Determinant of a Matrix

To find the determinant of a matrix, we have to know about minors and cofactors.

Minors: If A = [a_ij] be a matrix. Then, minor of an element a_ij is denoted by M_ij and defined as the determinant of the matrix formed by removing or deleting i^th row and j^th column from the matrix A.

Cofactors: Cofactor of a_ij is denoted by C_ij = $\left (-1 ^{i+j} \right )$ M_ij, where i and j are the row number and column number respectively. So, cofactor of any minor, is the minor itself or the opposite of the minor.

To understand more about minors and cofactors, lets take an example:
Let A = $\begin{bmatrix}
2&-5 \\
6 & -1
\end{bmatrix}$ be a 2x2 order matrix.

Then,
Minor of 2, M₁₁ = -1
Minor of -5, M₁₂ = 6
Minor of 6, M₂₁ = -5
Minor of -1, M₂₂ = 2

Cofactor of 2, C₁₁ = (-1)¹⁺¹ x M₁₁ = (-1)² x (-1) = -1
Cofactor of -5, C₁₂ = (-1)¹⁺² x M₁₂ = (-1)³ x (6) = -6
Cofactor of 6, C₂₁ = (-1)²⁺¹ x M₂₁ = (-1)³ x (-5) = 5
Cofactor of -1, C₂₂ = (-1)²⁺² x M₂₂ = (-1)⁴ x (2) = 2

There are some ways to find the determinant of the matrix. If the order of a given matrix is 2x2, then its easy to find the value of determinant. If the order of the matrix is more then two, then we use row expansion method or column expansion method.

In order to find the determinant of a 2x2 matrix, we should follow the steps given below:

Let A = $\begin{bmatrix}
a &b \\
c & d
\end{bmatrix}$
Then, det A = $\left | A \right |$ = $\begin{vmatrix}
a & b \\
c &d
\end{vmatrix}$

= $ad - bc$
So, we can say that the determinant of a 2x2 order matrix is the difference between the product of main diagonal elements and the product of non-diagonal elements.

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 a₁₁(element of R₁ row and C₁ column) by its cofactor i.e.
(-1)¹⁺¹ a₁₁ $\begin{vmatrix}
a_{22} &a_{23} \\
a_{32} &a_{33}
\end{vmatrix}$

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 a₂₁$\begin{vmatrix}
a_{12} &a_{13} \\
a_{32} &a_{33}
\end{vmatrix}$ + (-1)²⁺² a₂₂ $\begin{vmatrix}
a_{11} &a_{13} \\
a_{31} &a_{33}
\end{vmatrix}$+ (-1)²⁺³ a₂₃ $\begin{vmatrix}
a_{11} &a_{12} \\
a_{31} &a_{32}
\end{vmatrix}$

$\left | A \right |$ = -a₂₁( a₁₂a₃₃ - a₁₃a₃₂) + a₂₂( a₁₁a₃₃ - a₁₃a₃₁) - a₂₃ ( a₁₁a₃₂ - a₁₂a₃₁)

= -a₂₁a₁₂a₃₃ + a₂₁a₁₃a₃₂ + a₂₂a₁₁a₃₃ - a₂₂a₁₃a₃₁ - a₂₃a₁₁a₃₂ + a₂₃a₁₂a₃₁

=a₁₁a₂₂a₃₃ - a₁₁a₂₃a₃ - a₁₂ a₂₁a₃₃ + a₁₂a₂₃a₃₁ + a₁₃ a₂₁a₃₂ - a₁₃ a₂₂a₃₁ ----------------------------------------(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 a₁₁ $\begin{vmatrix}
a_{22} &a_{23} \\
a_{32} &a_{33}
\end{vmatrix}$ + (-1)²⁺¹ a₂₁ $\begin{vmatrix}
a_{12} &a_{13} \\
a_{32} &a_{33}
\end{vmatrix}$ + (-1)³⁺¹ a₃₁ $\begin{vmatrix}
a_{12} &a_{13} \\
a_{32} &a_{33}
\end{vmatrix}$

$\left | A \right |$ = a₁₁(a₂₂a₃₃ - a₂₃a₃₂) - a₂₁( a₁₂a₃₃ - a₁₃a₃₂) + a₃₁ ( a₁₂a₂₃ - a₁₃a₂₂)

= a₁₁a₂₂a₃₃ - a₁₁a₂₃a₃ - a₁₂ a₂₁a₃₃ + a₁₂a₂₃a₃₁ + a₁₃ a₂₁a₃₂ - a₁₃ a₂₂a₃₁ -----------------------------------(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

In order to find the determinant of a 4x4 matrix, we should follow the steps given below:

Let A = $\begin{vmatrix}
1 &0 &3 &-1 \\
2 &1 &0 &4 \\
-1 &5 &0 &2 \\
1 &1 &6 &0
\end{vmatrix}$

We select column 3 for expansion.

IAI = $3 \begin{vmatrix}
2 &1 &4 \\
-1 &5 &2 \\
1 &1 &0
\end{vmatrix} - 0 \begin{vmatrix}
1 &0 &-1\\
-1 &5 &2 \\
1 &1 &0
\end{vmatrix} + 0 \begin{vmatrix}
1 &0 &-1 \\
2 &1 &4 \\
1 &1 &0
\end{vmatrix} - 6\begin{vmatrix}
1 &0 &-1 \\
2 &1 &4\\
-1 &5 &2
\end{vmatrix}$

= $3 \begin{vmatrix}
2 &1 &4 \\
-1 &5 &2 \\
1 &1 &0
\end{vmatrix} + 0 + 0 - 6 \begin{vmatrix}
1 &0 &-1 \\
2 &1 &4\\
-1 &5 &2
\end{vmatrix}$

= $3B - 6C$ ............................................. (1)

Here, B = $\begin{vmatrix}
2 &1 &4 \\
-1 &5 &2 \\
1 &1 &0
\end{vmatrix}$ and C = $\begin{vmatrix}
1 &0 &-1 \\
2 &1 &4\\
-1 &5 &2
\end{vmatrix}$

Here, B and C are the determinants of order three. So, we can use expansion method for them.

Now, IBI = $\begin{vmatrix}
2 &1 &4 \\
-1 &5 &2 \\
1 &1 &0
\end{vmatrix}$

= $2 \begin{vmatrix}
5 & 2\\
1 &0
\end{vmatrix} - 1 \begin{vmatrix}
-1 &2\\
1 & 0
\end{vmatrix} + 4 \begin{vmatrix}
-1 &5 \\
1 & 1
\end{vmatrix}$

= $2(0 - 2) - 1(0 - 2) + 4(-1 - 5)$
= $-4 + 2 - 24$
= $-26$ ...................................................................(a)

Again, ICI = $\begin{vmatrix}
1 &0 &-1 \\
2 &1 &4\\
-1 &5 &2
\end{vmatrix}$

= $1 \begin{vmatrix}
1 & 4\\
5 &2
\end{vmatrix} - 0 \begin{vmatrix}
2 &4\\
-1 &2
\end{vmatrix} + (-1) \begin{vmatrix}
2 & 1\\
-1 &5
\end{vmatrix}$

= $1(2 - 20) - 0 - 1(10 + 1)$

= $-18 - 11$

= $-29$ .............................................................................(b)

Replacing B and C, by the use of (a) and (b), we get
IAI = $3B - 6C$
= $3(-26) - 6(-29)$
= $-96$
IAI = $-96$
Thus, finding the determinant of a 4x4 matrix can be done easily by following the above given steps.