Determinant is a number which is associated with every square matrix. It can be both positive or negative.
The expression a11a22 - a12 a21, associated with matrix A = $\begin{bmatrix}
a_{11} &a_{12} \\
a_{21} & a_{22}
\end{bmatrix}$
is defined as determinant of A and is denoted by det A or $\left | A \right |$. In linear algebra, we have various ways to define determinant of a matrix. The system of linear equations a11 X+ a12 Y = b11, a21 X + a22 Y = b21 can be written in matrix form as $\begin{bmatrix}
a_{11} &a_{12} \\
a_{21} & a_{22}
\end{bmatrix}$ $\begin{bmatrix}
X\\

Y\end{bmatrix}$ = $\begin{bmatrix}
b_{11}\\
b_{21}
\end{bmatrix}$ In order to find whether the above system has a unique solution or not, we have to determine the determinant value. If the determinant is
$\neq$ 0, then the system has an unique solution.

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

Minors:
If A = [aij] be a matrix. Then, minor of an element aij is denoted by Mij and defined as the determinant of the matrix formed by removing or deleting ith row and jth column from the matrix A.

Cofactors: Cofactor of aij is denoted by Cij = $\left (-1 ^{i+j} \right )$ Mij, 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.

Solved Example

Question: Evaluate $\begin{vmatrix}
x + 1 & x \\
1-x & x
\end{vmatrix}$
Solution:
We have $\begin{vmatrix}
x + 1 & x \\
1-x & x
\end{vmatrix}$ = $(x + 1) x - x(1 - x)$
= $x^{2} + x - x + x^{2}$
= $2x^{2}$


In order to find the determinant of a 3x3 matrix, we should follow the steps given below:

Let A3x3 = $\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

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| = $\begin{vmatrix}
-1 &1 &3 \\
3 &-1 &0 \\
1 &4 &0
\end{vmatrix}$

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

= $3(12 + 1) + 0 + 0$

= $39$

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.