Determinants Class 12 Notes- Chapter 4

What is Determinant of a matrix?

When a square matrix “A” of an order “n” is associated with a number, then it is titled as a determinant of the aforementioned matrix. The number involved in this square matrix can be a real number or a complex number.

Introduction to Determinants

Development of determinants took place when mathematicians were trying to solve a system of simultaneous linear equations.

$E.g.\left. \begin{matrix} {{a}_{1}}x+{{b}_{1}}y={{c}_{1}} \\ {{a}_{2}}x+{{b}_{2}}y={{c}_{2}} \\ \end{matrix} \right] \Rightarrow x=\frac{{{b}_{2}}{{c}_{1}}-{{b}_{1}}{{c}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\;\;\; and \;\; y=\frac{{{a}_{1}}{{c}_{2}}-{{a}_{2}}{{c}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}$⇒x=a1​b2​−a2​b1​b2​c1​−b1​c2​​andy=a1​b2​−a2​b1​a1​c2​−a2​c1​​

Mathematicians defined the symbol $\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\ \end{matrix} \right|$ as a determinant of order 2 and the four numbers arranged in row and column were called its elements. If we write the coefficients of the equations in the following form $\left| \begin{matrix} a & b \\ c & d \\ \end{matrix} \right|$ then such an arrangement is called a determinant. In a determinant, horizontal lines are known as rows and vertical lines are known as columns. The shape of every determinant is a square. If a determinant is of order n then it contains n rows and n columns.

E.g. $\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\ \end{matrix} \right|,\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|$,∣∣∣∣∣∣∣​a1​a2​a3​​b1​b2​b3​​c1​c2​c3​​∣∣∣∣∣∣∣​ are determinants of second and third order respectively.

For every square matrix A of order m x n, there exists a number associated with it called the determinant of a square matrix.

For a matrix of 1 x 1, the determinant is A = [a].

For a 2 x 2 matrix, $A=\left| \begin{matrix} {{a}} & {{b}} \\ {{c}} & {{d}} \\ \end{matrix} \right|$ the determinant is ad – bc.

In the case of a 3 x 3 matrix $A=\begin{bmatrix} a &b &c \\ d &e &f \\ g&h &i \end{bmatrix}$, the value of determinant is = a (ei − fh) − b (di − fg) + c (dh − eg).

Note:

(i) The number of elements in a determinant of order n is n2.

(ii) A determinant of order 1 is the number itself.

Properties of Determinants

• There will be no change in the value of determinant if the rows and columns are interchanged.
• Suppose any two rows or columns of a determinant are interchanged, then its sign changes.
• If any two rows or columns of a determinant are the same, then the determinant is 0.
• If any row or column of the determinant is multiplied by a variable k, then its value is multiplied by k.
• Say if some or all elements of a row or column are expressed as the sum of two or more terms, then the determinant can be expressed as the sum of two or more determinants.

In the field of mathematics, Determinants can be used for a myriad of different calculations such as:

• Finding the area of a Triangle.
• Obtaining the solutions of linear equations in two or three variables.
• Solve Linear equations by using the inverse of a matrix
• Getting the adjoint and the inverse of a square matrix
• 
  

• 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 $D=\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right|,$,
  minor of ${{a}_{12}}$ is denoted as ${{M}_{12}}=\left| \begin{matrix} {{a}_{21}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{33}} \\ \end{matrix} \right|$
  and so on.

  What are Cofactors?

  Cofactor of an element ${{a}_{i\,j}}$ is related to its minor as ${{C}_{i\,j}}={{\left( -1 \right)}^{i+j}}{{M}_{i\,j}},$Mij​, where ‘i’ denotes the ${{i}^{th}}$ row and ‘j’ denotes the ${{j}^{th}}$ column to which the element ${{a}_{i\,j}}$ belongs.
  Now we define the value of the determinant of order three in terms of ‘Minor’ and ‘Cofactor’ as
  $D={{a}_{11}}{{M}_{11}}-{{a}_{12}}{{M}_{12}}+{{a}_{13}}{{M}_{13}}\;\;\;\;\; or \;\;\;\;\;D={{a}_{11}}{{C}_{11}}-{{a}_{12}}{{C}_{12}}+{{a}_{13}}{{C}_{13}}$
  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) ${{a}_{11}}{{C}_{21}}+{{a}_{12}}{{C}_{22}}+{{a}_{13}}{{C}_{23}}=0,$ i.e. cofactor multiplied to different row/column elements results in zero value.

  Row and Column Operations of Determinants

  (a) ${{R}_{i}}\leftrightarrow {{R}_{j}}\;\;\;or\;\;\; {{C}_{i}}\leftrightarrow {{C}_{j}}, \;\;\;when \;\;\;\;i\ne j;$ This notation is used when we interchange ith row (or column) and jth row (or column).
  (b) ${{R}_{i}}\leftrightarrow {{C}_{i}};$ This converts the row into the corresponding column.
  (c) ${{R}_{i}}\to R{{k}_{i}}\;\;\;or\;\;\;{{C}_{i}}\to k{{C}_{i}};\,\,k\in R;$ This represents multiplication of ith row (or column) by k.
  (d) ${{R}_{i}}\to {{R}_{i}}k+{{R}_{j}}\;\;\;\;or\;\;\;Ci\to {{C}_{i}}k+{{C}_{j}};\left( i\ne j \right);$ This symbol is used to multiply ith row (or column) by k and adding the jth row (or column) to it.

• Practice Problems on How to Find Minors and Cofactors

  Question 1: Find the cofactor of a12 in the following $\left| \begin{matrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \\ \end{matrix} \right|$

  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

  ${{M}_{12}}=$ Minor of:
  

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

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

  
  

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

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

  Solution:

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

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

  Cofactor of $2= -1^{1+1}M_{11}=+3$ ${{M}_{12}}= \;\;Minor \;\;of \;\;element \left( -4 \right)=\left| \begin{matrix} 2…\,\,\,-4 \\ \,\,\,\,\,\,\,\,\,\,\,\,\vdots \\ 0\,\,\,\,\,\,\,\,\,3 \\ \end{matrix} \right|=0;\;\; Cofactor \;\;of \;\;\left( -4 \right)={{\left( -1 \right)}^{1+2}}{{M}_{12}}=\left( -1 \right)0=0$M12​=(−1)0=0 ${{M}_{21}}= Minor\;\;\; of\;\; element \left( 0 \right)=\left| \begin{matrix} 2\,\,\,-4 \\ \vdots \,\,\,\,\,\,\,\,\, \\ 0…\,\,3 \\ \end{matrix} \right|=-4;\;\;\; Cofactor\;\; of \;\;\left( 0 \right)={{\left( -1 \right)}^{2+1}}{{M}_{21}}=\left( -1 \right)\left( -4 \right)=4$M21​=(−1)(−4)=4 ${{M}_{22}}=\;\; Minor\; of\; element \left( 3 \right)=\left| \begin{matrix} 2\,\,\,\,-4 \\ \,\,\,\,\,\,\,\,\,\,\vdots \\ 0…\,\,\,3 \\ \end{matrix} \right|=2; \;Cofactor \;\;of \;\left( 3 \right)={{\left( -1 \right)}^{2+2}}{{M}_{22}}=+2$M22​=+2

  (ii) $\left| \begin{matrix} a & c \\ b & d \\ \end{matrix} \right|;\,{M}_{11}= Minor\; of\; element \;\left( a \right)=\left| \begin{matrix} a\,…\,c \\ \,\,\,\,\,\,\,\vdots \\ b\,\,\,\,\,d \\ \end{matrix} \right|=d;\; Cofactor\; of \;\left( a \right)={{\left( -1 \right)}^{1+1}}{{M}_{11}}={{\left( -1 \right)}^{2}}d=d$M11​=(−1)2d=d ${{M}_{12}}=\; Minor \;of\; element \;(c) =\left| \begin{matrix} a…c \\ \,\,\,\,\,\,\vdots \\ b\,\,\,\,d \\ \end{matrix} \right|=b;\; Cofactor \;of\; (c) ={{\left( -1 \right)}^{1+2}}{{M}_{12}}={{\left( -1 \right)}^{3}}b=-b$M12​=(−1)3b=−b ${{M}_{21}}=\; Minor\; of\; element\; \left( b \right)=\left| \begin{matrix} a\,\,\,\,\,c \\ \vdots \,\,\,\,\,\,\, \\ b\,…\,d \\ \end{matrix} \right|=c; \;Cofactor \;of \;\left( b \right)={{\left( -1 \right)}^{2+1}}{{M}_{21}}={{\left( -1 \right)}^{3}}c=-c$M21​=(−1)3c=−c ${{M}_{22}}=\; Minor\; of\; element \;\left( d \right)=\left| \begin{matrix} a\,\,\,\,\,c \\ \,\,\,\,\,\,\,\vdots \\ b…\,\,d \\ \end{matrix} \right|=a; \;Cofactor \;of \;\left( d \right)={{\left( -1 \right)}^{2+2}}{{M}_{22}}={{\left( -1 \right)}^{4}}a=a$M22​=(−1)4a=a

  
  

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

  Solution:

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

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

  The cofactors are:

  ${{A}_{11}}={{\left( -1 \right)}^{1+1}}{{M}_{11}}={{M}_{11}}=-17,{{A}_{12}}={{\left( -1 \right)}^{1+2}}{{M}_{12}}=-{{M}_{12}}=13,{{A}_{13}}={{\left( -1 \right)}^{1+3}}{{M}_{13}}={{M}_{13}}=-7$M11​=M11​=−17,A12​=(−1)1+2M12​=−M12​=13,A13​=(−1)1+3M13​=M13​=−7 ${{A}_{21}}={{\left( -1 \right)}^{2+1}}{{M}_{21}}=-{{M}_{21}}=-3,{{A}_{22}}={{\left( -1 \right)}^{2+2}}{{M}_{22}}={{M}_{22}}=-12,{{A}_{23}}={{\left( -1 \right)}^{2+3}}{{M}_{23}}=-{{M}_{23}}=6$M21​=−M21​=−3,A22​=(−1)2+2M22​=M22​=−12,A23​=(−1)2+3M23​=−M23​=6 ${{A}_{31}}={{\left( -1 \right)}^{3+1}}{{M}_{31}}={{M}_{31}}=-22,{{A}_{32}}={{\left( -1 \right)}^{3+2}}{{M}_{32}}=-{{M}_{32}}=-7,{{A}_{33}}={{\left( -1 \right)}^{3+3}}{{M}_{33}}={{M}_{33}}=10$M31​=M31​=−22,A32​=(−1)3+2M32​=−M32​=−7,A33​=(−1)3+3M33​=M33​=10