(iii) Interchange the identical rows and write a new matrix A₁.
(iii) Calculate det(A₁)
(iv) Applying property II, we can write: -det(A) = det(A₁)
(v) In the matrix A, we interchanged identical rows.
So A = A₁
(vi) Since A = A₁, the determinants will also be equal.
That is., det(A) = det(A₁)
(vii) Substituting the above value of det(A₁) into (iv), we get:
-det(A) = det(A)
• This is possible only if det(A) = 0

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Now we have a clear understanding about property III. Let us see an example. It can be written in 5 steps:

1. Let Δ = $\left |\begin{array}{r}                           
5    &{    6    }    &{    5    }    \\
-4    &{    3    }    &{   -4    }    \\
1    &{    9    }    &{    1    }    \\
\end{array}\right | $.
2. Let us evaluate Δ. We will expand along R₁.

4. In this example, C₁ and C₃ are identical.

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In the next section, we will see Property IV.

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