In order to find the determinant of a 3x3 matrix, we should follow the steps given below:
Let A
3x3 = $\begin{bmatrix}
a_{11} &a_{12} &a_{13} \\
a_{21} &a _{22} &a_{23} \\
a_{31} &a_{32} &a_{33}
\end{bmatrix} $ be a matrix.
Here, order of the matrix is 3x3. So, we can determine the determinant of the given matrix by expansion method. We have six ways of expansion, 3 corresponding to the three rows and 3 corresponding to the three columns.
So, $\left | A \right |$ = $\begin{vmatrix}
a_{11} &a_{12} &a_{13} \\
a_{21} &a_{22} &a_{23} \\
a_{31} &a_{32} &a _{33}
\end{vmatrix}$
Expansion along R1(First Row):
Step 1: Multiply a
11(element of R
1 row and C
1 column) by its cofactor i.e.
(-1)
1+1 a
11 $\begin{vmatrix}
a_{22} &a_{23} \\
a_{32} &a_{33}
\end{vmatrix}$
Step 2: Multiply a12 (element of R1 row and C2 column) by its cofactor
(-1)1+2 a12 $\begin{vmatrix}
a_{21} &a_{23} \\
a_{31} &a_{33}
\end{vmatrix}$
Step 3: Multiply a13 (element of R1 row and C3 column) by its cofactor
(-1)1+3 a13 $\begin{vmatrix}
a_{21} &a_{22} \\
a_{31} &a_{32}
\end{vmatrix}$
Step 4: So, the expansion of det A is the sum of the above three terms. So, we get
$\left | A \right |$ = (-1)1+1 a11 $\begin{vmatrix}
a_{22} &a_{23} \\
a_{32} &a_{33}
\end{vmatrix}$ + (-1)1+2 a12 $\begin{vmatrix}
a_{21} &a_{23} \\
a_{31} &a_{33}
\end{vmatrix}$ + (-1)1+3 a13 $\begin{vmatrix}
a_{21} &a_{22} \\
a_{31} &a_{32}
\end{vmatrix}$
$\left | A \right |$ = a11(a22a33 - a23a32) - a12 ( a21a33 - a23a31) + a13 ( a21a32 - a22a31)
= a11a22a33 - a11a23a3 - a12 a21a33 + a12a23a31 + a13 a21a32 - a13 a22a31 .....................................(i)
Expansion along R2(Second Row):
Again, IAI = $\begin{vmatrix}
a_{11} &a_{12} &a_{13} \\
a_{21} &a_{22} &a_{23} \\
a_{31} &a_{32} &a_{33}
\end{vmatrix}$
$\left | A \right |$ = (-1)
2+1 a
21$\begin{vmatrix}
a_{12} &a_{13} \\
a_{32} &a_{33}
\end{vmatrix}$ + (-1)
2+2 a
22 $\begin{vmatrix}
a_{11} &a_{13} \\
a_{31} &a_{33}
\end{vmatrix}$+ (-1)
2+3 a
23 $\begin{vmatrix}
a_{11} &a_{12} \\
a_{31} &a_{32}
\end{vmatrix}$
$\left | A \right |$ = -a
21( a
12a
33 - a
13a
32) + a
22( a
11a
33 - a
13a
31) - a
23 ( a
11a
32 - a
12a
31)
= -a
21a
12a
33 + a
21a
13a
32 + a
22a
11a
33 - a
22a
13a
31 - a
23a
11a
32 + a
23a
12a
31 =a
11a
22a
33 - a
11a
23a
3 - a
12 a
21a
33 + a
12a
23a
31 + a
13 a
21a
32 - a
13 a
22a
31 ----------------------------------------
(ii) Expansion along C1(First Column)
Again, IAI = $\begin{vmatrix}
a_{11} &a_{12} &a_{13} \\
a_{21} &a_{22} &a_{23} \\
a_{31} &a_{32} &a_{33}
\end{vmatrix}$
$\left | A \right |$ = (-1)
1+1 a
11 $\begin{vmatrix}
a_{22} &a_{23} \\
a_{32} &a_{33}
\end{vmatrix}$ + (-1)
2+1 a
21 $\begin{vmatrix}
a_{12} &a_{13} \\
a_{32} &a_{33}
\end{vmatrix}$ + (-1)
3+1 a
31 $\begin{vmatrix}
a_{12} &a_{13} \\
a_{32} &a_{33}
\end{vmatrix}$
$\left | A \right |$ = a
11(a
22a
33 - a
23a
32) - a
21( a
12a
33 - a
13a
32) + a
31 ( a
12a
23 - a
13a
22)
= a
11a
22a
33 - a
11a
23a
3 - a
12 a
21a
33 + a
12a
23a
31 + a
13 a
21a
32 - a
13 a
22a
31 -----------------------------------
(iii) Then, from (i), (ii) and (iii), it is clear that the value of the determinant of a 3 by 3 matrix is the same. So, we can use any expansion method either row expansion or column expansion method.
- We can use any expansion method, but for easier calculation, select that row or column which has maximum number of 0(zeros).
- While finding the determinant of a 3x3 matrix from expansion method, multiply by +1 or -1 instead of (-1)i+j according as i + j is odd or even
Solved Examples
Question 1: Calculate |A| = $\begin{vmatrix}
-1 &1 &2 \\
2 &-1 &0 \\
1 &3 &-1
\end{vmatrix}$
Solution:
For det A, we use row expansion method.
|A| = $(-1) \begin{vmatrix}
-1 & 0 \\
3 & -1
\end{vmatrix} - 1 \begin{vmatrix}
2 &0 \\
1 &-1
\end{vmatrix} + 2 \begin{vmatrix}
2 & -1 \\
1 & 3
\end{vmatrix}$
|A| = $(-1)(1 - 0) - 1(-2 - 0) + 2(6 + 1)$
= $-1 + 2 + 14$ = $15$
Question 2: Calculate |A| = $\begin{vmatrix}
-1 &1 &3 \\
3 &-1 &0 \\
1 &4 &0
\end{vmatrix}$
Solution:
To solve |A|, we select 3rd column for expansion since it has maximum zeros.
|A| = $\begin{vmatrix}
-1 &1 &3 \\
3 &-1 &0 \\
1 &4 &0
\end{vmatrix}$
= $3 \begin{vmatrix}
3 & -1\\
1&4
\end{vmatrix} - 0 \begin{vmatrix}
-1 & 1 \\
1 & 4
\end{vmatrix} + 0 \begin{vmatrix}
-1 & 1 \\
3 & -1
\end{vmatrix}$
= $3(12 + 1) + 0 + 0$
= $39$