If you are revising the Determinants Class 12 formulas before an exam, this sheet puts the NCERT chapter’s key results in one scannable place: determinants of order 1, 2 and 3, minors and cofactors, the area of a triangle, the adjoint and inverse of a matrix, and the matrix method for solving linear equations.
Each formula is grouped by topic with its symbol meanings, when-to-use guidance and worked examples using fresh numbers. For step-by-step derivations and explanations of why each result works, visit the Class 12 maths formulas hub.
Formulas at a Glance
Every formula on this page in one table. Symbol meanings and use-conditions follow in the sections below.
| Purpose (what you are finding) | Formula |
|---|---|
| Determinant of order 1 | \( |A| = |a_{11}| = a_{11} \) |
| Determinant of order 2 | \( |A| = a_{11}a_{22} – a_{12}a_{21} \) |
| Determinant of order 3 (expand along row 1) | \( |A| = a_1\begin{vmatrix} b_2 & c_2 \\ b_3 & c_3 \end{vmatrix} – b_1\begin{vmatrix} a_2 & c_2 \\ a_3 & c_3 \end{vmatrix} + c_1\begin{vmatrix} a_2 & b_2 \\ a_3 & b_3 \end{vmatrix} \) |
| Scalar multiple of a matrix (A = kB) | \( |A| = k^n|B| \), order \( n = 1, 2, 3 \) |
| Area of a triangle (take the absolute value) | \( \Delta = \frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \) |
| Equation of a line through two points (derived from area = 0) | \( \frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x & y & 1 \end{vmatrix} = 0 \) |
| Cofactor from a minor | \( A_{ij} = (-1)^{i+j}M_{ij} \) |
| Expansion of a determinant by cofactors | \( |A| = a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13} \) |
| Elements of one row times cofactors of another row | \( a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23} = 0 \) |
| Adjoint of a matrix of order 3 (transpose of the cofactor matrix) | \( \text{adj }A = \begin{bmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{bmatrix} \) |
| Adjoint of a matrix of order 2 | \( \text{adj}\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} = \begin{bmatrix} a_{22} & -a_{12} \\ -a_{21} & a_{11} \end{bmatrix} \) |
| Key adjoint relation | \( A(\text{adj }A) = (\text{adj }A)A = |A|I \) |
| Determinant of the adjoint (order n) | \( |\text{adj }A| = |A|^{n-1} \) |
| Determinant of a product of matrices | \( |AB| = |A||B| \) |
| Inverse of a matrix, only when |A| is not zero | \( A^{-1} = \frac{1}{|A|}\text{adj }A \) |
| Inverse of a product | \( (AB)^{-1} = B^{-1}A^{-1} \) |
| Linear system in matrix form | \( AX = B \) |
| Unique solution of AX = B (matrix method, when |A| is not zero) | \( X = A^{-1}B \) |
All Determinants Class 12 Formulas, Grouped by Topic
Determinant of a Matrix
For a square matrix \( A = [a_{ij}] \), the determinant is the single number associated with \( A \), written \( |A| \), \( \det A \) or \( \Delta \) (NCERT, pp. 77–78). Only square matrices have determinants, and \( |A| \) is read as “determinant of A”, not modulus of A.
Matrix of order 1 (NCERT, p. 102):
\[ |A| = |a_{11}| = a_{11} \]
Matrix of order 2 (NCERT, p. 102):
\[ |A| = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} = a_{11}a_{22} – a_{12}a_{21} \]
Matrix of order 3, expanded along \( R_1 \) (NCERT, p. 102):
\[ |A| = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = a_1\begin{vmatrix} b_2 & c_2 \\ b_3 & c_3 \end{vmatrix} – b_1\begin{vmatrix} a_2 & c_2 \\ a_3 & c_3 \end{vmatrix} + c_1\begin{vmatrix} a_2 & b_2 \\ a_3 & b_3 \end{vmatrix} \]
The same expansion in \( a_{ij} \) notation (NCERT, p. 78):
\[ |A| = a_{11}(a_{22}a_{33} – a_{32}a_{23}) – a_{12}(a_{21}a_{33} – a_{31}a_{23}) + a_{13}(a_{21}a_{32} – a_{31}a_{22}) \]
Expanding along any row or any column gives the same value, so choose the row or column with the most zeros (NCERT, p. 80). While expanding, multiply by +1 or −1 according as \( (i+j) \) is even or odd.
Scalar multiple property — if \( A = kB \), where A and B are square matrices of order n (NCERT, p. 80):
\[ |A| = k^n|B|, \quad n = 1, 2, 3 \]
Area of a Triangle
For a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), \( (x_3, y_3) \) (NCERT, pp. 82–83):
\[ \Delta = \frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \]
- Area is a positive quantity, so always take the absolute value of the determinant.
- Area = 0 if and only if the three points are collinear.
- If the area is given, use both positive and negative values of the determinant for calculation.
Equation of a line through two points — because three collinear points give area zero, putting a point \( P(x, y) \) on the line through \( (x_1, y_1) \) and \( (x_2, y_2) \) gives (NCERT, p. 83):
\[ \frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x & y & 1 \end{vmatrix} = 0 \]
Minors and Cofactors
The minor \( M_{ij} \) of an element \( a_{ij} \) is the determinant obtained by deleting the \( i \)-th row and \( j \)-th column (NCERT, p. 84). For a determinant of order \( n \) with \( n \geq 2 \), each minor is of order \( n – 1 \).
The cofactor of \( a_{ij} \), denoted \( A_{ij} \) (NCERT, p. 84):
\[ A_{ij} = (-1)^{i+j}M_{ij} \]
The factor \( (-1)^{i+j} \) is +1 when \( i + j \) is even and −1 when \( i + j \) is odd.
Expansion by cofactors — the value of a determinant equals the sum of the products of the elements of any row (or column) with the cofactors of the same row (or column). Along \( R_1 \) (NCERT, p. 86):
\[ |A| = a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13} \]
Cross-cofactor property — if elements of one row (or column) are multiplied by cofactors of a different row (or column), the sum is zero (NCERT, pp. 86–87):
\[ a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23} = 0 \]
Adjoint and Inverse of a Matrix
The adjoint of a square matrix A is the transpose of the matrix of cofactors (NCERT, p. 88):
\[ \text{adj }A = \begin{bmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{bmatrix} \]
For a matrix of order 2, the adjoint is quick: interchange the diagonal entries and change the signs of the other two (NCERT, p. 88):
\[ \text{adj}\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} = \begin{bmatrix} a_{22} & -a_{12} \\ -a_{21} & a_{11} \end{bmatrix} \]
Key relation between A and its adjoint (NCERT, p. 89):
\[ A(\text{adj }A) = (\text{adj }A)A = |A|I \]
A square matrix A is singular if \( |A| = 0 \) and non-singular if \( |A| \neq 0 \) (NCERT, p. 89).
Determinant of the adjoint, for A of order n (NCERT, p. 90):
\[ |\text{adj }A| = |A|^{n-1} \]
Determinant of a product (NCERT, p. 89):
\[ |AB| = |A||B| \]
The inverse — A is invertible if and only if A is non-singular (NCERT, p. 90):
\[ A^{-1} = \frac{1}{|A|}\text{adj }A, \quad |A| \neq 0 \]
Inverse of a product (NCERT, p. 92):
\[ (AB)^{-1} = B^{-1}A^{-1} \]
Derived quick facts: \( (A^{-1})^{-1} = A \) and \( \det(A^{-1}) = \frac{1}{\det A} \) (NCERT, p. 102; Exercise 4.4, Q18).
Solving Linear Systems Using the Inverse Matrix
A system \( a_1x + b_1y + c_1z = d_1 \), \( a_2x + b_2y + c_2z = d_2 \), \( a_3x + b_3y + c_3z = d_3 \) can be written as (NCERT, pp. 94–95):
\[ AX = B \]
where \( A = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), \( B = \begin{bmatrix} d_1 \\ d_2 \\ d_3 \end{bmatrix} \).
If A is non-singular, premultiply both sides by \( A^{-1} \) to get the unique solution (NCERT, p. 95):
\[ X = A^{-1}B \]
Consistency of the system \( AX = B \) (NCERT, p. 103):
| Condition | What it means for the system |
|---|---|
| \( |A| \neq 0 \) | Unique solution exists — the system is consistent. |
| \( |A| = 0 \) and \( (\text{adj }A)B \neq 0 \) | No solution exists — the system is inconsistent. |
| \( |A| = 0 \) and \( (\text{adj }A)B = 0 \) | The system may be consistent (infinitely many solutions) or inconsistent — examine the equations. |
What Each Symbol Means
Every symbol used in the formulas above, with its meaning and the nature of the quantity it represents.
| Symbol | What it means | Unit / nature |
|---|---|---|
| \( A, B \) | Square matrices of the same order (in products) | Array of numbers |
| \( a_{ij} \) | Element in the \( i \)-th row and \( j \)-th column of A | Real or complex number |
| \( |A| = \det A = \Delta \) | Determinant of A — the single number associated with A | Number; may be negative |
| \( k \) | Scalar in the relation \( A = kB \) | Real number |
| \( n \) | Order of the square matrix | Positive integer (1, 2, 3 in this chapter) |
| \( M_{ij} \) | Minor of \( a_{ij} \): determinant after deleting row \( i \), column \( j \) | Number (determinant of order \( n-1 \)) |
| \( A_{ij} \) | Cofactor of \( a_{ij} \) | Number \( (-1)^{i+j}M_{ij} \) |
| \( \text{adj }A \) | Adjoint of A — transpose of the cofactor matrix | Square matrix of order \( n \) |
| \( I \) | Identity matrix of order \( n \) | Square matrix with 1 on the diagonal, 0 elsewhere |
| \( A^{-1} \) | Inverse of A | Square matrix; exists only when \( |A| \neq 0 \) |
| \( (x_i, y_i) \) | Coordinates of the \( i \)-th vertex of the triangle | Coordinates (length units) |
| \( \Delta \) (area formula) | Area of the triangle | Square units |
| \( x, y, z \) | Variables of the linear system | Real numbers |
| \( X \) | Column matrix of variables \( \begin{bmatrix} x \\ y \\ z \end{bmatrix} \) | \( n \times 1 \) matrix |
| \( B \) | Column matrix of constants \( \begin{bmatrix} d_1 \\ d_2 \\ d_3 \end{bmatrix} \) | \( n \times 1 \) matrix |
| \( d_1, d_2, d_3 \) | Constant terms (right-hand side) of the equations | Real numbers |
When to Use Each Formula
A short decision guide — which formula to reach for, and the condition that must hold.
| Formula | Use it when… | Condition |
|---|---|---|
| Determinant of order 1 or 2 | You need the value of a small determinant — the building block of every expansion. | Always defined for a square matrix. |
| Order 3 expansion | You must evaluate a determinant of order 3. Expand along the row or column with the most zeros. | Any row or column gives the same value. |
| \( |A| = k^n|B| \) | A is a scalar multiple of B, for example \( A = 2B \). | \( n \) = order of the matrices. |
| Area formula | The coordinates of the three vertices are given. | Take \( |\Delta| \); area 0 means the points are collinear. |
| Area = 0 (line equation) | You need the equation of a line through two given points. | \( P(x, y) \) is any point on the line. |
| Minor and cofactor | You are building an adjoint or writing an expansion compactly. | Order \( n \geq 2 \). |
| Cofactor expansion | Computing any determinant as a sum of element × cofactor terms. | Multiply with cofactors of the same row or column. |
| Cross-cofactor sum = 0 | Verifying cofactors or proving \( A(\text{adj }A) = |A|I \). | Use cofactors of a different row or column. |
| adj A | First step towards finding the inverse. | Any square matrix. |
| \( A(\text{adj }A) = |A|I \) | Checking that a computed adjoint is correct. | Always true for square matrices. |
| \( |\text{adj }A| = |A|^{n-1} \) | Quick check of the adjoint’s determinant without computing all cofactors. | Order \( n \). |
| \( |AB| = |A||B| \) | Finding the determinant of a product matrix. | A and B square, of the same order. |
| Inverse formula | Finding \( A^{-1} \). | \( |A| \neq 0 \) — if \( |A| = 0 \) the matrix is singular and has no inverse. |
| \( X = A^{-1}B \) | Solving n linear equations in n unknowns by the matrix method. | \( |A| \neq 0 \) (unique solution exists). |
| Consistency conditions | Deciding whether a system has one, no, or infinitely many solutions. | Check \( |A| \) first, then \( (\text{adj }A)B \). |
NCERT Exercise Map
The chapter’s exercises drill the formulas in this order.
| Exercise | Formulas it practises |
|---|---|
| Exercise 4.1 | Evaluating determinants of order 1, 2 and 3; the scalar multiple rule \( |kA| = k^n|A| \). |
| Exercise 4.2 | Area of a triangle, collinearity, equation of a line using determinants. |
| Exercise 4.3 | Minors and cofactors of elements. |
| Exercise 4.4 | Adjoint, the relation \( A(\text{adj }A) = (\text{adj }A)A = |A|I \), inverse of a matrix. |
| Exercise 4.5 | Consistency of systems and solving them by the matrix method. |
Worked Examples
Example 1: Evaluate a determinant of order 3 along a row containing zero
Step 1: Evaluate \( \Delta = \begin{vmatrix} 2 & 3 & 1 \\ 0 & 4 & -1 \\ 5 & 0 & 2 \end{vmatrix} \).
Row 2 already contains a zero, so expand along \( R_2 \).
Step 2: Apply the cofactor expansion \( \Delta = -a_{21}M_{21} + a_{22}M_{22} – a_{23}M_{23} \).
\[ \Delta = -0\begin{vmatrix} 3 & 1 \\ 0 & 2 \end{vmatrix} + 4\begin{vmatrix} 2 & 1 \\ 5 & 2 \end{vmatrix} – (-1)\begin{vmatrix} 2 & 3 \\ 5 & 0 \end{vmatrix} \]
\[ \Delta = 0 + 4(2\cdot2 – 1\cdot5) + (2\cdot0 – 3\cdot5) = 4(4-5) + (-15) = -19 \]
Final answer: \( \Delta = -19 \).
Example 2: Find k when the triangle’s area is given (use both signs)
Step 1: The area of the triangle with vertices \( (k, 0) \), \( (6, 0) \), \( (0, 3) \) is 9 sq units.
Put the coordinates into the area formula.
\[ \Delta = \frac{1}{2}\begin{vmatrix} k & 0 & 1 \\ 6 & 0 & 1 \\ 0 & 3 & 1 \end{vmatrix} = \frac{1}{2}|k(0-3) – 0 + 1(6\cdot3 – 0)| = \frac{1}{2}|18 – 3k| \]
Step 2: Set \( \frac{1}{2}|18 – 3k| = 9 \), so \( |18 – 3k| = 18 \).
Because the area is given, the determinant may be positive or negative — solve both cases.
\[ 18 – 3k = 18 \Rightarrow k = 0; \qquad 18 – 3k = -18 \Rightarrow k = 12 \]
Final answer: \( k = 0 \) or \( k = 12 \). Check: for \( k = 12 \), the determinant is \( 18 – 36 = -18 \), and the area is \( \frac{1}{2}|-18| = 9 \) sq units.
Example 3: Solve a linear system using the inverse matrix
- Step 1: Write \( 3x + 4y = 18 \), \( 2x + 5y = 19 \) as \( AX = B \) with \( A = \begin{bmatrix} 3 & 4 \\ 2 & 5 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \end{bmatrix} \), \( B = \begin{bmatrix} 18 \\ 19 \end{bmatrix} \).
- Step 2: Check \( |A| = 3\cdot5 – 4\cdot2 = 15 – 8 = 7 \neq 0 \), so the inverse exists.
- Step 3: Form the adjoint — swap the diagonal entries and change the signs of the other two.
\[ \text{adj }A = \begin{bmatrix} 5 & -4 \\ -2 & 3 \end{bmatrix}, \qquad A^{-1} = \frac{1}{7}\begin{bmatrix} 5 & -4 \\ -2 & 3 \end{bmatrix} \]
Step 4: Premultiply both sides by \( A^{-1} \) to find \( X \).
\[ X = A^{-1}B = \frac{1}{7}\begin{bmatrix} 5 & -4 \\ -2 & 3 \end{bmatrix}\begin{bmatrix} 18 \\ 19 \end{bmatrix} = \frac{1}{7}\begin{bmatrix} 90 – 76 \\ -36 + 57 \end{bmatrix} = \frac{1}{7}\begin{bmatrix} 14 \\ 21 \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \end{bmatrix} \]
Final answer: \( x = 2 \), \( y = 3 \). Check in the original equations: \( 3(2) + 4(3) = 18 \) and \( 2(2) + 5(3) = 19 \).
Common Mistakes to Avoid
The errors students most often make while applying these determinant formulas, with a check you can run on your own answer.
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Reading \( |A| \) as absolute value (modulus) | \( |A| \) is the determinant — a single number that may be negative. | \( \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} = -2 \), not 2. |
| Wrong sign pattern in an order 3 expansion | Signs follow \( (-1)^{i+j} \), giving + − + / − + − / + − + starting at position (1, 1). | Expanding along \( R_1 \), the signs of \( a_{11}, a_{12}, a_{13} \) are +, −, +. |
| Forgetting the absolute value when an area is asked | Area is positive; use \( |\Delta| \). If the area is given, use both \( \pm \) values of the determinant. | A negative area answer means you dropped the modulus. |
| Writing adj A without transposing the cofactor matrix | adj A is the transpose of the cofactor matrix; for order 2, swap the diagonal and change the signs of the other two entries. | Multiply \( A \cdot \text{adj }A \): you must get \( |A| \) on the diagonal and 0 elsewhere. |
| Applying the inverse formula when \( |A| = 0 \) | An inverse exists if and only if \( |A| \neq 0 \); if \( |A| = 0 \) the matrix is singular. | Always compute \( |A| \) first — if it is zero, stop. |
| Multiplying \( A^{-1}B \) in the wrong order | Premultiply both sides: \( X = A^{-1}B \), never \( BA^{-1} \). | The product must come out as an \( n \times 1 \) matrix; the wrong order gives mismatched dimensions. |
Frequently Asked Questions
Is |A| the absolute value of A?
No. \( |A| \) is read as “determinant of A”, and it can be any real (or complex) number, including a negative one. Only the area of a triangle takes the absolute value of the determinant.
How do I check whether a matrix has an inverse?
Compute \( |A| \). If \( |A| \neq 0 \), the matrix is non-singular and \( A^{-1} = \frac{1}{|A|}\text{adj }A \). If \( |A| = 0 \), the matrix is singular and has no inverse.
When is the system AX = B called inconsistent?
When \( |A| = 0 \) and \( (\text{adj }A)B \neq 0 \), the system has no solution and is inconsistent. If \( |A| = 0 \) and \( (\text{adj }A)B = 0 \), the system may be consistent (infinitely many solutions) or inconsistent — you must examine the equations themselves.
Which row or column should I expand a determinant along?
The one with the most zeros. Every zero element contributes a zero term, so you compute fewer smaller determinants. Any row or column gives the same value, so use the easiest one.
Cross-check any formula against the chapter: open the official Rationalised NCERT chapter PDF (Determinants, Mathematics Part I). For other chapters and classes, browse the maths formula sheets index.
Reference: NCERT Class 12 Mathematics (Part I) textbook, chapter Determinants.
Explore Class 12 Maths Formulas
More for this chapter:
- Determinants Notes
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