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Class 12 Maths Notes

Determinants Class 12 Notes

Complete, exam-ready notes on determinants: expanding 2×2 and 3×3 determinants, their shortcut properties, adjoint and inverse, and using determinants to solve systems of linear equations — written for CBSE boards and JEE revision.

Class12SubjectMathematicsCoversCBSE · JEE

Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

What is this chapter about in one line?

Determinants are numbers computed from a square matrix that reveal whether it is invertible, help solve linear systems, and give the area of triangles.

Expansion of Determinants

Determinant

Every square matrix has a determinant: for A=(abcd)A = \begin{pmatrix} a & b\\ c & d \end{pmatrix}, ∣A∣=ad−bc|A| = ad - bc. A 3×3 determinant is expanded along a row or column using minors and cofactors.

∣abcdefghi∣=a(ei−fh)−b(di−fg)+c(dh−eg)\begin{vmatrix} a & b & c\\ d & e & f\\ g & h & i \end{vmatrix} = a(ei - fh) - b(di - fg) + c(dh - eg)
3×3 expansion

Properties of Determinants

  • |Aᵀ| = |A| — transposing does not change the determinant.
  • Interchanging two rows or columns changes the sign of the determinant.
  • If two rows or columns are identical or proportional, the determinant is 0.
  • A row can be scaled by multiplying the whole determinant by that factor.
  • Adding a multiple of one row to another leaves the determinant unchanged.

Speed trick

Before expanding by cofactors, use row operations to create a row with a single non-zero entry, then expand along it — it turns a 3×3 problem into a single 2×2.

Adjoint, Inverse and Singular Matrices

Adjoint and inverse

The adjoint adj(A) is the transpose of the cofactor matrix of A. For a square matrix, A−1=adj(A)∣A∣A^{-1} = \frac{\text{adj}(A)}{|A|}, so A⁻¹ exists if and only if ∣A∣≠0|A| \neq 0. A matrix with |A| = 0 is singular and has no inverse.

A−1=extadj(A)∣A∣,A adj(A)=∣A∣ I,∣AB∣=∣A∣∣B∣A^{-1} = \frac{ ext{adj}(A)}{|A|},\qquad A\,\text{adj}(A) = |A|\,I,\qquad |AB| = |A||B|
Inverse, adjoint property and product rule

Solving Linear Equations with Cramer's Rule

  • For a₁x + b₁y + c₁z = d₁, a₂x + b₂y + c₂z = d₂, a₃x + b₃y + c₃z = d₃, compute D = |coefficient matrix|.
  • x = Dₓ/D, y = D_y/D, z = D_z/D where Dₓ replaces the x-column with constants.
  • D ≠ 0 gives a unique solution; D = 0 with one of Dₓ, D_y, D_z ≠ 0 means no solution; all zero means infinite solutions.
x=DxD,y=DyD,z=DzDx = \frac{D_x}{D},\quad y = \frac{D_y}{D},\quad z = \frac{D_z}{D}
Cramer's rule

Area of a Triangle

Triangle area

The area of a triangle with vertices (x1,y1), (x2,y2), (x3,y3)(x_1,y_1),\ (x_2,y_2),\ (x_3,y_3) equals half the absolute value of the 3×3 determinant built from the coordinates. Points are collinear exactly when that determinant is zero.

Area=12∣∣x1y11x2y21x3y31∣∣\text{Area} = \frac{1}{2}\left| \begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1 \end{vmatrix} \right|
Area by determinant

Solved Examples

Example: Find |A| for A=(2357)A = \begin{pmatrix} 2 & 3\\ 5 & 7 \end{pmatrix}, and state whether A is invertible.

Solution: ∣A∣=(2)(7)−(3)(5)=14−15=−1≠0|A| = (2)(7) - (3)(5) = 14 - 15 = -1 \neq 0, so A is non-singular and invertible — the inverse is A−1=1−1(7−3−52)=(−735−2)A^{-1} = \frac{1}{-1}\begin{pmatrix} 7 & -3\\ -5 & 2 \end{pmatrix} = \begin{pmatrix} -7 & 3\\ 5 & -2 \end{pmatrix}.

Example: Using a property, show that ∣1ab+c1bc+a1ca+b∣=0\begin{vmatrix} 1 & a & b+c\\ 1 & b & c+a\\ 1 & c & a+b \end{vmatrix} = 0.

Solution: Add columns 2 and 3 to column 3. Each entry of column 3 becomes a + b + c, so column 3 becomes a constant multiple of column 1. A determinant with two proportional columns is zero.

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

2×2 determinant

∣abcd∣=ad−bc\begin{vmatrix} a & b\\ c & d \end{vmatrix} = ad - bc

Inverse via adjoint

A−1=adj(A)∣A∣A^{-1} = \frac{\text{adj}(A)}{|A|}

Adjoint identity

A adj(A)=∣A∣ IA\,\text{adj}(A) = |A|\,I

Product rule

∣AB∣=∣A∣∣B∣|AB| = |A||B|

Cramer's rule

x=Dx/D, y=Dy/Dx = D_x/D,\ y = D_y/D

Triangle area

Area=12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣\text{Area} = \frac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

  • A⁻¹ exists iff |A| ≠ 0.
  • |Aᵀ| = |A| and |AB| = |A||B|.
  • Identical or proportional rows ⇒ determinant = 0.
  • For uniqueness of a linear system, D ≠ 0.
  • Triangle area is half the absolute value of the coordinate determinant; zero area means collinear points.

FAQ

Common questions

What is a singular matrix?

A square matrix whose determinant is zero. It has no inverse and cannot be used to solve a system with a unique solution.

What is Cramer's rule?

A quick way to solve a linear system by ratios of determinants: x = Dₓ/D, y = D_y/D, z = D_z/D, where D is the coefficient determinant and Dₓ replaces the x-column with the constants.

When does a system of equations have no solution?

When the coefficient determinant D = 0 but at least one of Dₓ, D_y, D_z is non-zero. If all of them are also zero, the system has infinitely many solutions.

How do determinants give the area of a triangle?

The area is half the absolute value of the 3×3 determinant whose rows are (x₁,y₁,1), (x₂,y₂,1), (x₃,y₃,1). If the points are collinear, this determinant is zero.

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