# Determinants

A determinant is defined as a quantity which is obtained by adding the products of all elements in a square matrix. To find the determinant, a particular rule is followed. In this lesson, the concept of determinants is explained in detail along with solved examples, formulas, determinant types, and practice questions.

## 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) No. of elements in a determinant of order n are n2.

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

### Important Formulas for Determinants

(a) Determinant of order $3\times 3=\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|={{a}_{1}}\left| \begin{matrix} {{b}_{2}} & {{c}_{2}} \\ {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|-{{b}_{1}}\left| \begin{matrix} {{a}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{c}_{3}} \\ \end{matrix} \right|+{{c}_{1}}\left| \begin{matrix} {{a}_{2}} & {{b}_{2}} \\ {{b}_{3}} & {{b}_{3}} \\ \end{matrix} \right|$

(b) 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 a12 is denoted as ${{M}_{12}}=\left| \begin{matrix} {{a}_{21}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{33}} \\ \end{matrix} \right|$

and so on.

(c) Cofactor of an element ${{a}_{i\,j}}={{C}_{i\,j}}={{\left( -1 \right)}^{i+j}}{{M}_{i\,j}}$

### Evaluation of the Determinant using SARRUS Diagram

If $A=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right]$

is a square matrix of order 3, the below diagram is a Sarrus Diagram obtained by adjoining the first two columns on the right and draw dark and dotted lines as shown.

The value of the determinant is $\left( {{a}_{11}}{{a}_{22}}{{a}_{33}}+{{a}_{12}}{{a}_{23}}{{a}_{31}}+{{a}_{13}}{{a}_{21}}{{a}_{32}} \right)-\left( {{a}_{13}}{{a}_{22}}{{a}_{31}}+{{a}_{11}}{{a}_{23}}{{a}_{32}}+{{a}_{12}}{{a}_{21}}{{a}_{33}} \right).$

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.

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 12: 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.

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

Illustration 13: Show that $\left| \begin{matrix} {{a}^{2}}+{{x}^{2}} & ab-cx & ac+bx \\ ab+cx & {{b}^{2}}+{{x}^{2}} & bc-ax \\ ac-bx & bc+ax & {{c}^{2}}+{{x}^{2}} \\ \end{matrix} \right|={{\left| \begin{matrix} x & c & -b \\ -c & x & a \\ b & -a & x \\ \end{matrix} \right|}^{2}}.$

Solution:

By replacing all elements of L.H.S. to their respective cofactors and using determinant property we will obtain the required result.

Let D $=\left| \begin{matrix} x & c & -b \\ -c & x & a \\ b & -a & x \\ \end{matrix} \right|$

Co-factors of 1st row of D are ${{x}^{2}}+{{a}^{2}},ab+cx,ac-bx.$ Co-factors of 2nd row of D are$ab-cx,{{x}^{2}}+{{b}^{2}},ax+bc$ and co-factors of 3rd row of D are $ac+bx,bc-ax,{{x}^{2}}+{{c}^{2}}$

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

#### Practise This Question

For a parallelogram ABCD, side DC is base for a triangle which is connected to point E on side AB. If the area of parallelogram is 54 cm2. Find the area of ΔDEC