Minors and Cofactors in Matrix with Solved Examples Problems-Class 12 Math Determinants Notes Study Material free pdf download

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 the matrix and then the cofactors of the matrix. Below is a detailed explanation of “What minors and cofactors are”, along with steps to find them.

What Are Minors?

The minor of an element in a matrix is defined as the determinant obtained by deleting the row and column in which that element lies. For example, 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 a₁₂ is denoted as M₁₂.

Here,

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

What Are Cofactors?

Cofactor of an element a_ij 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 i^th row and ‘j’ denotes the j^th column to which the element a_ij 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 a 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 i^th row (or column) and j^th 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 the multiplication of i^th 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 i^th row (or column) by k and adding the j^th row (or column) to it.

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 the remaining elements, we can calculate the cofactor of a₁₂

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

M₁₂ = 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 (-3) = (-1)¹⁺² (-46) = -(-46) = 46

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

(i)

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

\(\begin{array}{l}\text{Cofactor of}\ 2= -1^{1+1}M_{11}=+3\\\end{array} \)

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

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

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

(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 the 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

Q1

Are the cofactor and minor the same?

No, the minor of an element in a matrix is defined as the determinant obtained by deleting the row and column in which that element lies. The cofactor of an element a_ij, is defined by C_ij = (-1)^i+j M_ij, where M_ij is minor of a_ij.

Q2

How many minors does a 3×3 matrix have?

A 3×3 matrix has 9 minors.

Q3

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

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

Q4

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

The cofactor of an element a_ij, is given by C_ij = (-1)^i+j M_ij, where M_ij is minor of a_ij.

Q5

Give any application of minors and cofactors of a matrix.

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

Q6

How to find the cofactor matrix?

First, we have to calculate the minors of all the elements of the matrix. This is done by deleting the row and column to 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 -1^i+j. If M_ij is the minor, then cofactor, C_ij = -1^i+j M_ij. Then, form the cofactor matrix with the obtained values.

Q7

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

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

Q8

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

A 2×2 matrix has 4 minors.