Determinants Class 12: NCERT Maths Part I Chapter 4 PDF

Looking for the NCERT determinants class 12 chapter? This page gives you the official Class 12 Mathematics Part I Chapter 4, Determinants, straight from NCERT — the full chapter PDF is linked right below.

Below the download, every idea the chapter teaches is explained in order: determinants of orders 1, 2 and 3, the area of a triangle, minors and cofactors, the adjoint and inverse of a matrix, and the matrix method for solving linear equations.

Download the Determinants Class 12 chapter PDF from NCERT

The file below is the official NCERT chapter file (lemh104.pdf), identical in content to the printed book distributed in the 2026-27 reprint. Open it to read the chapter exactly as printed — examples, exercises and summary included — the determinants class 12 chapter PDF is the one file you need for revision.

If you would rather browse the whole book, the official NCERT textbook page for Mathematics Part I lists every chapter of this book and links its file.

What the Determinants chapter contains at a glance

The table below is generated from the official chapter data and shows what the file holds — its sections, its exercises and its figures. Determinants is an equation-dense chapter rather than a diagram-dense one, so treat the exercise sets as the landmarks as you read.

What this chapter covers, section by section

The chapter moves in one direction: compute a determinant first, then use it. Knowing where each idea sits makes the chapter far easier to revise — if you can name the section, you can find the rule.

• 4.1 Introduction — starts from a system \(a_1x + b_1y = c_1\), \(a_2x + b_2y = c_2\) written as a matrix equation, and shows that the number \(a_1b_2 – a_2b_1\) decides whether a unique solution exists. That number is the determinant of the coefficient matrix. The chapter studies determinants up to order three only, with real entries.
• 4.2 Determinant — the formal definition of the determinant of a square matrix, then the determinants of order 1, 2 and 3, including expansion along any row or column.
• 4.3 Area of a Triangle — the determinant formula for the area of a triangle, with the absolute-value and plus-or-minus conventions.
• 4.4 Minors and Cofactors — the building blocks that let you write the expansion of a determinant in compact form.
• 4.5 Adjoint and Inverse of a Matrix — the adjoint, the condition for a square matrix to be invertible, and the inverse formula.
• 4.6 Applications of Determinants and Matrices — solving systems of linear equations by the matrix method (4.6.1), with the consistency condition that decides whether a solution exists.
• Summary and Historical Note — the chapter closes with a one-page summary of every result, and a note tracing determinants from Chinese and Japanese work on simultaneous equations through Laplace, Gauss, Binet and Cauchy.

Key concepts in the Determinants chapter

This section explains the chapter’s core ideas in plainer words, in the same order the book teaches them. Each part is anchored to its section number so you can follow along with the book open.

What a determinant is, and what it is not

A determinant is a single number computed from a square matrix \(A = [a_{ij}]\). The whole chapter is about what that number tells you: whether a system has a solution, how big a triangle is, and whether a matrix can be inverted.

Section 4.2 defines it formally: to every square matrix the determinant function assigns a unique number, written \(\det A\), \(|A|\) or \(\Delta\). Section 4.1 shows where the idea comes from — the number \(a_1b_2 – a_2b_1\) that decides whether a two-variable linear system has a unique solution.

The two printed remarks in section 4.2 are the test of whether you have understood the notation. First, \(|A|\) is read as determinant of A, never modulus of A — a determinant can be negative, while a modulus cannot. Second, only square matrices have determinants.

A matrix and a determinant are different objects, and the comparison is worth having clear before any formula work.

Determinants of order one and two

Order one is trivial: the determinant of \(A = [a]\) is just \(a\). Order two is the pattern everything else builds on — a \(2 \times 2\) determinant is the product of the main diagonal minus the product of the other diagonal.

\[ \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} = a_{11}a_{22} – a_{12}a_{21} \]

Why this shape? It is exactly the quantity \(a_1b_2 – a_2b_1\) from section 4.1: when it is non-zero, the corresponding linear system has one unique solution, and when it is zero, the system does not.

Exercise 4.1 Q1 and Q2 give you the same skill to practise. Q7 and Q8 are the find-x type: evaluate both determinants, equate them, and solve — the pattern shown in Example 5.

One rule from the book’s remarks that cuts work later: if \(A = kB\) for square matrices of order \(n\), then \(|A| = k^n|B|\). That is why Exercise 4.1 Q3 and Q4 ask you to verify \(|2A| = 4|A|\) and \(|3A| = 27|A|\).

The 3 × 3 determinant: expansion along a row or column

A \(3 \times 3\) determinant is evaluated by reducing it to \(2 \times 2\) ones. Pick a row or a column; for each element, take the \(2 \times 2\) determinant left after deleting that element’s row and column, attach the sign \((-1)^{i+j}\), and add the three products.

There are six ways to expand — along \(R_1, R_2, R_3, C_1, C_2\) or \(C_3\) — and all six give the same value. The book works out \(R_1\), \(R_2\) and \(C_1\) in full and leaves the other three as an exercise for you.

The signs follow \((-1)^{i+j}\), so the pattern across the top row is plus, minus, plus. The practical remark that matters most (section 4.2.3): expand along the row or column that contains the maximum number of zeros.

The find-x shortcut works the same way in reverse: write both determinants, equate them, and solve for \(x\), as in Example 5.

Area of a triangle using a determinant

The first real application of the chapter. Put the three vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) as the rows \((x, y, 1)\) of a determinant; half its absolute value is the area (section 4.3).

\[ \Delta = \frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \]

Three printed remarks carry the whole section. Area is positive, so always take the absolute value. If the area is given instead, use both the positive and the negative values of the determinant — this is why Exercise 4.2 Q5 has two answers. Three collinear points give area zero.

Zero area is the trick that finds lines, as Example 7 shows: let P(x, y) be a general point on line AB, then triangle ABP has zero area, so the determinant equals zero and solving it gives the line’s equation.

A quick original check: the triangle with vertices (1, 2), (4, 5), (3, 0) gives a determinant of \(-12\), hence area 6 square units.

\[ \frac{1}{2}\begin{vmatrix} 1 & 2 & 1 \\ 4 & 5 & 1 \\ 3 & 0 & 1 \end{vmatrix} = \frac{1}{2}|-12| = 6 \]

Minors and cofactors

Minors and cofactors compress the long expansion into one line, and they are the machinery behind the adjoint and inverse that come next.

Definition 1: the minor \(M_{ij}\) of an element \(a_{ij}\) is the determinant left after deleting the \(i\)-th row and \(j\)-th column. Definition 2: the cofactor \(A_{ij} = (-1)^{i+j}M_{ij}\).

A tiny original example: in \( \begin{vmatrix} 2 & -1 & 0 \\ 3 & 4 & 5 \\ 1 & -2 & 6 \end{vmatrix} \), deleting the second row and third column leaves \[ M_{23} = \begin{vmatrix} 2 & -1 \\ 1 & -2 \end{vmatrix} = (2)(-2) – (-1)(1) = -3, \qquad A_{23} = (-1)^{2+3}M_{23} = 3 \]

The payoff (section 4.4): a determinant equals the sum of the products of the elements of any row or column with their corresponding cofactors. Expanding \( \Delta = a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13} \) along \(R_1\) is the compact form of everything done earlier.

One note to remember: elements of one row multiplied by the cofactors of a different row sum to zero, as in \(a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23} = 0\). Example 11 verifies this with numbers.

Adjoint of a matrix, and the 2 × 2 shortcut

To divide by a matrix you multiply by its inverse, and the inverse is built from the adjoint. The most common slip here is missing the transpose.

Definition 3: the adjoint of A, written adj A, is the transpose of the cofactor matrix \([A_{ij}]\). Transpose means the \(i\)-th row becomes the \(i\)-th column — the book shows the pattern with the full \(3 \times 3\) grids.

Theorem 1: \(A(\text{adj }A) = (\text{adj }A)A = |A|I\). The verification uses exactly the two cofactor facts from the last section: matching products give \(|A|\), mismatched ones give zero, and the transpose is what lines up the matching positions.

The 2 × 2 shortcut (remark after Example 12): interchange \(a_{11}\) and \(a_{22}\), change the signs of \(a_{12}\) and \(a_{21}\).

Original check: for \(A = \begin{bmatrix} 4 & 5 \\ 2 & 3 \end{bmatrix}\), we have \(|A| = 2\), \[ \text{adj }A = \begin{bmatrix} 3 & -5 \\ -2 & 4 \end{bmatrix}, \qquad A(\text{adj }A) = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = 2I = |A|I \]

The inverse of a matrix

A matrix can be inverted exactly when its determinant is not zero. This single condition decides everything that follows.

Definition 4: a square matrix is singular if \(|A| = 0\). Definition 5: it is non-singular if \(|A| \neq 0\). Theorem 4: A is invertible if and only if A is non-singular, and then \[ A^{-1} = \frac{1}{|A|}\text{adj }A \]

Why the condition is necessary: the formula divides by \(|A|\), so \(|A| = 0\) would mean division by zero. The proof of Theorem 4 uses the identity matrix and the product rule \(|AB| = |A||B|\).

Three results from this section that reappear in the exercises: \(|AB| = |A||B|\) (Theorem 3), \(|\text{adj }A| = |A|^{n-1}\) for a matrix of order \(n\), and \((AB)^{-1} = B^{-1}A^{-1}\) — note the reversed order, verified in Example 14.

Example 15 shows a pattern that Exercise 4.4 Q13 to Q16 repeat: if a matrix satisfies a polynomial equation such as \(A^2 – 4A + I = O\), rearrange the equation to express \(A^{-1}\) without computing the adjoint at all.

Solving systems of linear equations: the matrix method

Every system of linear equations in two or three variables compresses into a single matrix equation \(AX = B\). Solving the system then means multiplying by \(A^{-1}\) on the left.

Section 4.6.1 sets it up: with A the coefficient matrix, X the column of unknowns and B the column of constants, the system \(a_1x + b_1y + c_1z = d_1\), and so on, becomes \(AX = B\). Case I: if \(|A| \neq 0\), then \(X = A^{-1}B\) is the unique solution.

Why the left? Because \(A^{-1}(AX) = (A^{-1}A)X = IX = X\). Multiplying on the right would give a different, wrong result — matrix multiplication does not commute.

Case II: if \(|A| = 0\), compute \((\text{adj }A)B\). If it is not the zero matrix, the system has no solution and is called inconsistent. If it is the zero matrix, the system may be consistent (infinitely many solutions) or inconsistent. The book’s note restricts this chapter to unique-solution systems, so your practice concentrates on Case I.

One original check of Case II: the system \(x + y = 3\), \(2x + 2y = 5\) has \(|A| = 0\), and \[ (\text{adj }A)B = \begin{bmatrix} 2 & -1 \\ -2 & 1 \end{bmatrix}\begin{bmatrix} 3 \\ 5 \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \end{bmatrix} \neq 0 \]

so the system has no solution. Exercise 4.5 Q7 to Q16 practise the matrix method, including three-variable systems and the word problem on the cost of onion, wheat and rice (Q16).

Reading the determinant layouts in the book

This chapter is symbol-dense rather than diagram-dense, so the objects worth learning to read are the determinant grids themselves. Once you see the rows and columns inside them, every expansion in the book is the same movement.

Every \(3 \times 3\) grid has rows \(R_1, R_2, R_3\) and columns \(C_1, C_2, C_3\). The element \(a_{ij}\) sits at the crossing of row \(i\) and column \(j\), and its minor is the \(2 \times 2\) grid you see after deleting that row and column — section 4.2.3 lays this out step by step.

Beware the printed expansions: the signs come from \((-1)^{i+j}\), so positions alternate between plus and minus. When you copy an expansion from the book, copy the signs first and the numbers second.

Beside the introduction the book prints the portrait of P.S. Laplace (1749–1827). His 1772 method of expanding a determinant in terms of complementary minors is exactly the expansion you practise in this chapter.

The closing Historical Note gives the context: Seki Kowa used the idea in 1683 while eliminating a quantity from two equations, Vandermonde is called the formal founder, Laplace and Lagrange developed expansion, Gauss used determinants in number theory, Binet and Cauchy gave the product theorem in 1812, and Jacobi’s work brought the word “determinant” its final acceptance.

Definitions you need to know

Use this table to check a term in seconds while you practise. Each definition is the book’s own, from sections 4.4 to 4.6.

Common mistakes in this chapter (and their corrections)

These are the errors the book’s own remarks and notes warn about — the chapter flags them because students actually make them. For each mistake, the middle column gives the rule and the last column a quick check.

Notice how many of these are sign errors. You can catch a sign slip in a \(3 \times 3\) determinant by expanding along a second row or column and comparing the two results.

Key formulas at a glance

One formula sheet for the chapter, with every symbol explained — the fastest thing to skim the night before.

Every formula above needs the right sign convention — that is where most wrong answers come from.

How to practise: exam notes for the chapter

Work through the exercise sets in the order the chapter builds them. Each set practises one skill, so the mapping below tells you what to expect before you start.

• Exercise 4.1 — evaluating \(2 \times 2\) and \(3 \times 3\) determinants; the \(|kA| = k^n|A|\) checks (Q3, Q4) and the find-x questions (Q7, Q8).
• Exercise 4.2 — area of a triangle, collinearity and line equations. Q5 is the trap: a given area means using both signs of the determinant, so you get two values of \(k\).
• Exercise 4.3 — minors and cofactors; evaluating a determinant using the cofactors of a chosen row or column, plus one objective question on which products define the determinant.
• Exercise 4.4 — adjoint, the verification \(A(\text{adj }A) = |A|I\), inverse of given matrices (Q5 to Q11), \((AB)^{-1} = B^{-1}A^{-1}\) (Q12), the polynomial-equation pattern (Q13 to Q16), and objective questions on \(|\text{adj }A|\) and \(\det(A^{-1})\) (Q17, Q18).
• Exercise 4.5 — consistency checks first (Q1 to Q6), then the matrix method (Q7 to Q16), finishing with the word problem on the cost of onion, wheat and rice (Q16).
• Miscellaneous Exercises — a mix of everything in the chapter, including a system solved through a given product (Misc Example 19) where the product turns out to be \(A^{-1}\).

The chapter itself restricts attention to systems with unique solutions only, so your practice here centres on Case I of the matrix method. Also note: textbook contents and the examinable syllabus are not always identical — check the current official syllabus before the exam.

Quick revision summary

This is the chapter’s own Summary, restated so you can revise without re-reading. Each bullet is a result the chapter finishes with.

• Determinant of order one is the element itself: \(|a_{11}| = a_{11}\).
• Determinant of order two is the difference of diagonal products: \(|A| = a_{11}a_{22} – a_{12}a_{21}\).
• A \(3 \times 3\) determinant expands along any row or column — all six ways give the same value.
• Area of a triangle \(= \frac{1}{2} \times\) absolute value of the determinant of the vertices; collinear points give area zero.
• A minor \(M_{ij}\) is the determinant after deleting row \(i\) and column \(j\); a cofactor adds the sign, \(A_{ij} = (-1)^{i+j}M_{ij}\).
• A determinant equals the sum of the products of elements of any row or column with corresponding cofactors; elements of one row times cofactors of another row sum to zero.
• adj A is the transpose of the cofactor matrix, and \(A(\text{adj }A) = (\text{adj }A)A = |A|I\).
• A square matrix is singular if \(|A| = 0\) and non-singular if \(|A| \neq 0\).
• A has an inverse if and only if A is non-singular, and \(A^{-1} = \frac{1}{|A|}\text{adj }A\).
• For \(AX = B\): \(|A| \neq 0\) gives the unique solution \(X = A^{-1}B\); \(|A| = 0\) with \((\text{adj }A)B \neq 0\) gives no solution; \((\text{adj }A)B = 0\) means the system may or may not be consistent.

Every step in this chapter hangs on one question: is \(|A|\) zero or not. If you remember only one thing, remember that.

Related chapters and resources

Section 4.1 says the chapter builds directly on Chapter 3, Matrices — a system of equations written as \(AX = B\) is the starting point of everything here. The next chapter in Mathematics Part I, Continuity and Differentiability, moves on to a different branch of calculus, and the chapter on differential equations later in the book draws on the matrix methods you learn here.

The chapter’s Historical Note connects determinants to work on simultaneous equations running from China and Japan through Laplace, Gauss, Binet and Cauchy.

For study help across this book, start with the Class 12 Mathematics notes; the Continuity and Differentiability notes are the natural next stop, and the Class 12 notes hub gathers all subjects, with the main CBSE notes index as the overall map. The differential equations Class 12 notes also apply these same matrix ideas.

Reference: NCERT Class 12 Mathematics Part I textbook, chapter 4, official edition on ncert.nic.in.

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Frequently asked questions on determinants

Why does the NCERT book say |A| is read as determinant of A and not modulus of A?

Because the vertical bars are the symbol for a determinant, and a determinant is a number that can be negative. A modulus is never negative. For example, \(|\begin{smallmatrix} -3 & 2 \\ 5 & 7 \end{smallmatrix}| = -31\) — a negative determinant — so reading it as a modulus would be wrong. The book states this in the remarks of section 4.2.

How do I find the inverse of a 2 × 2 matrix quickly?

Three steps. First compute \(|A| = ad – bc\); if it is zero, the matrix is singular and has no inverse. If it is non-zero, write adj A by interchanging \(a_{11}\) and \(a_{22}\) and changing the signs of \(a_{12}\) and \(a_{21}\), then multiply by \(\frac{1}{|A|}\).

For \(A = \begin{bmatrix} 4 & 5 \\ 2 & 3 \end{bmatrix}\), \(|A| = 2\) and adj \(A = \begin{bmatrix} 3 & -5 \\ -2 & 4 \end{bmatrix}\), so \(A^{-1} = \frac{1}{2}\begin{bmatrix} 3 & -5 \\ -2 & 4 \end{bmatrix}\).

When does a system of linear equations have no solution in the matrix method?

When the coefficient matrix A is singular, \(|A| = 0\). Then compute \((\text{adj }A)B\): if it is not the zero matrix, the system has no solution and is called inconsistent. If \((\text{adj }A)B\) is the zero matrix, the system may be consistent or inconsistent — the determinant alone cannot decide, and the chapter restricts itself to unique-solution systems.

How can determinants be used to find the equation of a line?

Because three collinear points form a triangle of area zero. Put a general point P(x, y) together with the two fixed points as rows \((x, y, 1)\), set the determinant equal to zero, and solve for y. The chapter’s Example 7 does exactly this to get the line through A(1, 3) and B(0, 0) as \(y = 3x\); Exercise 4.2 Q4 repeats the idea.

What is the difference between a matrix and a determinant?

A matrix is an array of numbers arranged in rows and columns; it has no single value. A determinant is one number calculated from a square matrix, written with vertical bars instead of brackets. Only square matrices have determinants, so you can talk about the determinant of a \(3 \times 3\) matrix but never of a \(2 \times 3\) matrix.

What does it mean for a matrix to be singular, and why does it stop the inverse from existing?

A square matrix A is singular when \(|A| = 0\) (Definition 4). The inverse formula \(A^{-1} = \frac{1}{|A|}\text{adj }A\) divides by \(|A|\), so \(|A| = 0\) would mean dividing by zero. The theorem the book states (Theorem 4) is that A is invertible if and only if A is non-singular.