Tuesday, April 2, 2024

20.4 - Property II

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

Property II
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    }    \\
c_1    &{    c_2    }    &{    c_3    }    \\
\end{array}\right | $.
2. We can write a new determinant Δ1 by interchanging any two rows (or columns). For example, by interchanging the second and third rows, we get:
Δ1 = $\left |\begin{array}{r}                           
a_1    &{    a_2    }    &{    a_3    }    \\
c_1    &{    c_2    }    &{    c_3    }    \\
b_1    &{    b_2    }    &{    b_3    }    \\
\end{array}\right |$.
3. Let us evaluate Δ. We will expand along R1.

4. Let us evaluate Δ1. We will expand along R1.


5. Let us compare Δ and Δ1. Both are written together below:


• Identical terms are given the same number. It is easy to see that, all six terms are identical. But the signs are opposite.

6. Let us compare (−Δ) and Δ1. Both are written together below:


• Identical terms are given the same number. It is easy to see that, all six terms are identical. The signs are also identical.
• So we get: −Δ = Δ1.

7. In the same way, the reader may check the result by interchanging any two columns.

8. Based on the above steps, we can write:
If any two rows (or columns) of a determinant are interchanged, then the sign of the determinant changes.

9. Suppose that:
    ♦ Ri and Rj represent the ith and jth rows respectively.
    ♦ Ci and Cj represent the ith and jth columns respectively.

• Then the process of interchanging the two rows can be represented as Ri ↔ Rj.
• Also, the process of interchanging the two columns can be represented as Ci ↔ Cj.


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

1. Let Δ = $\left |\begin{array}{r}                           
2    &{    -3    }    &{    5    }    \\
6    &{    0    }    &{    4    }    \\
1    &{    5    }    &{    7    }    \\
\end{array}\right | $.
2. Let us do C1 ↔ C3. We get:
Δ1 = $\left |\begin{array}{r}                           
5    &{    -3    }    &{    2    }    \\
4    &{    0    }    &{    6    }    \\
7    &{    5    }    &{    1    }    \\
\end{array}\right |$.
3. Let us evaluate Δ. We will expand along R2.


4. Let us evaluate Δ1. We will expand along C2.


5. Comparing the results in (3) and (4), we see that:
−Δ = Δ1

In the next section, we will see Property III.

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