# Determinants Class 12 Notes Maths Chapter 4 - CBSE

## Chapter:4

## What are Determinants ?

## Determinant Of Order 2

$$\text{If A =}\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \text{is a square matrix of order 2, then |A| =} \begin{vmatrix*}[r] a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix*}=a_{11} \space a_{22} – a_{12} \space a_{21}.$$

## Determinant Of Order 3

$$\text{If A =}\begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22}& a_{23}\\ a_{31} & a_{32}& a_{33} \end{bmatrix} \text{is a square matrix of order 3, then |A| =}(-1)^{1+1} a_{11} \begin{vmatrix*}[r] a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix*}+(-1)^{1+2}a_{12}\begin{vmatrix*}[r] a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix*}+ (-1)^{1+3}\space a_{13}\begin{vmatrix*}[r] a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix*}$$

## Minors

_{ij}lies. Minor of an element a

_{ij}is denoted by M

_{ij}.

## Cofactors

_{ij}, denoted by A

_{ij}is defined by A

_{ij}= (-1)

^{i+j}M

_{ij}, where M

_{ij}is minor of a

_{ij}.

## Area Of A Triangle

$$\text{The area of a triangle whose vertices are }(\text{x}_1, \text{y}_1),(\text{x}_2, \text{y}_2)\space \text{and} \space(\text{x}_3 ,\text{y}_3) is :\space \Delta=\frac{1}{2}\begin{vmatrix*}[r] \text{x}_{1} & \text{y}_{1} & 1 \\ \text{x}_{2} & \text{y}_{2} & 1\\ \text{x}_{3} & \text{y}_{3} & 1 \end{vmatrix*}$$

## Adjoint And Inverse Of A Square Matrix

_{ij}]n×n is defined as the transpose of the matrix [A

_{ij}]n×n, where A

_{ij}is the cofactor of the element aij. Adjoint of the matrix A is denoted by adj A. If A is a square matrix of order m, and if there exists another square matrix B of the same order m, such that AB = BA = I, then B is called the inverse matrix of A and denoted by A

^{-1}.

## Consistency

A system of equations is said to be consistent if its solution (one or more) exists.

## Inconsistency

A system of equations is said to be inconsistent if its solution does not exist.

## Square Matrix

A square matrix A has inverse if and only if A is non- singular ,

$$\text{A}^{-1} = \frac{1}{|\text{A}|}\space\text{(adj A)} $$

## Solution Of Equation

^{-1}B, Where |A|

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