Monday, April 1, 2024

20.3 - Property I

In the previous section, we saw how to obtain the determinant of order 3. In this section, we will see some properties of determinants.

• The properties will help us to obtain maximum number of zeroes in a row/column.
• When zeroes are obtained in this way, the evaluation of the determinant will become easier.
• The properties are applicable to determinants of any order. But for our present discussion, we will consider order 3 only.

Property I
This can be written in 8 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 rows and columns. That is.,
Δ1 = $\left |\begin{array}{r}                           
a_1    &{    b_1    }    &{    c_1    }    \\
a_2    &{    b_2    }    &{    c_2    }    \\
a_3    &{    b_3    }    &{    c_3    }    \\
\end{array}\right |$.
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. The signs are also identical.
• So we get: Δ = Δ1.

6. Based on the above steps, we can write:
The determinant remains unchanged if it's rows and columns are interchanged.

7. Recall that, transpose of a matrix is obtained by interchanging rows and columns of that matrix. So we can write:
If A is a square matrix, then det(A) = det(A')

8. If Ri and Ci represent the ith row and ith column respectively, then the process of interchanging the rows and columns can be represented as Ri ↔ Ci.


Now we have a clear understanding about property I. 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. By interchanging rows and columns, we get:
Δ1 = $\left |\begin{array}{r}                           
2    &{    6    }    &{    1    }    \\
-3    &{   0    }    &{    5    }    \\
5    &{    4    }    &{    7    }    \\
\end{array}\right |$.
3. Let us evaluate Δ. We will expand along R2.


4. Let us evaluate Δ1. We will expand along R2.


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


In the next section, we will see Property II.

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