Chapter 4 · Determinants · Go for it

Determinants, the quick way

A determinant is one number made from a square matrix. Learn ten moves and you can answer most questions in this chapter.

Start here

The chapter in one minute

What a determinant is, and what the exam asks.

A determinant is one number worked out from a square matrix. You write it between straight bars, like $|A|$ ⁠.

That one number tells you a lot. If $|A| \ne 0$ ⁠, the matrix has an inverse, and a system of equations with matrix $A$ has exactly one solution.

Half a determinant also gives the area of a triangle. Take its size, because an area is never negative.

There are no long proofs here. The marks come from careful arithmetic: the right signs, the right order, and every line written down. Most exam questions on this chapter are one of ten kinds, and each kind has a fixed recipe.

10question types

13formula cards

+ − +the sign pattern behind every cofactor

1check before every inverse: is

|A| \ne 0

⁠?

Formulas

The formulas you must know

Learn these cards. Every question in the chapter uses them.

2\times2

determinant

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

Main diagonal product minus the other diagonal product.

3\times3

determinant

\begin{aligned} \Delta &= a_{11}A_{11} + a_{12}A_{12} \\ &\quad + a_{13}A_{13} \end{aligned}

Each entry of a row times its cofactor, added. Any row or column gives the same answer.

Cofactor

A_{ij} = (-1)^{i + j}M_{ij}

$M_{ij}$ is the minor: cover row $i$ and column $j$ ⁠, and find the determinant that is left. The signs go + − + along row 1, then − + −, then + − +.

Area of a triangle

\Delta = \frac12\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}

The area is $|\Delta|$ square units. Three points on one line give zero.

A number times a matrix

|kA| = k^n|A|

$n$ is the order of $A$ ⁠. For a $3\times3$ matrix, $|2A| = 8|A|$ ⁠.

A product

|AB| = |A|\,|B|

For square matrices of the same order.

Adjoint of a

2\times2

\operatorname{adj}\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}

Swap the two diagonal entries, and change the sign of the other two.

Adjoint and

|A|

\begin{aligned} &A(\operatorname{adj}A) = (\operatorname{adj}A)A \\ &= |A|\,I \end{aligned}

$\operatorname{adj}A$ is the matrix of cofactors, transposed: the cofactors of row 1 go down column 1.

Inverse

A^{-1} = \frac{1}{|A|}\operatorname{adj}A

Only when $|A| \ne 0$ ⁠. A matrix with determinant 0 has no inverse.

Determinant of the adjoint

|\operatorname{adj}A| = |A|^{n - 1}

$n$ is the order. For a $3\times3$ matrix, $|\operatorname{adj}A| = |A|^2$ ⁠.

Determinant of the inverse

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

True for every invertible matrix.

Inverse of a product

(AB)^{-1} = B^{-1}A^{-1}

For invertible $A$ and $B$ of the same order. Reverse the order.

Matrix method

X = A^{-1}B

The one solution of $AX = B$ when $|A| \ne 0$ ⁠.

Question types

The question types and how to crack each one

For each type: how to spot it, what to do, and one real NCERT question worked the exam way.

1. Evaluate a determinant: cross for $2\times2$ ⁠, expand for $3\times3$

Spot it: "evaluate" and a square block of numbers between straight bars.

For a $2\times2$ ⁠: main diagonal product minus the other diagonal product.
For a $3\times3$ ⁠: choose the row or column with the most zeros.
Take each entry, times its sign, times the $2\times2$ left when you cover its row and column. The signs are + − + for row 1 or 3 (and column 1 or 3), and − + − for row 2 (and column 2).
Add the three terms.

Example 3. Evaluate

\Delta = \begin{vmatrix} 1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0 \end{vmatrix}

Expanding along $C_3$ ⁠, which has two zeros:

\begin{aligned} \Delta &= 4\begin{vmatrix} -1 & 3 \\ 4 & 1 \end{vmatrix} - 0\begin{vmatrix} 1 & 2 \\ 4 & 1 \end{vmatrix} \\ &\quad + 0\begin{vmatrix} 1 & 2 \\ -1 & 3 \end{vmatrix} \\ &= 4(-1 - 12) - 0 + 0 \\ &= 4(-13) \end{aligned}

∴ $\Delta = -52$

See a full exam answer · Go deeper: see it visually · Basics: the + − + sign pattern

In past papers (6): click to see

Board 2026 Q4 · Board 2026 Q5 · Board 2026 Q33(b) · Board 2024 Q12 · Board 2024 Q19 · Sample 2025-26 Q6

2. Find $x$ ⁠: two determinants set equal

Spot it: two determinants with an equals sign between them, and $x$ inside.

Work out each side with the $2\times2$ rule.
Set the two answers equal and solve for $x$ ⁠.
If you reach $x^2$ equal to a positive number, there are two answers: plus and minus.

Exercise 4.1 Q7 (i). Find $x$ if

\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}

\begin{aligned} &\Rightarrow\ 2 - 20 = 2x^2 - 24 \\ &\Rightarrow\ 2x^2 = 6 \\ &\Rightarrow\ x^2 = 3 \end{aligned}

∴ $x = \pm\sqrt3$

See a full exam answer · Go deeper: see it visually · Basics: the algebra you need

In past papers (1): click to see

Sample 2022-23 Q12

3. Area of a triangle: half a determinant

Spot it: three corners (vertices) of a triangle, and the word "area".

Write each vertex as a row: $x$ ⁠, $y$ ⁠, 1.
$\Delta$ is half of that determinant. Expand it.
The area is $|\Delta|$ ⁠, in square units. It is never negative.

Example 6. Find the area of the triangle with vertices $(3, 8)$ ⁠, $(-4, 2)$ and $(5, 1)$ ⁠.

\Delta = \frac12\begin{vmatrix} 3 & 8 & 1 \\ -4 & 2 & 1 \\ 5 & 1 & 1 \end{vmatrix}

Expanding along $R_1$ ⁠:

\begin{aligned} \Delta &= \frac12\big[3(2 - 1) \\ &\quad - 8(-4 - 5) \\ &\quad + 1(-4 - 10)\big] \\ &= \frac12(3 + 72 - 14) \\ &= \frac{61}{2} \end{aligned}

∴ $\text{Area} = \frac{61}{2}$ sq. units

See a full exam answer · Go deeper: see it visually · Basics: determinant as area

4. Area given, or points on one line: use $\pm$ or 0

Spot it: the area is given and a vertex has a $k$ in it; or "collinear"; or "find the equation of the line using determinants".

Area given: set half the determinant equal to plus or minus the area.
Solve both equations. You usually get two values.
Collinear points: show the determinant is 0. The line through two points: use $(x, y)$ as the third point and set the determinant equal to 0.

Exercise 4.2 Q3 (i). Find $k$ if the area of the triangle is 4 sq. units and the vertices are $(k, 0)$ ⁠, $(4, 0)$ ⁠, $(0, 2)$ ⁠.

\frac12\begin{vmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix} = \pm 4

Expanding along $C_2$ ⁠:

\begin{aligned} &\frac12\big[-2(k - 4)\big] = \pm 4 \\ &\Rightarrow\ 4 - k = \pm 4 \end{aligned}

∴ $k = 0$ or $k = 8$

See a full exam answer · Go deeper: see it visually · Basics: determinant as area

In past papers (2): click to see

Sample 2024-25 Q6 · Sample 2023-24 Q3

5. Minors and cofactors: cover, work out, add the sign

Spot it: the words "minor" or "cofactor", or "using cofactors, evaluate".

Minor $M_{ij}$ ⁠: cover row $i$ and column $j$ ⁠, and work out the $2\times2$ that is left.
Cofactor: the minor with its sign from the + − + pattern.
To evaluate with cofactors, multiply each entry of the row by its own cofactor, and add.

Exercise 4.3 Q3. Using cofactors of the second row, evaluate

\Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix}

\begin{aligned} A_{21} &= -\begin{vmatrix} 3 & 8 \\ 2 & 3 \end{vmatrix} \\ &= -(9 - 16) = 7 \\ A_{22} &= \begin{vmatrix} 5 & 8 \\ 1 & 3 \end{vmatrix} \\ &= 15 - 8 = 7 \\ A_{23} &= -\begin{vmatrix} 5 & 3 \\ 1 & 2 \end{vmatrix} \\ &= -(10 - 3) = -7 \end{aligned}

Expanding along $R_2$ ⁠:

\begin{aligned} \Delta &= 2(7) + 0(7) + 1(-7) \\ &= 14 + 0 - 7 \end{aligned}

∴ $\Delta = 7$

See a full exam answer · Go deeper: see it visually · Basics: the + − + sign pattern

6. Inverse of a matrix: $|A|$ ⁠, then the adjoint, then divide

Spot it: "find the inverse" or "find $A^{-1}$ ".

Find $|A|$ ⁠. If it is 0, there is no inverse. If not, write " $|A| \ne 0$ ⁠, so $A^{-1}$ exists".
Find every cofactor. For a $3\times3$ there are nine.
Write $\operatorname{adj}A$ ⁠: the cofactors of row 1 go down column 1, and so on.
Divide: $A^{-1} = \frac{1}{|A|}\operatorname{adj}A$ ⁠.

Exercise 4.4 Q5. Find the inverse of

A = \begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}

\begin{aligned} |A| &= 2(3) - 4(-2) \\ &= 6 + 8 = 14 \end{aligned}

Since $|A| \ne 0$ ⁠, $A^{-1}$ exists.

Cofactors: $A_{11} = 3$ ⁠, $A_{12} = -4$ ⁠, $A_{21} = 2$ ⁠, $A_{22} = 2$ ⁠.

\operatorname{adj}A = \begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}

∴ $A^{-1} = \frac{1}{14}\begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}$

See a full exam answer · Go deeper: see it visually · Basics: why the inverse divides by $|A|$

In past papers (4): click to see

Board 2025 Q1 · Board 2023 Q33(a) · Sample 2026-27 Q26 · Sample 2025-26 Q5

7. A matrix equation, then "hence find $A^{-1}$ "

Spot it: "show that $A^2 - 5A + 7I = O$ " (or a similar equation), then "hence find $A^{-1}$ ".

Work out $A^2$ ⁠, then the left side, entry by entry. Show that you get $O$ ⁠.
Write " $|A| \ne 0$ ", then multiply the whole equation by $A^{-1}$ ⁠: $A^2$ becomes $A$ ⁠, $A$ becomes $I$ ⁠, and $I$ becomes $A^{-1}$ ⁠.
Make $A^{-1}$ the subject and work it out.

Exercise 4.4 Q13. Show that $A^2 - 5A + 7I = O$ ⁠, and hence find $A^{-1}$ ⁠, for

A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}

A^2 = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}

\begin{aligned} &A^2 - 5A + 7I \\ &= \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} \\ &\quad + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = O \end{aligned}

As $|A| = 7 \ne 0$ ⁠, multiply by $A^{-1}$ ⁠:

A - 5I + 7A^{-1} = O

A^{-1} = \frac17(5I - A)

∴ $A^{-1} = \frac17\begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$

See a full exam answer · Go deeper: see it visually · Basics: matrix facts the chapter uses

In past papers (2): click to see

Board 2023 Q1 · Sample 2020-21 Q20(b)

8. Consistency: does the system have a solution?

Spot it: "examine the consistency of the system of equations".

Write the system as $AX = B$ ⁠.
If $|A| \ne 0$ ⁠, the system is consistent, with exactly one solution.
If $|A| = 0$ ⁠, find $(\operatorname{adj}A)B$ ⁠. If it is not $O$ ⁠, the system is inconsistent: no solution.
If $(\operatorname{adj}A)B = O$ too, the system may have infinitely many solutions or none.

Exercise 4.5 Q3. Examine the consistency of

\begin{aligned} x + 3y &= 5 \\ 2x + 6y &= 8 \end{aligned}

This is $AX = B$ with

\begin{aligned} A &= \begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix} \\ B &= \begin{bmatrix} 5 \\ 8 \end{bmatrix} \end{aligned}

|A| = 6 - 6 = 0

So $A$ is singular, and we find $(\operatorname{adj}A)B$ ⁠.

\begin{aligned} &(\operatorname{adj}A)B \\ &= \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}\begin{bmatrix} 5 \\ 8 \end{bmatrix} \\ &= \begin{bmatrix} 6 \\ -2 \end{bmatrix} \ne O \end{aligned}

∴ The system is inconsistent.

See a full exam answer · Go deeper: see it visually · Basics: two lines that cross, are parallel or are the same

9. Solve equations by the matrix method

Spot it: "solve, using the matrix method", with two or three equations. A word problem (prices, three numbers) is the same: write the equations first.

Write $AX = B$ ⁠, showing $A$ ⁠, $X$ and $B$ ⁠.
Find $|A|$ and write " $|A| \ne 0$ ⁠, so there is a unique solution".
Find $\operatorname{adj}A$ and $A^{-1}$ ⁠, exactly as in type 6.
Work out $X = A^{-1}B$ and read off $x$ ⁠, $y$ (and $z$ ⁠).

Exercise 4.5 Q7. Solve by the matrix method

\begin{aligned} 5x + 2y &= 4 \\ 7x + 3y &= 5 \end{aligned}

\begin{aligned} A &= \begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix} \\ X &= \begin{bmatrix} x \\ y \end{bmatrix} \\ B &= \begin{bmatrix} 4 \\ 5 \end{bmatrix} \end{aligned}

$|A| = 15 - 14 = 1$ ⁠. As $|A| \ne 0$ ⁠, the solution is unique.

\begin{aligned} A^{-1} &= \frac{1}{|A|}\operatorname{adj}A \\ &= \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix} \end{aligned}

\begin{aligned} X &= A^{-1}B \\ &= \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}\begin{bmatrix} 4 \\ 5 \end{bmatrix} \\ &= \begin{bmatrix} 12 - 10 \\ -28 + 25 \end{bmatrix} \end{aligned}

∴ $x = 2$ ⁠, $y = -3$

See a full exam answer · Go deeper: see it visually

In past papers (11): click to see

Board 2026 Q33(a) · Board 2025 Q35 · Board 2024 Q32(a) · Board 2023 Q33(b) · Sample 2026-27 Q38 · Sample 2025-26 Q32 · Sample 2024-25 Q33 · Sample 2023-24 Q34 · Sample 2022-23 Q35 · Sample 2020-21 Q36(a) · Sample 2020-21 Q36(b)

10. One-mark questions on $|kA|$ ⁠, $|\operatorname{adj}A|$ and $|A^{-1}|$

Spot it: an MCQ about a matrix you are not given, only its order.

Find the card that fits: $|kA|$ ⁠, $|AB|$ ⁠, $|\operatorname{adj}A|$ or $|A^{-1}|$ ⁠.
Put in the order $n$ of the matrix.
Choose the option. In an MCQ only the right letter earns the mark.

Exercise 4.4 Q17 (MCQ). $A$ is a non-singular $3\times3$ matrix. Then $|\operatorname{adj}A|$ is (A) $|A|$ (B) $|A|^2$ (C) $|A|^3$ (D) $3|A|$ ⁠.

Use the card with $n = 3$ ⁠:

\begin{aligned} |\operatorname{adj}A| &= |A|^{3 - 1} \\ &= |A|^2 \end{aligned}

Answer: (B) $|A|^2$

See a full exam answer · Go deeper: see it visually

In past papers (15): click to see

Board 2025 Q6 · Board 2025 Q15 · Board 2023 Q2 · Board 2023 Q3 · Sample 2026-27 Q5 · Sample 2025-26 Q4 · Sample 2024-25 Q1 · Sample 2024-25 Q4 · Sample 2023-24 Q2 · Sample 2023-24 Q10 · Sample 2023-24 Q13 · Sample 2022-23 Q2 · Sample 2022-23 Q13 · Sample 2020-21 Q5(b) · Sample 2020-21 Q20(a)

Shortcuts

Shortcuts and quick tips

Small habits that save time and marks.

Expand along the row or column with the most zeros. A zero entry needs no working at all.
The sign of a cofactor depends only on its place, never on the entry: + − + in row 1, − + − in row 2, + − + in row 3.
Put brackets round negative entries before you multiply: $2(-1) - 4(-5) = 18$ ⁠.
For a $2\times2$ inverse, use the shortcut: swap the diagonal, change the sign of the other two, and divide by $|A|$ ⁠.
Check an inverse: $AA^{-1} = I$ ⁠. For a $2\times2$ it takes a minute.
Check a solution: put your $x$ ⁠, $y$ ⁠, $z$ back into every equation.
If the area determinant comes out negative, the area is its size: $|-15| = 15$ ⁠.
For a $3\times3$ matrix, $|3A| = 27|A|$ ⁠, not $3|A|$ ⁠. The number is raised to the power of the order.
In an MCQ on $|kA|$ ⁠, $|\operatorname{adj}A|$ or $|A^{-1}|$ ⁠, use the cards. No long working is needed.
If the question gives a product $AB = I$ ⁠, then $A^{-1} = B$ ⁠, with no cofactors. If $AB = kI$ with $k \ne 0$ ⁠, then $A^{-1} = \frac1k B$ ⁠.

Watch out

Mistakes that cost marks

Each one is common, and each has an easy fix.

Making a determinant positive → $|A|$ is a determinant, not a size, and can be negative. Only the area takes the modulus.
Using the minor as the cofactor → add the sign: $A_{12} = -M_{12}$ ⁠.
Forgetting to transpose the cofactors → in $\operatorname{adj}A$ ⁠, the cofactors of row 1 go down column 1.
Writing $|2A| = 2|A|$ for a $3\times3$ matrix → it is $|2A| = 8|A|$ ⁠.
Using only plus when the area is given → use $\pm$ ⁠, or you may lose a value of $k$ ⁠.
Finding $A^{-1}$ without looking at $|A|$ → first write " $|A| \ne 0$ ⁠, so $A^{-1}$ exists". It carries half a mark.
Writing $X = BA^{-1}$ → the order matters: $X = A^{-1}B$ ⁠.
Saying " $|A| = 0$ ⁠, so no solution" → first find $(\operatorname{adj}A)B$ ⁠. If it is not $O$ ⁠, there is no solution. If it is $O$ ⁠, there may be infinitely many solutions or none.

Marks

How the examiner gives marks

From CBSE's own marking schemes. Write these lines and the marks follow.

Matrix method (a 5-mark question in CBSE sample papers): writing the system as $AX = B$ ⁠, with $A$ ⁠, $X$ and $B$ shown, earns marks.
Finding $|A|$ is half a mark. The line " $|A| \ne 0$ ⁠, so $A^{-1}$ exists" is another half mark.
The adjoint carries the most: one scheme gives it one and a half marks. Work out every cofactor with its sign.
Then $A^{-1}$ ⁠, the line $X = A^{-1}B$ ⁠, the product and the values of $x$ ⁠, $y$ ⁠, $z$ each earn their own part. The final answer is worth up to 1 mark.
Write conclusions in words, such as "so the system has a unique solution". End an area with "sq. units".
A slip is penalised only once, so keep going. An MCQ earns 1 mark for the right option only.

Practice

Practise these first

Easiest first. Together they cover every type above.

Exercise 4.1 Q8 (MCQ) two equal determinants give $x^2 = 36$
Exercise 4.1 Q5 four $3\times3$ determinants: expand along the zeros
Exercise 4.2 Q1 areas of three triangles
Exercise 4.2 Q4 the equation of a line: area 0
Exercise 4.3 Q2 minors and cofactors of a $3\times3$
Exercise 4.4 Q18 (MCQ) the determinant of an inverse
Example 15 a matrix equation gives $A^{-1}$
Exercise 4.4 Q10 a $3\times3$ inverse: nine cofactors
Exercise 4.5 Q5 consistency when $|A| = 0$
Exercise 4.5 Q12 three equations by the matrix method
Example 19 a given product gives $A^{-1}$ ⁠, then solve
Exercise 4.5 Q16 a word problem: prices by the matrix method

You can do this

Ten moves, one at a time. Work through the practice list, check every sign, and you are ready for the exam.

Ready to go deeper? Open the interactive lesson

Past papers

34 CBSE board and sample-paper questions on Determinants, each with a full-marks exam answer. Practise them →