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 D=a11a12a13a21a22a23a31a32a33,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|,

minor of a12{{a}_{12}} is denoted as M12=a21a23a31a33{{M}_{12}}=\left| \begin{matrix} {{a}_{21}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{33}} \\ \end{matrix} \right|

and so on.

What are Cofactors?

Cofactor of an element aij{{a}_{i\,j}} is related to its minor as Cij=(1)i+jMij,{{C}_{i\,j}}={{\left( -1 \right)}^{i+j}}{{M}_{i\,j}}, where ‘i’ denotes the ith{{i}^{th}} row and ‘j’ denotes the jth{{j}^{th}} column to which the element aij{{a}_{i\,j}} belongs.

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

D=a11M11a12M12+a13M13          or          D=a11C11a12C12+a13C13D={{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}}

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) a11C21+a12C22+a13C23=0,{{a}_{11}}{{C}_{21}}+{{a}_{12}}{{C}_{22}}+{{a}_{13}}{{C}_{23}}=0, i.e. cofactor multiplied to different row/column elements results in zero value.

Row and Column Operations of Determinants

(a) RiRj      or      CiCj,      when        ij;{{R}_{i}}\leftrightarrow {{R}_{j}}\;\;\;or\;\;\; {{C}_{i}}\leftrightarrow {{C}_{j}}, \;\;\;when \;\;\;\;i\ne j; This notation is used when we interchange ith row (or column) and jth row (or column).

(b) RiCi;{{R}_{i}}\leftrightarrow {{C}_{i}}; This converts the row into the corresponding column.

(c) RiRki      or      CikCi;kR;{{R}_{i}}\to R{{k}_{i}}\;\;\;or\;\;\;{{C}_{i}}\to k{{C}_{i}};\,\,k\in R; This represents multiplication of ith row (or column) by k.

(d) RiRik+Rj        or      CiCik+Cj;(ij);{{R}_{i}}\to {{R}_{i}}k+{{R}_{j}}\;\;\;\;or\;\;\;Ci\to {{C}_{i}}k+{{C}_{j}};\left( i\ne j \right); 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 235604157\left| \begin{matrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \\ \end{matrix} \right|

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

M12={{M}_{12}}= Minor of:

a12(3)=235604157a_{12}(-3)=\begin{vmatrix} 2… &-3… &5 \\ & \vdots & \\ 6& 0 &4 \\ & \vdots & \\ 1 &5 &-7 \end{vmatrix} = 6 (-7) – 4(1) = -42 – 4 = -46

Cofactor of (3)=(1)1+2(46)=(46)=46\left( -3 \right)={{\left( -1 \right)}^{1+2}}\left( -46 \right)=-\left( -46 \right)=46

 

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

(i)2403          (ii)acbd(i) \left| \begin{matrix} 2 & -4 \\ 0 & 3 \\ \end{matrix} \right| \;\;\;\;\; (ii) \left| \begin{matrix} a & c \\ b & d \\ \end{matrix} \right|

Solution:

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

(i) 2403  M11=Minor        of      element(2)=2403=3\left| \begin{matrix} 2 & -4 \\ 0 & 3 \\ \end{matrix} \right|\;M_{11}= Minor\;\;\;\; of\;\;\; element\left( 2 \right)=|\begin{matrix} 2…-4 \\ \vdots \\ 0\,\,\,\,\,\,\,3 \\ \end{matrix}|=3

Cofactor of 2=11+1M11=+32= -1^{1+1}M_{11}=+3 M12=    Minor    of    element(4)=2403=0;    Cofactor    of    (4)=(1)1+2M12=(1)0=0{{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 M21=Minor      of    element(0)=2403=4;      Cofactor    of    (0)=(1)2+1M21=(1)(4)=4{{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 M22=    Minor  of  element(3)=2403=2;  Cofactor    of  (3)=(1)2+2M22=+2{{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

(ii) acbd;M11=Minor  of  element  (a)=acbd=d;  Cofactor  of  (a)=(1)1+1M11=(1)2d=d\left| \begin{matrix} a & c \\ b & d \\ \end{matrix} \right|;\,{M}_{11}= Minor\; of\; element \;\left( a \right)=\left| \begin{matrix} a\,…\,c \\ \,\,\,\,\,\,\,\vdots \\ b\,\,\,\,\,d \\ \end{matrix} \right|=d;\; Cofactor\; of \;\left( a \right)={{\left( -1 \right)}^{1+1}}{{M}_{11}}={{\left( -1 \right)}^{2}}d=d M12=  Minor  of  element  (c)=acbd=b;  Cofactor  of  (c)=(1)1+2M12=(1)3b=b{{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 M21=  Minor  of  element  (b)=acbd=c;  Cofactor  of  (b)=(1)2+1M21=(1)3c=c{{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 M22=  Minor  of  element  (d)=acbd=a;  Cofactor  of  (d)=(1)2+2M22=(1)4a=a{{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

 

Question 3: Find the minor and cofactor of each element of the determinant 223145213\left| \begin{matrix} 2 & -2 & 3 \\ 1 & 4 & 5 \\ 2 & 1 & -3 \\ \end{matrix} \right|

Solution:

By eliminating the row and column of an element, the determinant of remaining elements is the minor of the element, i.e. Mi×j{{M}_{i\times j}} and by using formula (1)i+jMi×j{{\left( -1 \right)}^{i+j}}{{M}_{i\times j}} we will get the cofactor of the element.

The minors are M11=4513=17,    M12=1523=13,    M13=1421=7{{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 M21=2313=3,      M22=2323=12,      M23=2221=6,{{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, M31=2345=22,      M32=2315=7,      M33=2214=10{{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

The cofactors are:

A11=(1)1+1M11=M11=17,A12=(1)1+2M12=M12=13,A13=(1)1+3M13=M13=7{{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 A21=(1)2+1M21=M21=3,A22=(1)2+2M22=M22=12,A23=(1)2+3M23=M23=6{{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 A31=(1)3+1M31=M31=22,A32=(1)3+2M32=M32=7,A33=(1)3+3M33=M33=10{{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

1 Comment

  1. Hurbhookun Leckraj Sharma

    It was very helpful for the guidance.