Determinant of a Matrix

Determinant is a scalar value that can be calculated from the elements of a square matrix. It  is an arrangement of numbers in the form  abcd\left| \begin{matrix} a & b \\ c & d \\ \end{matrix} \right| . Determinant for a 3×3 matrix is determined by a1b1c1a2b2c2a3b3c3\begin{vmatrix} a_{1} & b_{1} & c_{1}\\ a_{2}& b_{2} & c_{2}\\ a_{3}& b_{3} & c_{3} \end{vmatrix} = a1(b2c3-b3c2)-b1(a2c3-a3c2)+c1(a2b3-a3b2). In this article, we come across properties and multiplication of determinants.

 

Introduction to Determinants

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

E.g.a1x+b1y=c1a2x+b2y=c2]x=b2c1b1c2a1b2a2b1      and    y=a1c2a2c1a1b2a2b1E.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}}}

Mathematicians defined the symbol a1b1a2b2\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 abcd\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. a1b1a2b2,a1b1c1a2b2c2a3b3c3\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| 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=abcdA=\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=[abcdefghi]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.

Contents in Determinants

Illustration 1: Expand 325914235\left| \begin{matrix} 3 & 2 & 5 \\ 9 & -1 & 4 \\ 2 & 3 & -5 \\ \end{matrix} \right|

by Sarrus rules.

Solution:

By using Sarrus rule i.e. Δ=(a11a22a33+a12a23a31+a13a21a32)(a13a22a31+a11a23a32+a12a21a33)\Delta =\left( {{a}_{11}}{{a}_{22}}{{a}_{33}}+{{a}_{12}}{{a}_{23}}{{a}_{31}}+{{a}_{13}}{{a}_{21}}{{a}_{32}} \right)-\left( {{a}_{13}}{{a}_{22}}{{a}_{31}}+{{a}_{11}}{{a}_{23}}{{a}_{32}}+{{a}_{12}}{{a}_{21}}{{a}_{33}} \right)we can expand the given determinant.

Determinant Example Question

Here, Δ=325914235Δ=1536+90+16+135+10=230\Delta =\left| \begin{matrix} 3 & 2 & 5 \\ 9 & -1 & 4 \\ 2 & 3 & -5 \\ \end{matrix} \right|\Rightarrow \Delta =15-36+90+16+135+10=230

 

Illustration 2: Evaluate the determinant : x2x+1x1x+1x+1\left| \begin{matrix} {{x}^{2}}-x+1 & x-1 \\ x+1 & x+1 \\ \end{matrix} \right|

Solution:

By using determinant expansion formula we can get the result.

We have x2x+1x1x+1x+1=(x2x+1)(x+1)(x+1)(x1)=x3+x2x2x+x+1x2+1=x3x2+2\left| \begin{matrix} {{x}^{2}}-x+1 & x-1 \\ x+1 & x+1 \\ \end{matrix} \right|=\left( {{x}^{2}}-x+1 \right)\left( x+1 \right)-\left( x+1 \right)\left( x-1 \right)={{x}^{3}}+{{x}^{2}}-{{x}^{2}}-x+x+1-{{x}^{2}}+1={{x}^{3}}-{{x}^{2}}+2

Symmetric and Skew Symmetric Determinants

Symmetric Determinant

A determinant is called a Symmetric Determinant if aij=aji,i,j  e.g.  [ahghbfgfc{{a}_{i\,j}}={{a}_{j\,i}},\forall \,i,j \;e.g.\; [\left| \begin{matrix} a & h & g \\ h & b & f \\ g & f & c \\ \end{matrix} \right|

Skew Symmetric Determinant

A determinant is called a skew symmetric determinant if aij=ajii,j{{a}_{i\,j}}=-{{a}_{j\,i}}\forall i,\,j for every element.

E.g. 031305150\left| \begin{matrix} 0 & 3 & -1 \\ -3 & 0 & 5 \\ 1 & -5 & 0 \\ \end{matrix} \right|

Note: (i) det |A| = 0 ⇒A is singular matrix (ii) det | A | ≠ 0 ⇒A is non-singular matrix

Multiplication of Two Determinants

(a) Multiplication of two second order determinants as follows: (as R to C method)

a1b1a2b2×l1m1l2m2=a1l1+b1l2a1m1+b1m2a2l1+b2l2a2m1+b2m2\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\ \end{matrix} \right|\times \left| \begin{matrix} {{l}_{1}} & {{m}_{1}} \\ {{l}_{2}} & {{m}_{2}} \\ \end{matrix} \right|=\left| \begin{matrix} {{a}_{1}}{{l}_{1}}+{{b}_{1}}{{l}_{2}} & {{a}_{1}}{{m}_{1}}+{{b}_{1}}{{m}_{2}} \\ {{a}_{2}}{{l}_{1}}+{{b}_{2}}{{l}_{2}} & {{a}_{2}}{{m}_{1}}+{{b}_{2}}{{m}_{2}} \\ \end{matrix} \right|

(b) Multiplication of two third order determinants is defined

a1b1c1a2b2c2a3b3c3×l1m1n1l2m2n2l3m3n3\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|\times \left| \begin{matrix} {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\ {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\ {{l}_{3}} & {{m}_{3}} & {{n}_{3}} \\ \end{matrix} \right|

(as R to C method)

=a1l1+b1l2+c1l3a1m1+b1m2+c1m3a1n1+b1n2+c1n3a2l1+b2l2+c2l3a2m1+b2m2+c2m3a2n1+b2n2+c2n3a3l1+b3l2+c3l3a3m1+b3m2+c3m3a3n1+b3n2+c3n3=\left| \begin{matrix} {{a}_{1}}{{l}_{1}}+{{b}_{1}}{{l}_{2}}+{{c}_{1}}{{l}_{3}} & {{a}_{1}}{{m}_{1}}+{{b}_{1}}{{m}_{2}}+{{c}_{1}}{{m}_{3}} & {{a}_{1}}{{n}_{1}}+{{b}_{1}}{{n}_{2}}+{{c}_{1}}{{n}_{3}} \\ {{a}_{2}}{{l}_{1}}+{{b}_{2}}{{l}_{2}}+{{c}_{2}}{{l}_{3}} & {{a}_{2}}{{m}_{1}}+{{b}_{2}}{{m}_{2}}+{{c}_{2}}{{m}_{3}} & {{a}_{2}}{{n}_{1}}+{{b}_{2}}{{n}_{2}}+{{c}_{2}}{{n}_{3}} \\ {{a}_{3}}{{l}_{1}}+{{b}_{3}}{{l}_{2}}+{{c}_{3}}{{l}_{3}} & {{a}_{3}}{{m}_{1}}+{{b}_{3}}{{m}_{2}}+{{c}_{3}}{{m}_{3}} & {{a}_{3}}{{n}_{1}}+{{b}_{3}}{{n}_{2}}+{{c}_{3}}{{n}_{3}} \\ \end{matrix} \right|

Note:

(i) The two determinants to be multiplied must be of the same order.

(ii) To get the Tmn{{T}_{mn}} (term in the mth row nth column) in the product, Take the mth row of the 1st determinant and multiply it by the corresponding terms of the nth column of the 2nd determinant and add.

(iii) This method is the row by column multiplication rule for the product of 2 determinants of the nth order determinant.

(iv) IfΔ′ is the determinant formed by replacing the elements of a Δ of order n by their corresponding co-factors then Δ=Δn1.\Delta ‘={{\Delta }^{n-1}}. (Δ)(\Delta ‘) is called the reciprocal determinant).

 

Illustration 3: Reduce the power of the determinant 0cbc0aba02{{\left| \begin{matrix} 0 & c & b \\ c & 0 & a \\ b & a & 0 \\ \end{matrix} \right|}^{2}}

to 1.

Solution:

By multiplying the given determinant two times we get the determinant as required.

0cbc0aba02=0cbc0aba00cbc0aba0b2+c2abacbac2+a2bccacba2+b2{{\left| \begin{matrix} 0 & c & b \\ c & 0 & a \\ b & a & 0 \\ \end{matrix} \right|}^{2}}=\left| \begin{matrix} 0 & c & b \\ c & 0 & a \\ b & a & 0 \\ \end{matrix} \right|\left| \begin{matrix} 0 & c & b \\ c & 0 & a \\ b & a & 0 \\ \end{matrix} \right|\Rightarrow \left| \begin{matrix} {{b}^{2}}+{{c}^{2}} & ab & ac \\ ba & {{c}^{2}}+{{a}^{2}} & bc \\ ca & cb & {{a}^{2}}+{{b}^{2}} \\ \end{matrix} \right|

 

 

Determinant of cofactors of D is

Dc=x2+a2ab+cxacbxabcxx2+b2ax+bcac+bxbcaxx2+c2=a2+x2abcxacbxab+cxb2+x2bcaxacbxax+bcx2+c2=D2{{D}^{c}}=\left| \begin{matrix} {{x}^{2}}+{{a}^{2}} & ab+cx & ac-bx \\ ab-cx & {{x}^{2}}+{{b}^{2}} & ax+bc \\ ac+bx & bc-ax & {{x}^{2}}+{{c}^{2}} \\ \end{matrix} \right|=\left| \begin{matrix} {{a}^{2}}+{{x}^{2}} & ab-cx & ac-bx \\ ab+cx & {{b}^{2}}+{{x}^{2}} & bc-ax \\ ac-bx & ax+bc & {{x}^{2}}+{{c}^{2}} \\ \end{matrix} \right|={{D}^{2}}

(Row interchanging into columns) =xcbcxabax2    (DC=D2,D  is  third  orderdeterminant)= {{\left| \begin{matrix} x & c & -b \\ -c & x & a \\ b & -a & x \\ \end{matrix} \right|}^{2}} \;\;({{D}^{C}}={{D}^{2}},D \;is \;third\; order determinant)

Hence a2+x2abcxac+bxab+cxb2+x2bcaxacbxbc+axc2+x2=xcbcxabax2\left| \begin{matrix} {{a}^{2}}+{{x}^{2}} & ab-cx & ac+bx \\ ab+cx & {{b}^{2}}+{{x}^{2}} & bc-ax \\ ac-bx & bc+ax & {{c}^{2}}+{{x}^{2}} \\ \end{matrix} \right|={{\left| \begin{matrix} x & c & -b \\ -c & x & a \\ b & -a & x \\ \end{matrix} \right|}^{2}}