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:
    ♦ R_i and R_j represent the i^th and j^th rows respectively.
    ♦ C_i and C_j represent the i^th and j^th columns respectively.

• Then the process of interchanging the two rows can be represented as R_i ↔ R_j.
• Also, the process of interchanging the two columns can be represented as C_i ↔ C_j.

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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. 

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In the next section, we will see Property III.

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