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Minors and Cofactors

Minors and cofactors are two of the most important concepts in matrices as they are crucial in finding the adjoint and the inverse of a matrix. To find the determinants of a large square matrix (like 4×4), it is important to find the minors of that matrix and then the cofactors of that matrix. Below is a detailed explanation on “what are minors and cofactors” along with steps to find them.

All Topics in Determinants

What are Minors?

Minor of an element in a matrix is defined as the determinant obtained by deleting the row and column in which that element lies. e.g. in the determinant

\(\begin{array}{l}D=\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right|,\end{array} \)

minor of

\(\begin{array}{l}{{a}_{12}}\end{array} \)
is denoted as
\(\begin{array}{l}{{M}_{12}}=\left| \begin{matrix} {{a}_{21}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{33}} \\ \end{matrix} \right|\end{array} \)

and so on.

What are Cofactors?

Cofactor of an element

\(\begin{array}{l}{{a}_{i\,j}}\end{array} \)
is related to its minor as
\(\begin{array}{l}{{C}_{i\,j}}={{\left( -1 \right)}^{i+j}}{{M}_{i\,j}},\end{array} \)
where ‘i’ denotes the
\(\begin{array}{l}{{i}^{th}}\end{array} \)
row and ‘j’ denotes the
\(\begin{array}{l}{{j}^{th}}\end{array} \)
column to which the element
\(\begin{array}{l}{{a}_{i\,j}}\end{array} \)
belongs.

Now we define the value of the determinant of order three in terms of ‘Minor’ and ‘Cofactor’ as

\(\begin{array}{l}D={{a}_{11}}{{M}_{11}}-{{a}_{12}}{{M}_{12}}+{{a}_{13}}{{M}_{13}}\;\;\;\;\; or \;\;\;\;\;D={{a}_{11}}{{C}_{11}}-{{a}_{12}}{{C}_{12}}+{{a}_{13}}{{C}_{13}}\end{array} \)

Note: (a) A determinant of order 3 will have 9 minors and each minor will be a determinant of order 2 and a determinant of order 4 will have 16 minors and each minor will be determinant of order 3.

(b)

\(\begin{array}{l}{{a}_{11}}{{C}_{21}}+{{a}_{12}}{{C}_{22}}+{{a}_{13}}{{C}_{23}}=0,\end{array} \)
i.e. cofactor multiplied to different row/column elements results in zero value.

Row and Column Operations of Determinants

(a)

\(\begin{array}{l}{{R}_{i}}\leftrightarrow {{R}_{j}}\;\;\;or\;\;\; {{C}_{i}}\leftrightarrow {{C}_{j}}, \;\;\;when \;\;\;\;i\ne j;\end{array} \)
 This notation is used when we interchange ith row (or column) and jth row (or column).

(b)

\(\begin{array}{l}{{R}_{i}}\leftrightarrow {{C}_{i}};\end{array} \)
This converts the row into the corresponding column.

(c)

\(\begin{array}{l}{{R}_{i}}\to R{{k}_{i}}\;\;\;or\;\;\;{{C}_{i}}\to k{{C}_{i}};\,\,k\in R;\end{array} \)
This represents multiplication of ith row (or column) by k.

(d)

\(\begin{array}{l}{{R}_{i}}\to {{R}_{i}}k+{{R}_{j}}\;\;\;\;or\;\;\;Ci\to {{C}_{i}}k+{{C}_{j}};\left( i\ne j \right);\end{array} \)
This symbol is used to multiply ith row (or column) by k and adding the jth row (or column) to it.

Also Read:

Practice Problems on How to Find Minors and Cofactors

Question 1: Find the cofactor of a12 in the following

\(\begin{array}{l}\left| \begin{matrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \\ \end{matrix} \right|\end{array} \)

Solution:

In this problem we have to find the cofactor of a12 therefore eliminate all the elements of the first row and the second column and by obtaining the determinant of remaining elements we can calculate the cofactor of a12

Here a12= Element of first row and second column = –3

\(\begin{array}{l}{{M}_{12}}=\end{array} \)
Minor of a12 
\(\begin{array}{l}\begin{vmatrix} 6 & 4\\ 1& -7 \end{vmatrix}\end{array} \)
= 6 (-7) – 4(1)

= -42 – 4 = -46

Cofactor of

\(\begin{array}{l}\left( -3 \right)={{\left( -1 \right)}^{1+2}}\left( -46 \right)=-\left( -46 \right)=46\end{array} \)

Question 2: Write the minors and cofactors of the elements of the following determinants:

\(\begin{array}{l}(i) \left| \begin{matrix} 2 & -4 \\ 0 & 3 \\ \end{matrix} \right| \;\;\;\;\; (ii) \left| \begin{matrix} a & c \\ b & d \\ \end{matrix} \right|\end{array} \)

Solution:

By eliminating row and column of an element, the remaining is the minor of the element.

(i)

\(\begin{array}{l}\left| \begin{matrix} 2 & -4 \\ 0 & 3 \\ \end{matrix} \right|\;\\M_{11}= Minor\;\;\;\; of\;\;\; element\left( 2 \right)=3\end{array} \)

Cofactor of

\(\begin{array}{l}2= -1^{1+1}M_{11}=+3\\\end{array} \)
\(\begin{array}{l}{{M}_{12}}= \;\;Minor \;\;of \;\;element \left( -4 \right)=\left| \begin{matrix} 2…\,\,\,-4 \\ \,\,\,\,\,\,\,\,\,\,\,\,\vdots \\ 0\,\,\,\,\,\,\,\,\,3 \\ \end{matrix} \right|=0;\;\; \\Cofactor \;\;of \;\;\left( -4 \right)={{\left( -1 \right)}^{1+2}}{{M}_{12}}=\left( -1 \right)0=0\\\end{array} \)
\(\begin{array}{l}{{M}_{21}}= Minor\;\;\; of\;\; element \left( 0 \right)=\left| \begin{matrix} 2\,\,\,-4 \\ \vdots \,\,\,\,\,\,\,\,\, \\ 0…\,\,3 \\ \end{matrix} \right|=-4;\;\;\; \\Cofactor\;\; of \;\;\left( 0 \right)={{\left( -1 \right)}^{2+1}}{{M}_{21}}=\left( -1 \right)\left( -4 \right)=4\\\end{array} \)
\(\begin{array}{l}{{M}_{22}}=\;\; Minor\; of\; element \left( 3 \right)=\left| \begin{matrix} 2\,\,\,\,-4 \\ \,\,\,\,\,\,\,\,\,\,\vdots \\ 0…\,\,\,3 \\ \end{matrix} \right|=2; \;\\Cofactor \;\;of \;\left( 3 \right)={{\left( -1 \right)}^{2+2}}{{M}_{22}}=+2\end{array} \)

(ii)

\(\begin{array}{l}\left| \begin{matrix} a & c \\ b & d \\ \end{matrix} \right|;\,{M}_{11}= Minor\; of\; element \;\left( a \right)=d;\\\; Cofactor\; of \;\left( a \right)={{\left( -1 \right)}^{1+1}}{{M}_{11}}={{\left( -1 \right)}^{2}}d=d\\\end{array} \)
\(\begin{array}{l}{{M}_{12}}=\; Minor \;of\; element \;(c) =\left| \begin{matrix} a…c \\ \,\,\,\,\,\,\vdots \\ b\,\,\,\,d \\ \end{matrix} \right|=b;\; Cofactor \;of\; (c) ={{\left( -1 \right)}^{1+2}}{{M}_{12}}={{\left( -1 \right)}^{3}}b=-b\\\end{array} \)
\(\begin{array}{l}{{M}_{21}}=\; Minor\; of\; element\; \left( b \right)=\left| \begin{matrix} a\,\,\,\,\,c \\ \vdots \,\,\,\,\,\,\, \\ b\,…\,d \\ \end{matrix} \right|=c; \;Cofactor \;of \;\left( b \right)={{\left( -1 \right)}^{2+1}}{{M}_{21}}={{\left( -1 \right)}^{3}}c=-c\\\end{array} \)
\(\begin{array}{l}{{M}_{22}}=\; Minor\; of\; element \;\left( d \right)=\left| \begin{matrix} a\,\,\,\,\,c \\ \,\,\,\,\,\,\,\vdots \\ b…\,\,d \\ \end{matrix} \right|=a; \;Cofactor \;of \;\left( d \right)={{\left( -1 \right)}^{2+2}}{{M}_{22}}={{\left( -1 \right)}^{4}}a=a\end{array} \)

Question 3: Find the minor and cofactor of each element of the determinant

\(\begin{array}{l}\left| \begin{matrix} 2 & -2 & 3 \\ 1 & 4 & 5 \\ 2 & 1 & -3 \\ \end{matrix} \right|\end{array} \)

Solution:

By eliminating the row and column of an element, the determinant of remaining elements is the minor of the element, i.e.

\(\begin{array}{l}{{M}_{i\times j}}\end{array} \)
and by using formula
\(\begin{array}{l}{{\left( -1 \right)}^{i+j}}{{M}_{i\times j}}\end{array} \)
we will get the cofactor of the element.

The minors are

\(\begin{array}{l}{{M}_{11}}=\left| \begin{matrix} 4 & 5 \\ 1 & -3 \\ \end{matrix} \right|=-17, \;\;{{M}_{12}}=\left| \begin{matrix} 1 & 5 \\ 2 & -3 \\ \end{matrix} \right|=-13,\;\; {{M}_{13}}=\left| \begin{matrix} 1 & 4 \\ 2 & 1 \\ \end{matrix} \right|=-7\end{array} \)
\(\begin{array}{l}{{M}_{21}}=\left| \begin{matrix} -2 & 3 \\ 1 & -3 \\ \end{matrix} \right|=3, \;\;\;{{M}_{22}}=\left| \begin{matrix} 2 & 3 \\ 2 & -3 \\ \end{matrix} \right|=-12,\;\;\; {{M}_{23}}=\left| \begin{matrix} 2 & -2 \\ 2 & 1 \\ \end{matrix} \right|=6,\end{array} \)
\(\begin{array}{l}{{M}_{31}}=\left| \begin{matrix} -2 & 3 \\ 4 & 5 \\ \end{matrix} \right|=-22, \;\;\;{{M}_{32}}=\left| \begin{matrix} 2 & 3 \\ 1 & 5 \\ \end{matrix} \right|=7, \;\;\;{{M}_{33}}=\left| \begin{matrix} 2 & -2 \\ 1 & 4 \\ \end{matrix} \right|=10\end{array} \)

The cofactors are:

\(\begin{array}{l}{{A}_{11}}={{\left( -1 \right)}^{1+1}}{{M}_{11}}={{M}_{11}}=-17,{{A}_{12}}={{\left( -1 \right)}^{1+2}}{{M}_{12}}=-{{M}_{12}}=13,{{A}_{13}}={{\left( -1 \right)}^{1+3}}{{M}_{13}}={{M}_{13}}=-7\end{array} \)
\(\begin{array}{l}{{A}_{21}}={{\left( -1 \right)}^{2+1}}{{M}_{21}}=-{{M}_{21}}=-3,{{A}_{22}}={{\left( -1 \right)}^{2+2}}{{M}_{22}}={{M}_{22}}=-12,{{A}_{23}}={{\left( -1 \right)}^{2+3}}{{M}_{23}}=-{{M}_{23}}=-6\end{array} \)
\(\begin{array}{l}{{A}_{31}}={{\left( -1 \right)}^{3+1}}{{M}_{31}}={{M}_{31}}=-22,{{A}_{32}}={{\left( -1 \right)}^{3+2}}{{M}_{32}}=-{{M}_{32}}=-7,{{A}_{33}}={{\left( -1 \right)}^{3+3}}{{M}_{33}}={{M}_{33}}=10\end{array} \)

Frequently Asked Questions

Are cofactor and minor the same?

No. Minor of an element in a matrix is defined as the determinant obtained by deleting the row and column in which that element lies. Cofactor of an element aij, is defined by Cij = (-1)i+j Mij, where Mij is minor of aij.

How many minors does a 3×3 matrix have?

A 3×3 matrix has 9 minors.

How do you find the minor of an element in a matrix?

Minor of an element is calculated by finding the determinant obtained by deleting the row and column in which that element lies.

Give the formula to find the cofactor of an element in a matrix.

Cofactor of an element aij, is given by Cij = (-1)i+j Mij, where Mij is minor of aij.

Give any application of minors and cofactors of a matrix.

We use minors and cofactors to find the adjoint and inverse of matrices.

How to find the cofactor matrix?

First we have to calculate the minors of the all the elements of the matrix. This is done by deleting the row and column which the elements belong and then finding the determinant by considering the remaining elements. Then find the cofactor of the elements. It is done by multiplying the minor of the element with -1i+j. If Mij is the minor, then cofactor, Cij = -1i+j Mij. Then form the cofactor matrix with the obtained values.

Give the easy method to find adjoint matrix of a 2×2 matrix.

First we interchange the elements on the main diagonal. (a11 and a22). Then put negative sign for the elements at a12 and a21 position. Resulting matrix is the adjoint of the given 2×2 matrix.

What is the number of minors in a 2×2 matrix?

A 2×2 matrix has 4 minors.

1 Comment

  1. Hurbhookun Leckraj Sharma

    It was very helpful for the guidance.