# Minors And Cofactors Of A Determinant

For any square matrix A of order n, the determinant of the sub matrix obtained by eliminating the ith row and jth column of the matrix A is known as the minor Mij of the element aijof the given matrix.

The order of the minor obtained is (n – 1), as we are deleting a row and a column of the matrix A.

Let us take an example to understand the concept of minors in a better way,

Example: Let A =

$\begin{bmatrix} 3 & 4 & 1 & -6\\ 7 & -8 & 0 & 2\\ 7 & 6 & 4 & 1\\ -1 & 3 & 2 & 0 \end{bmatrix}$

find all the possible minors for the matrix $A$.

Solution: Matrix A is of the order 4 × 4, so the minors will be of order 3 × 3. Let us start deleting the first row and first column and then continue so on.

By deleting 1st row and 1st column, we get the minor for element M11, and so on, we have

M11= $\begin{vmatrix} -8 & 0 & 2\cr 6 & 4 & 1\cr 3 & 2 & 0 \end{vmatrix}$= -8 (0 – 2) + 2 (12 – 12) = 16

M12$\begin{vmatrix} 7 & 0 & 2\cr 7 & 4 & 1\cr -1 & 2 & 0 \end{vmatrix}$= 7(0 – 2) + 2(14 + 4) = -14 + 36 = 22

M13= $\begin{vmatrix} 7 & -8 & 2 \cr 7 & 6 & 1\cr -1 & 3 & 0 \end{vmatrix}$= 7 (0-3) + 8 (0 + 1) + 2(21 + 6)

M14= $\begin{vmatrix} 7 & -8 & 0\cr 7 & 6 & 4\cr -1 & 3 & 2 \end{vmatrix}$= 7(12 – 12) + 8 (14 + 4) = 8 (18) = 144

M21= $\begin{vmatrix} 4 & 1 & -6\cr 6 & 4 & 1\cr 3 & 2 & 0 \end{vmatrix}$= 4(-2) -1(-3) -6 (12 – 12) = -8 + 3 =-5

M22= $\begin{vmatrix} 3 & 1 & -6\cr 7 & 4 & 1\cr -1 & 2 & 0 \end{vmatrix}$ = 3 (0-2) – 1 (0 + 1) -6(14 +4) = -6 -1 -108 = – 115

M23=$\begin{vmatrix} 3 & 4 & -6 \cr 7 & 6 & 1 \cr -1 & 3 & 0 \end{vmatrix}$= 3 (0 – 3) – 4 (0 + 1) – 6(21 + 6) = -9 -4 -162 = -175

M24= $\begin{vmatrix} 3 & 4 & 1\cr 7 & 6 & 4\cr -1 & 3 & 2 \end{vmatrix}$ = 3 (12 – 12) -4 (14 + 4) + 1(21 + 6) = 72 + 27 = 99

M31$\begin{vmatrix} 4 & 1 & -6\cr -8 & 0 & 2\cr 3 & 2 & 0 \end{vmatrix}$= 4 (0-4) -1 (0-6) -6(-16 – 0) = -16 + 6 + 96 = 86

M32= $\begin{vmatrix} 3 & 1 & -6\cr 7 & 0 & 2\cr -1 & 2 & 0 \end{vmatrix}$= 3(0 – 4) -1 (0 + 2) -6(14 – 0) = -12 -2 – 72

M33= $\begin{vmatrix} 3 & 4 & -6\cr 7 & -8 & 2\cr -1 & 3 & 0 \end{vmatrix}$= 3(0 – 6) – 4 (0 + 2) -6(21 – 8) = -18 – 8 – 78 = -104

M34= $\begin{vmatrix} 3 & 4 & 1\cr 7 & -8 & 0\cr -1 & 3 & 2 \end{vmatrix}$= 3(-16 – 0) – 4 (14 + 0) – 1(21 – 8) = -48 – 56 – 13 = -117

M41= $\begin{vmatrix} 4 & 1 & -6\cr -8 & 0 & 2\cr 6 & 4 & 1 \end{vmatrix}$= 4(0 – 8) – 1 (-8 – 12) – 6(-32 – 0) = -32 + 20 + 192 = 180

M42=$\begin{vmatrix} 3 & 1 & -6\cr 7 & 0 & 2\cr 7 & 4 & 1 \end{vmatrix}$ = 3(0 – 8) – 1 (7 – 14) – 6(28 – 0) = -24 + 7 -168 = -185

M43= $\begin{vmatrix} 3 & 4 & -6\cr 7 & -8 & 2\cr 7 & 4 & 1 \end{vmatrix}$= 3(-8 – 8) – 4(7 – 14) – 6(28 + 56) = -48 + 28 – 504 = -524

M44= $\begin{vmatrix} 3 & 4 & 1\cr 7 & -8 & 0\cr 7 & 6 & 4 \end{vmatrix}$= 3(-32 – 0) – 4(28 – 0) – 1(42 + 56) = -93 – 112 – 98 = -303

These are the possible minors for the given matrix.

## Cofactor:

The cofactor of a square matrix A of order (n × n) is equal to

$\large \mathbf{(-1)^{i + j}M_{ij}}$,

where $M_{ij}$ represents the minor of the element $a_{ij}$of the matrix A.

The notation to represent a cofactor is $C_{ij}$.

Steps to find the cofactor matrix:

1. Find the minor $M_{ij}$ of the given matrix.
2.  Then add the values of i, j and put this value as power to -1 .
3. Multiply this obtained value which would either be +1 or – 1 to the value of the minor $M_{ij}$.

The value obtained is the cofactor of that element.

$C_{ij}$ = $(-1)^{i + j}M_{ij}$

The trick to find the value of cofactors is by putting positive and negative signs alternatively in the determinant which is easier and takes less than a minute to solve.

Thus for a 2 × 2 matrix if the minor matrix is given by $\begin{bmatrix} a & b \cr c & d \cr \end{bmatrix}$
then the sign that the elements of the cofactor matrix will take is represented by $\begin{vmatrix} + & – \cr – & + \end{vmatrix}$
.

The cofactor matrix is given by

$\begin{bmatrix} a & -b\cr -c & d \end{bmatrix}$
.

Similarly,a matrix of order 3 is given by

$\begin{bmatrix} a & b & c\cr d & e & f\cr g & h & i \end{bmatrix}$
then sign of the elements of the cofactor matrix is given as $\begin{bmatrix} + & – & + \cr – & + & – \cr + & – & + \end{bmatrix}$
and the cofactor matrix is given as

$\begin{bmatrix} ei-hf & -(di-gf) & dh-ge\cr -(bi-ch) & ai-gc & -(ah-gb)\cr bf-ec & -(af-dc) & (ae-db) \end{bmatrix}$
.

The same rule of alternate negative and positive signs follows for higher order matrix.

Let us have a look at an example to make it clearer,

Example: Let the matrix A be given by:

$\begin{bmatrix} -8 & 0 & 2\\ 6 & 4 & 1\\ 3 & 2 & 0 \end{bmatrix}$

Find the value of the all the cofactors.

Solution: The minors are given by,

M11= -2

M12= -3

M13= 0

M21= -4

M22 = -6

M23 = -16

M31 = -8

M32 = -20

M33 = -32

As the minors are now known, we apply the formula $C_{ij}$ =$(-1)^{i+j}M_{ij}$  to find out the cofactors.

The cofactor matrix is given by:

$\begin{bmatrix} -2 & 3 & 0\cr 4 & -6 & 16\cr -8 & 20 & -32 \end{bmatrix}$

We are thorough with what are minors and cofactors. To know more about determinants visit us online on www.byjus.com