Determinants Class 12: Notes, Matrix Method and Solved Questions

What Determinants is about

Weightage: This chapter is part of Unit II, Algebra, which carries 10 of the 80 theory marks together with Matrices.Source: CBSE Mathematics syllabus 2026-27, Unit II Algebra

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 2.1 marks per paper. See the questions

Key concepts

Determinant of a 2 × 2 and 3 × 3 matrix

NCERT section 4.2

For $A=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]$, $|A|=a_{11}{a}_{22}-a_{12}{a}_{21}$.

For a 3 × 3 matrix, expand along the first row:

$|A|=a_{11}(a_{22}{a}_{33}-a_{23}{a}_{32})-a_{12}(a_{21}{a}_{33}-a_{23}{a}_{31})+a_{13}(a_{21}{a}_{32}-a_{22}{a}_{31})$

You may expand along any row or column (with signs $+-+$ alternating as in a chessboard); the value is the same. Choose the row or column with the most zeros to save work.

Useful results for an n × n matrix A: $|kA|=k^n|A|$; $|A^{\prime }|=|A|$; $|AB|=|A||B|$. A matrix with $|A|=0$ is singular; otherwise non-singular.

Worked example

Evaluate $\left|\begin{array}{ccc}2& -1& 3\\ 1& 1& 1\\ 1& -1& 1\end{array}\right|$. [2 marks]

Try it first, then show the solutionSolution

1. $=2(1\cdot 1-1\cdot (-1))-(-1)(1\cdot 1-1\cdot 1)+3(1\cdot (-1)-1\cdot 1)$
2. $=2(2)+1(0)+3(-2)=4+0-6=-2$

Answer: −2

Area of a triangle using determinants

NCERT section 4.3

The area of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ is

$\Delta =\frac{1}{2}\left|\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|\right|$

Take the absolute value, since area is positive. If an area is given, the determinant can be either + or − twice the area, so you get two equations (and often two answers).

Three points are collinear if and only if this determinant is zero. This also gives the equation of the line through two points $(x_1,y_1)$ and $(x_2,y_2)$: $\left|\begin{array}{ccc}x& y& 1\\ x_1& y_1& 1\\ x_2& y_2& 1\end{array}\right|=0$.

Worked example

Find the area of the triangle with vertices (1, 2), (4, 6) and (7, 2). [2 marks]

Try it first, then show the solutionSolution

1. $\left|\begin{array}{ccc}1& 2& 1\\ 4& 6& 1\\ 7& 2& 1\end{array}\right|=1(6-2)-2(4-7)+1(8-42)=4+6-34=-24$
2. Area $=\frac{1}{2}|-24|=12$ square units

Answer: 12 square units

Minors and cofactors

NCERT section 4.4

The minor $M_{ij}$ of the element $a_{ij}$ is the determinant left after deleting the ith row and jth column. The cofactor is $A_{ij}=(-1{)}^{i+j}{M}_{ij}$: the minor with a sign from the chessboard pattern

$\left[\begin{array}{ccc}+& -& +\\ -& +& -\\ +& -& +\end{array}\right]$

Two facts:

• The determinant equals the sum of the products of the elements of any row (or column) with their own cofactors: $|A|=a_{11}{A}_{11}+a_{12}{A}_{12}+a_{13}{A}_{13}$.
• The sum of the products of the elements of one row with the cofactors of another row is zero: $a_{11}{A}_{21}+a_{12}{A}_{22}+a_{13}{A}_{23}=0$.

Worked example

For $A=\left[\begin{array}{ccc}1& 1& 1\\ 2& -1& 1\\ 1& 2& -1\end{array}\right]$, find the cofactors of the first row and hence |A|. [2 marks]

Try it first, then show the solutionSolution

1. $A_{11}=+\left|\begin{array}{cc}-1& 1\\ 2& -1\end{array}\right|=1-2=-1$; $A_{12}=-\left|\begin{array}{cc}2& 1\\ 1& -1\end{array}\right|=-(-2-1)=3$; $A_{13}=+\left|\begin{array}{cc}2& -1\\ 1& 2\end{array}\right|=4+1=5$
2. $|A|=1(-1)+1(3)+1(5)=7$

Answer: Cofactors −1, 3, 5; |A| = 7

Adjoint of a matrix

NCERT section 4.5

The adjoint of a square matrix A is the transpose of the matrix of its cofactors: $\text{adj }A=[A_{ij}{]}^{\prime }$.

For a 2 × 2 matrix there is a shortcut: if $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$, then $\text{adj }A=\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$ (swap the diagonal, change the signs of the other two).

Key theorem: $A(\text{adj }A)=(\text{adj }A)A=|A|I$. It follows from the two cofactor facts: each diagonal entry of the product is a row times its own cofactors ($|A|$), and each off-diagonal entry is a row times another row's cofactors (0).

Consequences for an n × n matrix: $|\text{adj }A|=|A{|}^{n-1}$ (for n = 3, $|\text{adj }A|=|A{|}^2$).

Worked example

Find adj A for $A=\left[\begin{array}{cc}3& -2\\ 5& 4\end{array}\right]$ and verify $A(\text{adj }A)=|A|I$. [2 marks]

Try it first, then show the solutionSolution

1. $\text{adj }A=\left[\begin{array}{cc}4& 2\\ -5& 3\end{array}\right]$; $|A|=12+10=22$
2. $A(\text{adj }A)=\left[\begin{array}{cc}12+10& 6-6\\ 20-20& 10+12\end{array}\right]=\left[\begin{array}{cc}22& 0\\ 0& 22\end{array}\right]=22I$ ✓

Answer: $\text{adj }A=\left[\begin{array}{cc}4& 2\\ -5& 3\end{array}\right]$; product = 22I

Inverse of a matrix by the adjoint method

NCERT section 4.5

A square matrix A is invertible if and only if it is non-singular ($|A|\ne 0$), and then

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

Steps: (1) find |A| and check it is non-zero; (2) find all nine cofactors; (3) write the cofactor matrix and transpose it to get adj A; (4) divide by |A|; (5) check one or two entries of $A{A}^{-1}=I$.

Related results: $|A^{-1}|=\frac{1}{|A|}$; $(AB{)}^{-1}=B^{-1}{A}^{-1}$; $(A^{\prime }{)}^{-1}=(A^{-1}{)}^{\prime }$.

Worked example

Find $A^{-1}$ for $A=\left[\begin{array}{ccc}1& 1& 1\\ 2& -1& 1\\ 1& 2& -1\end{array}\right]$. [3 marks]

Try it first, then show the solutionSolution

1. |A| = 7 ≠ 0, so $A^{-1}$ exists.
2. Cofactors: row 1: −1, 3, 5; row 2: $A_{21}=-(-1-2)=3$, $A_{22}=(-1-1)=-2$, $A_{23}=-(2-1)=-1$; row 3: $A_{31}=(1+1)=2$, $A_{32}=-(1-2)=1$, $A_{33}=(-1-2)=-3$
3. $\text{adj }A=\left[\begin{array}{ccc}-1& 3& 2\\ 3& -2& 1\\ 5& -1& -3\end{array}\right]$ (transpose of the cofactor matrix)
4. $A^{-1}=\frac{1}{7}\left[\begin{array}{ccc}-1& 3& 2\\ 3& -2& 1\\ 5& -1& -3\end{array}\right]$

Answer: $A^{-1}=\frac{1}{7}\left[\begin{array}{ccc}-1& 3& 2\\ 3& -2& 1\\ 5& -1& -3\end{array}\right]$

Consistency of a system of linear equations

NCERT section 4.6

Write the system as $AX=B$, where A is the coefficient matrix, X the column of unknowns and B the column of constants. A system is consistent if it has at least one solution and inconsistent if it has none.

For a square matrix A:

• If $|A|\ne 0$: the system is consistent with a unique solution $X=A^{-1}B$.
• If $|A|=0$, find $(\text{adj }A)B$:
• if $(\text{adj }A)B\ne O$, the system is inconsistent (no solution);
• if $(\text{adj }A)B=O$, the system may have infinitely many solutions or none (it is consistent with infinitely many solutions when the equations are multiples of each other, as in NCERT's examples).

Worked example

Examine the consistency of $2x+3y=5$ and $4x+6y=7$. [2 marks]

Try it first, then show the solutionSolution

1. $A=\left[\begin{array}{cc}2& 3\\ 4& 6\end{array}\right]$, $|A|=12-12=0$
2. $\text{adj }A=\left[\begin{array}{cc}6& -3\\ -4& 2\end{array}\right]$; $(\text{adj }A)B=\left[\begin{array}{c}30-21\\ -20+14\end{array}\right]=\left[\begin{array}{c}9\\ -6\end{array}\right]\ne O$
3. So the system is inconsistent (the lines are parallel).

Answer: Inconsistent: no solution

Solving equations by the matrix method

NCERT section 4.6

For a system with $|A|\ne 0$:

1. Write A, X, B.
2. Find |A| (non-zero).
3. Find adj A and $A^{-1}=\frac{1}{|A|}\text{adj }A$.
4. Compute $X=A^{-1}B$ by multiplying.
5. Check the solution in one equation.

Sometimes the question gives a product, such as $AB=6I$, which tells you $A^{-1}=\frac{1}{6}B$ directly. Equations in $\frac{1}{x},\frac{1}{y},\frac{1}{z}$ are solved by substituting $u=\frac{1}{x}$ and so on. Word problems (prices, numbers of students) are set up as systems first.

Worked example

Solve by the matrix method: x + y + z = 6, 2x − y + z = 3, x + 2y − z = 2. [5 marks]

Try it first, then show the solutionSolution

1. $A=\left[\begin{array}{ccc}1& 1& 1\\ 2& -1& 1\\ 1& 2& -1\end{array}\right]$, $X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$, $B=\left[\begin{array}{c}6\\ 3\\ 2\end{array}\right]$; |A| = 7 ≠ 0, so a unique solution exists.
2. $A^{-1}=\frac{1}{7}\left[\begin{array}{ccc}-1& 3& 2\\ 3& -2& 1\\ 5& -1& -3\end{array}\right]$ (from the cofactors)
3. $X=A^{-1}B=\frac{1}{7}\left[\begin{array}{c}-6+9+4\\ 18-6+2\\ 30-3-6\end{array}\right]=\frac{1}{7}\left[\begin{array}{c}7\\ 14\\ 21\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$
4. Check: 2(1) − 2 + 3 = 3 ✓

Answer: x = 1, y = 2, z = 3

Maths in real life: Determinants

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Evaluating Determinants
• Adjoint and Inverse of a Matrix
• Solving Linear Equations by Matrix Method

Determinants formula sheet

2 × 2 determinant

$\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=ad-bc$

Expansion by a row

$|A|=a_{11}{A}_{11}+a_{12}{A}_{12}+a_{13}{A}_{13}$

Cofactor

$A_{ij}=(-1{)}^{i+j}{M}_{ij}$

Area of a triangle

$\Delta =\frac{1}{2}\left|\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|\right|$

Zero for collinear points.

Scalar multiple

$|kA|=k^n|A|(A\text{ of order }n)$

Product and transpose

$|AB|=|A||B|,|A^{\prime }|=|A|$

Adjoint

$\text{adj }A=[A_{ij}{]}^{\prime },A(\text{adj }A)=(\text{adj }A)A=|A|I$

Determinant of adjoint

$|\text{adj }A|=|A{|}^{n-1}$

Inverse

$A^{-1}=\frac{1}{|A|}\text{adj }A,|A|\ne 0$

Determinant of inverse

$|A^{-1}|=\frac{1}{|A|}$

Matrix method

$AX=B\Rightarrow X=A^{-1}B(|A|\ne 0)$

How the board asks Determinants

The question patterns to expect, and the method that works for each.

Tricks to remember Determinants

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Chessboard signs

  Cofactor signs follow a chessboard starting with + in the top-left: + − + / − + − / + − +.

  Use it when: Every cofactor.

• 2 × 2 adjoint: swap and flip

  Swap the diagonal entries, flip the signs of the other two.

  Use it when: Fast 2 × 2 inverses.

• Transpose before you divide

  Cofactor matrix → TRANSPOSE → adjoint → divide by |A|. Forgetting the transpose is the most common error.

  Use it when: 3 × 3 inverses.

• k to the n

  Multiplying an n × n matrix by k multiplies its determinant by kⁿ, because every one of the n rows is scaled.

  Use it when: |kA| MCQs.

• adj power is n − 1

  |adj A| = |A|^(n−1): for 3 × 3, square it; for 2 × 2, it is just |A|.

  Use it when: |adj A| MCQs.

Determinants: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Determinants

The slips examiners see most often, and what to write instead.

• ✗Mistake: Forgetting to transpose the cofactor matrix

  Writing cofactors in place.

  ✓Do this: adj A = (cofactor matrix)′. Check: the (1, 2) entry of adj A is the cofactor $A_{21}$.

  Usually costs: Usually 2-3 marks (wrong inverse and solution).

• ✗Mistake: Wrong sign on a cofactor

  Skipping $(-1{)}^{i+j}$.

  ✓Do this: Write the chessboard of signs beside the matrix before computing.

  Usually costs: Every later step.

• ✗Mistake: Not checking |A| ≠ 0 before finding the inverse

  Rushing to cofactors.

  ✓Do this: Write '|A| = … ≠ 0, so A⁻¹ exists' as the first line; it carries a mark.

  Usually costs: 1 mark.

• ✗Mistake: Writing $|kA|=k|A|$

  Treating it like a matrix.

  Wrong working

  A is 3 × 3, |A| = 2: |3A| = 6

  Right working

  |3A| = 3³ × 2 = 54

  ✓Do this: Use $k^n$.

  Usually costs: 1 mark.

• ✗Mistake: Using Cramer's rule or determinant properties

  Older syllabi.

  ✓Do this: Cramer's rule and properties of determinants are not in the 2026-27 syllabus; use expansion and the matrix method.

  Usually costs: Method marks.

NCERT exercise guide: Determinants

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Determinants: previous year questions

• Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 2.1 marks per paper; 13 questions in all, counting OR alternatives. In 3 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Determinants

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

Every Determinants question, by year

Practice questions on Determinants

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   If A is a 2 × 2 matrix with |A| = 3, then |adj A| is:
   (a) 3
   (b) 9
   (c) 1/3
   (d) 6

   Hint

   $|A{|}^{n-1}$ with n = 2.

   Answer

   (a) 3

2. Q2 · 1 mark · MCQBoard level

   If |A| = 5 for a 3 × 3 matrix A, then $|A^{-1}|$ is:
   (a) 5
   (b) 1/5
   (c) 25
   (d) 1/25

   Hint

   $|A^{-1}|=1/|A|$.

   Answer

   (b) 1/5

3. Q3 · 1 mark · Very short answerBoard level

   Evaluate $\left|\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right|$.

   Hint

   Use ${\cos }^2+{\sin }^2=1$.

   Answer

   ${\cos }^2\theta +{\sin }^2\theta =1$

4. Q4 · 1 mark · Assertion-reasonHarder board level

   Assertion (A): The points (1, 2), (3, 6) and (5, 10) are collinear.
   Reason (R): Three points are collinear if the determinant giving the area of the triangle formed by them is zero.
   (a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

   Hint

   All lie on y = 2x.

   Answer

   (a); the determinant is $1(6-10)-2(3-5)+1(30-30)=0$.

5. Q5 · 2 marks · Very short answerHarder board level

   If A and B are 3 × 3 matrices with |A| = 2 and |B| = −3, find |2AB| and $|A^{\prime }{B}^{-1}|$.

   Hint

   $|kM|=k^3|M|$.

   Answer

   $|2AB|=8\times 2\times (-3)=-48$; $|A^{\prime }{B}^{-1}|=2\times \left(-\frac{1}{3}\right)=-\frac{2}{3}$.

6. Q6 · 2 marks · Very short answerHarder board level

   Find the inverse of $\left[\begin{array}{cc}3& -2\\ 5& 4\end{array}\right]$ and verify it.

   Hint

   Swap and flip; |A| = 22.

   Answer

   $A^{-1}=\frac{1}{22}\left[\begin{array}{cc}4& 2\\ -5& 3\end{array}\right]$; $A{A}^{-1}=\frac{1}{22}\left[\begin{array}{cc}22& 0\\ 0& 22\end{array}\right]=I$.

7. Q7 · 2 marks · Very short answerHarder board level

   Examine the consistency of x + 2y = 3, 2x + 4y = 6. If consistent, describe the solutions.

   Hint

   (adj A)B.

   Answer

   |A| = 0; adj A = [[4, −2], [−2, 1]]; (adj A)B = [12 − 12, −6 + 6] = O. The second equation is twice the first, so the system is consistent with infinitely many solutions: x = 3 − 2y for any y.

8. Q8 · 3 marks · Short answerHarder board level

   Find $A^{-1}$ for $A=\left[\begin{array}{ccc}1& 2& 0\\ 0& 1& 1\\ 1& 0& 2\end{array}\right]$ and verify that $A^{-1}A=I$ for the first row.

   Hint

   |A| = 1(2 − 0) − 2(0 − 1) + 0 = 4.

   Answer

   |A| = 4. Cofactors: row 1: 2, 1, −1; row 2: −4, 2, 2; row 3: 2, −1, 1. adj A = $\left[\begin{array}{ccc}2& -4& 2\\ 1& 2& -1\\ -1& 2& 1\end{array}\right]$, $A^{-1}=\frac{1}{4}\text{adj }A$. Row 1 of $A^{-1}A$: $\frac{1}{4}[2+0+2,4-4+0,0-4+4]=[1,0,0]$ ✓.

9. Q9 · 5 marks · Long answerHarder board level

   The sum of three numbers is 12. The first number plus twice the second is 11, and the first number is 2 less than the third. Form the equations and solve them by the matrix method.

   Hint

   x + y + z = 12, x + 2y = 11, x − z = −2.

   Answer

   $A=\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 0\\ 1& 0& -1\end{array}\right]$, $B=\left[\begin{array}{c}12\\ 11\\ -2\end{array}\right]$, $|A|=1(-2)-1(-1)+1(-2)=-3\ne 0$. Cofactors: row 1: −2, 1, −2; row 2: 1, −2, 1; row 3: −2, 1, 1. $\text{adj }A=\left[\begin{array}{ccc}-2& 1& -2\\ 1& -2& 1\\ -2& 1& 1\end{array}\right]$. $X=-\frac{1}{3}\left[\begin{array}{c}-24+11+4\\ 12-22-2\\ -24+11-2\end{array}\right]=-\frac{1}{3}\left[\begin{array}{c}-9\\ -12\\ -15\end{array}\right]=\left[\begin{array}{c}3\\ 4\\ 5\end{array}\right]$. The numbers are 3, 4 and 5 (check: 3 + 4 + 5 = 12, 3 + 8 = 11, 3 − 5 = −2).

10. Q10 · 2 marks · Very short answerHOTS / competency

    If $A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$, show that $|A^n|=1$ for every natural n, and find adj A and $A^{-1}$.

    Hint

    $|A^n|=|A{|}^n$.

    Answer

    |A| = 1, so $|A^n|=1^n=1$. adj A = [[1, −1], [0, 1]] and $A^{-1}$ = adj A since |A| = 1.

11. Q11 · 3 marks · Short answerHOTS / competency

    Show that the matrix $A=\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]$ satisfies $A^2-4A+I=O$, and use this to find $A^{-1}$. Check with the adjoint method.

    Hint

    Multiply by $A^{-1}$.

    Answer

    $A^2=\left[\begin{array}{cc}7& 12\\ 4& 7\end{array}\right]$; $A^2-4A+I=\left[\begin{array}{cc}7-8+1& 12-12\\ 4-4& 7-8+1\end{array}\right]=O$. So $A^{-1}=4I-A=\left[\begin{array}{cc}2& -3\\ -1& 2\end{array}\right]$. Adjoint method: |A| = 1, adj A = [[2, −3], [−1, 2]] ✓.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Write the formula for the inverse using the adjoint.Show answer

   $A^{-1}=\frac{1}{|A|}\text{adj }A$, provided $|A|\ne 0$.

2. 2.If A is 3 × 3 and |A| = 4, find |adj A|.Show answer

   $4^2=16$.

3. 3.If A is 2 × 2 and |A| = 5, find |3A|.Show answer

   $9\times 5=45$.

4. 4.When are three points collinear?Show answer

   When the area determinant is zero.

5. 5.Cofactor sign of $a_{23}$?Show answer

   $(-1{)}^5=-1$.

6. 6.Adjoint of $\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$?Show answer

   $\left[\begin{array}{cc}4& -2\\ -3& 1\end{array}\right]$.

7. 7.When does AX = B have a unique solution?Show answer

   When |A| ≠ 0.

8. 8.If |A| = 0 and (adj A)B ≠ O, what kind of system is it?Show answer

   Inconsistent (no solution).

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you evaluate a 3 × 3 determinant by expanding along the easiest row?
• ☐Can you find |kA|, |adj A| and |A⁻¹| from |A|?
• ☐Can you find the area of a triangle and the value of k for a given area?
• ☐Can you test three points for collinearity?
• ☐Can you find all nine cofactors of a 3 × 3 matrix without sign errors?
• ☐Can you write adj A as the transpose of the cofactor matrix?
• ☐Can you find $A^{-1}$ and verify $A{A}^{-1}=I$?
• ☐Can you decide whether a system is consistent using |A| and (adj A)B?
• ☐Can you solve a 3-variable system by $X=A^{-1}B$?
• ☐Can you set up a word problem as AX = B?

Determinants: frequently asked questions