Chapters

Determinants

NCERT Class 12 Mathematics — Chapter 4

All Mathematical truths are relative and conditional. — C. P. Steinmetz

4.1 Introduction

For a system

a1x+b1y=c1,a2x+b2y=c2a_1x+b_1y=c_1,\qquad a_2x+b_2y=c_2

a unique solution exists when a1b2−a2b1≠0a_1b_2-a_2b_1\mathrel{\htmlClass{course-neq}{\char"2260}} 0. That number is the determinant of

A=[a1b1a2b2]A=\begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix}

This chapter treats determinants up to order 33, minors, cofactors, adjoint, inverse, and solving linear systems.

∣A∣|A| means determinant of AA, not modulus. Only square matrices have determinants.


4.2 Determinant

Order 11

If A=[a]A=[a], then ∣A∣=a|A|=a.

Order 22

∣a11a12a21a22∣=a11a22−a21a12\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix}=a_{11}a_{22}-a_{21}a_{12}

Order 33 (expansion along a row or column)

Expanding along R1R_1:

∣A∣=a11(a22a33−a32a23)−a12(a21a33−a31a23)+a13(a21a32−a31a22)|A|=a_{11}(a_{22}a_{33}-a_{32}a_{23})-a_{12}(a_{21}a_{33}-a_{31}a_{23})+a_{13}(a_{21}a_{32}-a_{31}a_{22})

Expanding along any row or column gives the same value. Prefer a row/column with many zeros.

If A=kBA=kB and both are n×nn\times n, then ∣A∣=kn∣B∣|A|=k^n|B|.


4.3 Properties (used constantly)

  • ∣A′∣=∣A∣|A'|=|A|
  • Interchanging two rows (or columns) changes the sign of ∣A∣|A|
  • If two rows (or columns) are identical, ∣A∣=0|A|=0
  • kk times a row (or column) multiplies the determinant by kk
  • Adding a multiple of one row to another does not change ∣A∣|A|

4.4 Minors, cofactors, adjoint, inverse

Minor MijM_{ij}: determinant after deleting row ii and column jj.

Cofactor: Aij=(−1)i+jMijA_{ij}=(-1)^{i+j}M_{ij}

Adjoint: adj⁡A\operatorname{adj} A is the transpose of the cofactor matrix.

A(adj⁡A)=(adj⁡A)A=∣A∣IA(\operatorname{adj} A)=(\operatorname{adj} A)A=|A|I

AA is singular if ∣A∣=0|A|=0, non-singular if ∣A∣≠0|A|\mathrel{\htmlClass{course-neq}{\char"2260}} 0.

AA has an inverse if and only if AA is non-singular, and

A−1=1∣A∣adj⁡AA^{-1}=\frac{1}{|A|}\operatorname{adj} A


4.5 System of linear equations

Write AX=BAX=B. Unique solution:

X=A−1Bwhen ∣A∣≠0X=A^{-1}B\qquad\text{when }|A|\mathrel{\htmlClass{course-neq}{\char"2260}} 0

  • ∣A∣≠0|A|\mathrel{\htmlClass{course-neq}{\char"2260}} 0: unique solution
  • ∣A∣=0|A|=0 and (adj⁡A)B≠0(\operatorname{adj} A)B\mathrel{\htmlClass{course-neq}{\char"2260}} 0: no solution (inconsistent)
  • ∣A∣=0|A|=0 and (adj⁡A)B=0(\operatorname{adj} A)B=0: may or may not be consistent

Chapter Summary

  • ∣A∣|A| is defined only for square matrices
  • Expand along any row or column
  • Inverse exists   ⟺  \iff ∣A∣≠0|A|\mathrel{\htmlClass{course-neq}{\char"2260}} 0
  • A−1=1∣A∣adj⁡AA^{-1}=\dfrac{1}{|A|}\operatorname{adj} A
  • Unique solution of AX=BAX=B is X=A−1BX=A^{-1}B when ∣A∣≠0|A|\mathrel{\htmlClass{course-neq}{\char"2260}} 0

Exercises (NCERT)

  • Exercise 4.1 — evaluating determinants
  • Later exercises — properties, adjoint, inverse, linear systems
  • Miscellaneous Exercise