Determinant of a Matrix

What is it for?

Linear algebra deals with the determinant, it is computed using the elements of a square matrix.It can be considered as the scaling factor for the transformation of a matrix. Useful in solving a system of linear equation, calculating the inverse of a matrix and calculus operations.

Symbol

The determinant of a matrix is represented by two vertical lines or simply by writing det and writing the matrix name. eg. |A|, det(A), det A

Calculating the Determinant

To find a Determinant of a matrix, for every square matrix  \( [A]_{n×n} \) there exists a determinant to the matrix such that it represents a unique value given by applying some determinant finding techniques.

Determinant 0f A Matrix

For 2 x 2 Matrix

Therefore for a  2 × 2 matrix the determinant can be represented as Δ

\( Δ = det A = \begin{bmatrix}
a_{11} & a_{12}\cr
a_{21} & a_{22}
\end{bmatrix}\)

Note down the difference between the representation of a matrix and a determinant. In the case of a matrix we enclose the value in a square bracket whereas in case of a determinant we enclose it in between two lines.

Determinant of 1 × 1 matrix,

If [A] = [a]  then its determinant is given as |a|  which is equal to the value enclosed in the matrix.

The value of thedeterminant of a 2 × 2 matrix can be given as

det A = \( a_{11} × a_{22} – a_{21} × a_{21} \)

Let us take an example to understand this very clearly,

Example 1: The matrix  is given by, A = \(\begin{bmatrix}
3 & -1\cr
4 & 3
\end{bmatrix}\)
 Find the value of |A|,

Solution: We know the determinant can be calculated as:

|A| = (3 × 3) – (-1 × 4 )

|A| = 13

Thus, the value of determinant of a matrix is a unique value in nature.

The determinant of a 3 × 3 matrix is written as
det A = \(\left| \begin{matrix}
a_{11} & a_{12} & a_{13}\cr
a_{21} & a_{22} & a_{23} \cr
a_{31} & a_{32} & a_{33}
\end{matrix}\right|\)

The value of the determinant can be found out by expansion of the matrix along any row.

det A = \( a_{11}\)\( \begin{bmatrix}
a_{22} & a_{23}\cr
a_{32} & a_{33}
\end{bmatrix} \)
\( a_{12}\)\( \begin{bmatrix}
a_{21} & a_{23}\cr
a_{31} & a_{33}
\end{bmatrix} \)
+ \( a_{13}\)\( \begin{bmatrix}
a_{21} & a_{22}\cr
a_{31} & a_{32}
\end{bmatrix} \)

Minor:

The sub matrices \( \begin{bmatrix}
a_{22} & a_{23}\cr
a_{32} & a_{33}
\end{bmatrix} \)
,\( \begin{bmatrix}
a_{21} & a_{23}\cr
a_{31} & a_{33}
\end{bmatrix} \)
 and \( \begin{bmatrix}
a_{21} & a_{22}\cr
a_{31} & a_{32}
\end{bmatrix} \)
 are known as the minors of the determinants.

The minor \( M_{ij} \) of the element \( a_{ij} \)  of a matrix A of order n × n is defined as the determinant of the sub matrix of order (n-1). The minors are obtained by eliminating the \(i^{th} \) and\(j^{th} \) row and column respectively.

⇒ det A = \( a_{11} ( a_{22}a_{33} – a_{23}a_{32}) – a_{12} (a_{21}a_{33} – a_{23}a_{31}) + a_{13} ( a_{21}a_{32} – a_{22}a_{31}) \)

For 3 x 3 Matrix

Let us see an example to find out the determinant of a  3 × 3 matrix,

Example: The matrix  is given by, A = \( \begin{bmatrix}
4 & -3 & 5\cr
1 & 0 & 3\cr
-1 & 5 & 2
\end{bmatrix} \)
Find |A| .

Solution: To find the determinant of [A], let us expand the determinant along row 1.

Therefore, det A =

⇒ |A| = 4 \(\left|\begin{matrix}
0 & 3\cr
5 & 2
\end{matrix} \right|\)
– ( -3)  \(\left|\begin{matrix}
1 & 3\cr
-1 & 2
\end{matrix} \right|\)
+ 5  \(\left|\begin{matrix}
1 & 0\cr
-1 & 5
\end{matrix} \right|\)
<

⇒ |A| = 4 (0 – 15) + 3(2+3) + 5(5-0) ⇒ |A| = -20

Hence the determinant of a 3 × 3  matrix has a unique value.

It should be noted that the determinant is tried to be expanded along the row which has the maximum number of zeroes to make the calculations easy.

We have learned what determinants are and how to find the determinant of a given matrix. Please visit www.byjus.com to learn more about determinants and other concepts. We make learning a unique experience for you same as every determinant has a unique value.


Practise This Question

The set of intelligent students in a class is