Determinant of a Matrix

 

Introduction to Determinants

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

\(E.g. \left. \begin{matrix} {{a}_{1}}x+{{b}_{1}}y={{c}_{1}} \\ {{a}_{2}}x+{{b}_{2}}y={{c}_{2}} \\ \end{matrix} \right]\Rightarrow x=\frac{{{b}_{2}}{{c}_{1}}-{{b}_{1}}{{c}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\;\;\; and \;\; y=\frac{{{a}_{1}}{{c}_{2}}-{{a}_{2}}{{c}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\)

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. Its value was taken as \(\left( {{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}} \right)\) which is the same as the denominator. 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. \(\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\ \end{matrix} \right|,\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|\)

are determinants of second and third order respectively.

Note:

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

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

Contents in Determinants

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

by Sarrus rules.

Solution:

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

Determinant Example Question

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

 

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

Solution:

By using determinant expansion formula we can get the result.

We have \(\left| \begin{matrix} {{x}^{2}}-x+1 & x-1 \\ x+1 & x+1 \\ \end{matrix} \right|=\left( {{x}^{2}}-x+1 \right)\left( x+1 \right)-\left( x+1 \right)\left( x-1 \right)={{x}^{3}}+{{x}^{2}}-{{x}^{2}}-x+x+1-{{x}^{2}}+1={{x}^{3}}-{{x}^{2}}+2\)

Symmeteric and Skew Symmetric Determinants

Symmetric Determinant

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

Skew Symmetric Determinant

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

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

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)

\(\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\ \end{matrix} \right|\times \left| \begin{matrix} {{l}_{1}} & {{m}_{1}} \\ {{l}_{2}} & {{m}_{2}} \\ \end{matrix} \right|=\left| \begin{matrix} {{a}_{1}}{{l}_{1}}+{{b}_{1}}{{l}_{2}} & {{a}_{1}}{{m}_{1}}+{{b}_{1}}{{m}_{2}} \\ {{a}_{2}}{{l}_{1}}+{{b}_{2}}{{l}_{2}} & {{a}_{2}}{{m}_{1}}+{{b}_{2}}{{m}_{2}} \\ \end{matrix} \right|\)

(b) Multiplication of two third order determinants is defined

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

(as R to C method)

\(=\left| \begin{matrix} {{a}_{1}}{{l}_{1}}+{{b}_{1}}{{l}_{2}}+{{c}_{1}}{{l}_{3}} & {{a}_{1}}{{m}_{1}}+{{b}_{1}}{{m}_{2}}+{{c}_{1}}{{m}_{3}} & {{a}_{1}}{{n}_{1}}+{{b}_{1}}{{n}_{2}}+{{c}_{1}}{{n}_{3}} \\ {{a}_{2}}{{l}_{1}}+{{b}_{2}}{{l}_{2}}+{{c}_{2}}{{l}_{3}} & {{a}_{2}}{{m}_{1}}+{{b}_{2}}{{m}_{2}}+{{c}_{2}}{{m}_{3}} & {{a}_{2}}{{n}_{1}}+{{b}_{2}}{{n}_{2}}+{{c}_{2}}{{n}_{3}} \\ {{a}_{3}}{{l}_{1}}+{{b}_{3}}{{l}_{2}}+{{c}_{3}}{{l}_{3}} & {{a}_{3}}{{m}_{1}}+{{b}_{3}}{{m}_{2}}+{{c}_{3}}{{m}_{3}} & {{a}_{3}}{{n}_{1}}+{{b}_{3}}{{n}_{2}}+{{c}_{3}}{{n}_{3}} \\ \end{matrix} \right|\)

Note:

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

(ii) To get the \({{T}_{mn}}\) (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 nrd order determinant.

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

 

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

to 1.

Solution:

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

\({{\left| \begin{matrix} 0 & c & b \\ c & 0 & a \\ b & a & 0 \\ \end{matrix} \right|}^{2}}=\left| \begin{matrix} 0 & c & b \\ c & 0 & a \\ b & a & 0 \\ \end{matrix} \right|\left| \begin{matrix} 0 & c & b \\ c & 0 & a \\ b & a & 0 \\ \end{matrix} \right|\Rightarrow \left| \begin{matrix} {{b}^{2}}+{{c}^{2}} & ab & ac \\ ba & {{c}^{2}}+{{a}^{2}} & bc \\ ca & cb & {{a}^{2}}+{{b}^{2}} \\ \end{matrix} \right|\)

 

 

Determinant of cofactors of D is

\({{D}^{c}}=\left| \begin{matrix} {{x}^{2}}+{{a}^{2}} & ab+cx & ac-bx \\ ab-cx & {{x}^{2}}+{{b}^{2}} & ax+bc \\ ac+bx & bc-ax & {{x}^{2}}+{{c}^{2}} \\ \end{matrix} \right|=\left| \begin{matrix} {{a}^{2}}+{{x}^{2}} & ab-cx & ac-bx \\ ab+cx & {{b}^{2}}+{{x}^{2}} & bc-ax \\ ac-bx & ax+bc & {{x}^{2}}+{{c}^{2}} \\ \end{matrix} \right|={{D}^{2}}\)

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

Hence \(\left| \begin{matrix} {{a}^{2}}+{{x}^{2}} & ab-cx & ac+bx \\ ab+cx & {{b}^{2}}+{{x}^{2}} & bc-ax \\ ac-bx & bc+ax & {{c}^{2}}+{{x}^{2}} \\ \end{matrix} \right|={{\left| \begin{matrix} x & c & -b \\ -c & x & a \\ b & -a & x \\ \end{matrix} \right|}^{2}}\)