In the previous section, we saw how to obtain the determinant by expansion along the first row R1. In this section, we will see expansion along the second row.
Expansion along R2Step 1:
• Take the first element of R2, which is a21.
• Delete the row in which a21 is situated. This is indicated by the yellow rectangle in fig,20.3(a) below:
 |
| Fig.20.3 |
• Now write the following three items:
(i) The 2 × 2 determinant obtained by deleting the row and column.
(ii) Element which is under consideration. Here it is a21.
(iii) (-1)s, where s is the sum of the suffices of a21.
So (-1)s = (-1)2+1.
• Finally, multiply the three items together. We get:
$(-1)^{2+1} \times a_{21} \times \left|\begin{array}{r}
a_{12} &{ a_{13} } \\
a_{32} &{ a_{33} } \\
\end{array}\right|$
Step 2:
• Take the second element of R2, which is a22.
• Delete the row in which a22 is situated. This is indicated by the yellow rectangle in fig.20.3(b) above.
• Delete the column in which a22 is situated. This is indicated by the green rectangle in fig.20.3(b) above.
• Now write the following three items:
(i) The 2 × 2 determinant obtained by deleting the row and column.
(ii) Element which is under consideration. Here it is a22.
(iii) (-1)s, where s is the sum of the suffices of a22.
So (-1)s = (-1)2+2.
• Finally, multiply the three items together. We get:
$(-1)^{2+2} \times a_{22} \times \left|\begin{array}{r}
a_{11} &{ a_{13} } \\
a_{31} &{ a_{33} } \\
\end{array}\right|$
Step 3:
• Take the third element of R2, which is a23.
• Delete the row in which a23 is situated. This is indicated by the yellow rectangle in fig,20.3(c) above.
• Delete the column in which a23 is situated. This is indicated by the green rectangle in fig,20.3(c) above.
• Now write the following three items:
(i) The 2 × 2 determinant obtained by deleting the row and column.
(ii) Element which is under consideration. Here it is a23.
(iii) (-1)s, where s is the sum of the suffices of a23.
So (-1)s = (-1)2+3.
• Finally, multiply the three items together. We get:
$(-1)^{2+3} \times a_{23} \times \left|\begin{array}{r}
a_{11} &{ a_{12} } \\
a_{31} &{ a_{32} } \\
\end{array}\right|$
Step 4:
•
This is the final step. Here we add the results obtained in the above three steps.
•
The sum thus obtained is the determinant of A. We can write:
◼ The process by which we apply the above four steps to find the determinant of order 3, is known as expansion along R2.
• We saw two expansions:
♦ Expansion along R1
♦ Expansion along R2
Both gave the same result.
•
In fact, there is a total of six expansions (corresponding to the three rows and three columns):
♦ Expansion along R1
♦ Expansion along R2
♦ Expansion along R3
♦ Expansion along C1
♦ Expansion along C2
♦ Expansion along C3
• All six will give the same result.
Let us try one more:
Expansion along C3
Step 1:
• Take the first element of C3, which is a13.
• Delete the column in which a13 is situated. This is indicated by the green rectangle in fig,20.4(a) below:
 |
| Fig.20.4 |
• Now write the following three items:
(i) The 2 × 2 determinant obtained by deleting the column and row.
(ii) Element which is under consideration. Here it is a13.
(iii) (-1)s, where s is the sum of the suffices of a13.
So (-1)s = (-1)1+3.
• Finally, multiply the three items together. We get:
$(-1)^{1+3} \times a_{13} \times \left|\begin{array}{r}
a_{21} &{ a_{22} } \\
a_{31} &{ a_{32} } \\
\end{array}\right|$
Step 2:
• Take the second element of C3, which is a23.
• Delete the column in which a23 is situated. This is indicated by the green rectangle in fig,20.4(b) above.
• Delete the row in which a23 is situated. This is indicated by the yellow rectangle in fig,20.4(b) above.
• Now write the following three items:
(i) The 2 × 2 determinant obtained by deleting the row and column.
(ii) Element which is under consideration. Here it is a23.
(iii) (-1)s, where s is the sum of the suffices of a23.
So (-1)s = (-1)2+3.
• Finally, multiply the three items together. We get:
$(-1)^{2+3} \times a_{23} \times \left|\begin{array}{r}
a_{11} &{ a_{12} } \\
a_{31} &{ a_{32} } \\
\end{array}\right|$
Step 3:
• Take the third element of C3, which is a33.
• Delete the column in which a33 is situated. This is indicated by the green rectangle in fig,20.3(c) above.
• Delete the row in which a33 is situated. This is indicated by the yellow rectangle in fig,20.3(c) above.
• Now write the following three items:
(i) The 2 × 2 determinant obtained by deleting the row and column.
(ii) Element which is under consideration. Here it is a33.
(iii) (-1)s, where s is the sum of the suffices of a33.
So (-1)s = (-1)3+3.
• Finally, multiply the three items together. We get:
$(-1)^{3+3} \times a_{33} \times \left|\begin{array}{r}
a_{11} &{ a_{12} } \\
a_{21} &{ a_{22} } \\
\end{array}\right|$
Step 4:
•
This is the final step. Here we add the results obtained in the above three steps.
•
The sum thus obtained is the determinant of A. We can write:
◼ The process by which we apply the above four steps to find the determinant of order 3, is known as expansion along C3.
• We saw three expansions:
♦ Expansion along R1
♦ Expansion along R2
♦ Expansion along C3
All of then gave the same result.
• We can use any one of the six expansions that we mentioned earlier. The reader may write the 4 steps for R3, C1, C2 and become convinced about this fact.
•
Based on the above discussion, we can write two points:
(i) We must always use the expansion along that row/column which has the maximum number of zeroes.
(ii) Consider the term (-1)s. Instead of calculating (-1)s, we can put:
♦ 1 in the place of (-1)s, if s is even
♦ -1 in the place of (-1)s, if s is odd.
Now we will see a special case. It can be written in steps:
1. Consider two matrices A = $\left[\begin{array}{r} 3 &{ 3 } \\
6 &{ 0 } \\
\end{array}\right]
$ and B = $\left[\begin{array}{r}
1 &{ 1 } \\
2 &{ 0 } \\
\end{array}\right]$
•
We see that A = 3B
2. Let us calculate the two determinants:
♦ |A| = 3(0) − 6(3) = −18
♦ |B| = 1(0) − 2(1) = −2
3. We see that:
|A| = 9(-2) = 32|B|
•
Note that, the exponent of 3 is 2. This 2 is the order of both A and B.
4. This can be written in general form as:
If A = kB, then |A| = kn|B|, where n is the order of the two matrices A and B.
5. The formula written in (4) is applicable for the following three cases:
(i) When both A and B are matrices of order 1.
(ii) When both A and B are matrices of order 2.
(iii) When both A and B are matrices of order 3.
• We have seen the method for calculating determinant, when the given matrix is of the order 2 or 3.
•
What about the determinant of a matrix, whose order is 1?
•
Answer is simple:
For a 1 × 1 matrix [a], the determinant is a.
In the next section, we will see some solved examples related to determinants.
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