Determinants ISC Class 12: Notes, Properties and Solved Questions

What Determinants is about

Weightage: Determinants is part of the unit Algebra, which carries 10 of the 80 marks in the ISC Mathematics, shared with 1 other chapter. Questions on it can appear in any section of the paper.Source: CISCE ISC Mathematics syllabus, examination year 2027, distribution of marks

Key concepts

Evaluating determinants; minors and cofactors

For a 2 × 2 matrix, $\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=ad-bc$. The minor $M_{ij}$ of an element is the determinant left after deleting its row and column; its cofactor is $C_{ij}=(-1{)}^{i+j}{M}_{ij}$ (signs follow the chessboard pattern + − +). A 3 × 3 determinant equals the sum of the products of the elements of any one row (or column) with their own cofactors: $|A|=a_{11}{C}_{11}+a_{12}{C}_{12}+a_{13}{C}_{13}$. Choose the row or column with most zeros. The sum of products of elements of one row with the cofactors of another row is 0.

Worked example

Evaluate $\left|\begin{array}{ccc}1& 2& 3\\ 0& 4& 5\\ 1& 0& 6\end{array}\right|$ and find the minor and cofactor of the element 2.

Try it first, then show the solutionSolution

1. Expand along R₁: $1(24-0)-2(0-5)+3(0-4)=24+10-12=22$.
2. Element 2 is $a_{12}$: $M_{12}=\left|\begin{array}{cc}0& 5\\ 1& 6\end{array}\right|=-5$; $C_{12}=(-1{)}^3(-5)=5$.

Answer: |A| = 22; $M_{12}=-5$, $C_{12}=5$

Properties of determinants

(1) |A′| = |A| (rows and columns can be interchanged). (2) Interchanging two rows (or columns) changes the sign. (3) If two rows (or columns) are identical or proportional, the determinant is 0. (4) A factor common to a row (or column) can be taken out. (5) If each element of a row is a sum of two terms, the determinant splits into two. (6) Adding a multiple of one row (or column) to another, $R_i\to R_i+k{R}_j$, does not change the value. (7) |AB| = |A||B| and $|kA|=k^n|A|$ for an n × n matrix. Strategy: create a row or column of equal entries or zeros, take out common factors, then expand.

Worked example

Without expanding, show that $\left|\begin{array}{ccc}1& a& b+c\\ 1& b& c+a\\ 1& c& a+b\end{array}\right|=0$.

Try it first, then show the solutionSolution

1. Apply $C_3\to C_3+C_2$: the third column becomes a + b + c in every row.
2. Take out (a + b + c): the third column becomes 1, 1, 1, identical to the first column.
3. A determinant with two identical columns is 0.

Answer: The determinant is 0

Factorising determinants

Many results ask you to show that a determinant equals a product of factors. Use column (or row) subtraction to create common factors: in $\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|$, $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$ produce entries b − a, c − a and $b^2-a^2=(b-a)(b+a)$, from which (b − a) and (c − a) come out. Keep the order of factors and signs careful; check with simple numbers at the end.

Worked example

Show that $\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|=(a-b)(b-c)(c-a)$.

Try it first, then show the solutionSolution

1. $R_2\to R_2-R_1$, $R_3\to R_3-R_1$: rows become (0, b − a, (b − a)(b + a)) and (0, c − a, (c − a)(c + a)).
2. Take out (b − a) and (c − a): $(b-a)(c-a)\left|\begin{array}{ccc}1& a& a^2\\ 0& 1& b+a\\ 0& 1& c+a\end{array}\right|$.
3. Expand along $C_1$: $(b-a)(c-a)[(c+a)-(b+a)]=(b-a)(c-a)(c-b)=(a-b)(b-c)(c-a)$.

Answer: Proved (check: a = 2, b = 3, c = 5 gives 6 both ways)

Area of a triangle and collinearity

For vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$: $\Delta =\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$, and the area is |Δ| (the determinant itself can be negative, so it equals ±2 × area). Three points are collinear if and only if this determinant is 0. When the area is given and a coordinate is unknown, write the determinant = ±2 × area and solve both equations.

Worked example

Find the area of the triangle with vertices (1, 2), (4, 6), (7, 2), and check whether (1, 1), (2, 3), (3, 5) are collinear.

Try it first, then show the solutionSolution

1. Expanding the determinant gives $\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|=\frac{1}{2}|1(6-2)+4(2-2)+7(2-6)|=\frac{1}{2}|4+0-28|=12$.
2. Second set: $1(3-5)+2(5-1)+3(1-3)=-2+8-6=0$, so the determinant is 0.

Answer: Area 12 square units; the points are collinear

Adjoint and inverse of a matrix

The adjoint adj A is the transpose of the matrix of cofactors. Key identity: $A(\text{adj}A)=(\text{adj}A)A=|A|I$. Hence, if |A| ≠ 0, $A^{-1}=\frac{1}{|A|}\text{adj}A$. For a 2 × 2 matrix $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$, adj A = $\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$ (swap the diagonal, change signs of the others). Properties for an n × n matrix: $|\text{adj}A|=|A{|}^{n-1}$, $|A^{-1}|=\frac{1}{|A|}$, adj(AB) = (adj B)(adj A), $|kA|=k^n|A|$.

Worked example

Find the inverse of $A=\left[\begin{array}{ccc}1& 2& 3\\ 0& 4& 5\\ 1& 0& 6\end{array}\right]$.

Try it first, then show the solutionSolution

1. |A| = 22. Cofactors: $C_{11}=24,C_{12}=5,C_{13}=-4$; $C_{21}=-12,C_{22}=3,C_{23}=2$; $C_{31}=-2,C_{32}=-5,C_{33}=4$.
2. adj A (transpose) $=\left[\begin{array}{ccc}24& -12& -2\\ 5& 3& -5\\ -4& 2& 4\end{array}\right]$
3. $A^{-1}=\frac{1}{22}\left[\begin{array}{ccc}24& -12& -2\\ 5& 3& -5\\ -4& 2& 4\end{array}\right]$ (check: A(adj A) = 22I).

Answer: $A^{-1}=\frac{1}{22}\left[\begin{array}{ccc}24& -12& -2\\ 5& 3& -5\\ -4& 2& 4\end{array}\right]$

Solving equations by the matrix method (Martin's rule)

Write the system $a_{1}x+b_{1}y+c_{1}z=d_1$, etc., as AX = B with A the coefficient matrix, X = [x, y, z]′ and B = [d₁, d₂, d₃]′. If |A| ≠ 0, multiply by $A^{-1}$: $X=A^{-1}B=\frac{1}{|A|}(\text{adj}A)B$, giving the unique solution. Always check the solution in one of the original equations.

Worked example

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

Try it first, then show the solutionSolution

1. $A=\left[\begin{array}{ccc}1& 1& 1\\ 2& -1& 1\\ 1& 2& -1\end{array}\right]$, |A| = 1(1 − 2) − 1(−2 − 1) + 1(4 + 1) = −1 + 3 + 5 = 7 ≠ 0.
2. adj A $=\left[\begin{array}{ccc}-1& 3& 2\\ 3& -2& 1\\ 5& -1& -3\end{array}\right]$; (adj A)B = $\left[\begin{array}{c}-6+9+4\\ 18-6+2\\ 30-3-6\end{array}\right]=\left[\begin{array}{c}7\\ 14\\ 21\end{array}\right]$.
3. X = $\frac{1}{7}$[7, 14, 21]′ = [1, 2, 3]′. Check: 1 + 4 − 3 = 2 ✓.

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

Consistency of a system

A system is consistent if it has at least one solution. For AX = B: (1) if |A| ≠ 0, it is consistent with a unique solution; (2) if |A| = 0 and (adj A)B ≠ O, it is inconsistent (no solution); (3) if |A| = 0 and (adj A)B = O, it may be consistent with infinitely many solutions or inconsistent; check by elimination. In two variables, |A| = 0 means the lines are parallel (no solution) or coincident (infinitely many). Whether a unique solution exists depends only on the coefficients, not on the constants.

Worked example

Is the system x + 2y = 3, 2x + 4y = 7 consistent?

Try it first, then show the solutionSolution

1. $|A|=\left|\begin{array}{cc}1& 2\\ 2& 4\end{array}\right|=0$.
2. adj A $=\left[\begin{array}{cc}4& -2\\ -2& 1\end{array}\right]$; (adj A)B $=\left[\begin{array}{c}12-14\\ -6+7\end{array}\right]=\left[\begin{array}{c}-2\\ 1\end{array}\right]\ne O$.

Answer: Inconsistent: the lines are parallel and there is no solution

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

• Solving Equations by Matrix Method (Martin's Rule)
• Properties of Determinants
• Adjoint and Inverse of a Matrix

Determinants formula sheet

2 × 2 determinant

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

Cofactor and expansion

$C_{ij}=(-1{)}^{i+j}{M}_{ij},|A|={\sum }_j{a}_{ij}{C}_{ij}$

Any row or column.

Product and scalar rules

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

n = order.

Area of a triangle

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

Area = |Δ|; collinear if the determinant is 0.

Adjoint identity

$A(\text{adj}A)=(\text{adj}A)A=|A|I$

Inverse

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

|A| ≠ 0.

Adjoint properties

$|\text{adj}A|=|A{|}^{n-1},|A^{-1}|=\frac{1}{|A|},\text{adj}(AB)=(\text{adj}B)(\text{adj}A)$

Matrix method

$AX=B\Rightarrow X=A^{-1}B=\frac{1}{|A|}(\text{adj}A)B$

Unique solution when |A| ≠ 0.

Consistency

$|A|\ne 0:unique;|A|=0,(\text{adj}A)B\ne O:none$

|A| = 0 and (adj A)B = O: infinitely many or none.

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 alternate like a chessboard starting with + at the top left: + − + / − + − / + − +.

  Use it when: Writing cofactors quickly.

• 2 × 2 adjoint: swap and flip

  Swap the diagonal entries, change the signs of the off-diagonal ones.

  Use it when: Fast 2 × 2 inverses.

• Adjoint is transposed

  Cofactor matrix first, then transpose: 'cofactors lie down to become the adjoint'.

  Use it when: Avoiding the most common inverse error.

• k comes out n times

  |kA| = kⁿ|A|: each of the n rows gives up a factor k.

  Use it when: Section A properties.

• Zero test

  |A| ≠ 0: one answer. |A| = 0: look at (adj A)B: non-zero means no answer.

  Use it when: Consistency questions.

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 D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 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: Expanding when the question says 'without expanding'

  CISCE examiners' comments, 2024: in a determinant to be evaluated without expansion, candidates made errors applying the properties or ignored the instruction.

  Wrong working

  Expands the 3 × 3 determinant term by term.

  Right working

  Use row/column operations (e.g. multiply rows to create a common column, or $C_i\to C_i+C_j$) to get identical columns, then state the property.

  ✓Do this: Name each operation you use; the marks are for the method.

  Usually costs: Up to the whole question.

• ✗Mistake: Errors in row and column operations

  CISCE examiners' comments, 2025: in a 'show the determinant is 0' question, candidates made mistakes in row and column operations, multiplied in several places and erred when expanding.

  Wrong working

  Replaces $R_1$ by $2R_1+R_2$ without dividing by 2 afterwards.

  Right working

  Use only $R_i\to R_i+k{R}_j$ (value unchanged); if you multiply a row by k, divide the determinant by k.

  ✓Do this: Write each operation above the arrow.

  Usually costs: 1-2 marks.

• ✗Mistake: Not forming the equations or making cofactor errors

  CISCE examiners' comments, 2024: in a furniture word problem, candidates could not form the system of equations in three variables and made computational errors in finding the inverse.

  Wrong working

  Reads the table by rows when the variables correspond to columns, or forgets cofactor signs.

  Right working

  One equation per resource (row), one variable per product (column); apply the chessboard signs, then transpose.

  ✓Do this: Label rows and columns before writing A; check A(adj A) = |A|I on one entry.

  Usually costs: 2-4 marks.

• ✗Mistake: Thinking constants decide uniqueness

  CISCE examiners' comments, 2025: for x + y = λ, 5x + ky = 2, candidates did not see that only the coefficient determinant decides whether the solution is unique.

  Wrong working

  Unique solution depends on both k and λ.

  Right working

  |A| = k − 5 depends only on k; unique solution iff k ≠ 5.

  ✓Do this: Uniqueness: look at |A| only; consistency when |A| = 0: then look at B.

  Usually costs: 1 mark.

• ✗Mistake: Forgetting to transpose the cofactor matrix

  A frequent slip in finding the inverse: the matrix of cofactors is used directly as the adjoint.

  Wrong working

  adj A = matrix of cofactors.

  Right working

  adj A = (matrix of cofactors)′.

  ✓Do this: Check one entry of A(adj A): it should give |A| on the diagonal and 0 elsewhere.

  Usually costs: 2-3 marks.

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 two rows of a determinant are identical, its value is:
   (a) 1
   (b) 0
   (c) −1
   (d) the product of the diagonal entries

   Hint

   Swap the two rows.

   Answer

   (b) 0

2. Q2 · 1 mark · MCQBoard level

   The system 3x + 2y = 5, 6x + ky = 7 has no unique solution when k equals:
   (a) 2
   (b) 3
   (c) 4
   (d) 6

   Hint

   |A| = 3k − 12.

   Answer

   (c) 4

3. Q3 · 1 mark · MCQBoard level

   If A is 3 × 3 and |A| = 3, then |adj A| is:
   (a) 3
   (b) 6
   (c) 9
   (d) 27

   Hint

   $|A{|}^{n-1}$.

   Answer

   (c) 9

4. Q4 · 1 mark · Very short answerBoard level

   Write the minor and cofactor of the element 5 in $\left|\begin{array}{ccc}1& 2& 3\\ 0& 4& 5\\ 1& 0& 6\end{array}\right|$.

   Hint

   5 is $a_{23}$.

   Answer

   $M_{23}=\left|\begin{array}{cc}1& 2\\ 1& 0\end{array}\right|=-2$; $C_{23}=(-1{)}^5(-2)=2$.

5. Q5 · 2 marks · Short answerBoard level

   Evaluate $\left|\begin{array}{ccc}1& 2& 3\\ 0& 4& 5\\ 1& 0& 6\end{array}\right|$ by expanding along the first column.

   Hint

   Only two non-zero entries.

   Answer

   1(24 − 0) − 0 + 1(10 − 12) = 22

6. Q6 · 2 marks · Short answerHarder board level

   Show that (1, 1), (2, 3), (3, 5) are collinear and find the line through them.

   Hint

   Zero-area determinant.

   Answer

   Determinant = 0, so collinear; line: y = 2x − 1.

7. Q7 · 2 marks · Short answerHarder board level

   Find k if the area of the triangle with vertices (k, 0), (4, 0), (0, 2) is 4 square units.

   Hint

   Two signs.

   Answer

   k = 0 or 8

8. Q8 · 2 marks · Short answerHOTS / competency

   A 3 × 3 matrix has |A| = 5. Find |adj A|, |A⁻¹| and |2A⁻¹|.

   Hint

   $|kB|=k^3|B|$.

   Answer

   25, 1/5, 8/5

9. Q9 · 3 marks · Short answerHarder board level

   Using properties, show that $\left|\begin{array}{ccc}1& a& b+c\\ 1& b& c+a\\ 1& c& a+b\end{array}\right|=0$.

   Hint

   $C_3\to C_3+C_2$.

   Answer

   $C_3\to C_3+C_2$ gives a column (a + b + c)(1, 1, 1); taking out the factor leaves two identical columns, so the value is 0.

10. Q10 · 3 marks · Short answerHarder board level

    Show that $\left|\begin{array}{ccc}a& b& c\\ a^2& b^2& c^2\\ bc& ca& ab\end{array}\right|=(a-b)(b-c)(c-a)(ab+bc+ca)$.

    Hint

    Multiply columns by a, b, c to make the last row abc.

    Answer

    Multiply $C_1,C_2,C_3$ by a, b, c and divide by abc: rows become (a², b², c²), (a³, b³, c³), (abc, abc, abc). Take abc out of $R_3$ (cancels): $\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ a^3& b^3& c^3\\ 1& 1& 1\end{array}\right|$. $C_2-C_1$, $C_3-C_1$ and factorising give (a − b)(b − c)(c − a)(ab + bc + ca). (Check: a = 2, b = 3, c = 5 gives 186 both ways.)

11. Q11 · 3 marks · Short answerHarder board level

    Examine the consistency of (a) x + 2y = 3, 2x + 4y = 7 and (b) x + y + z = 1, x + 2y + 3z = 2, x + 3y + 5z = 3.

    Hint

    |A| = 0 in both; then test.

    Answer

    (a) |A| = 0, (adj A)B = [−2, 1]′ ≠ O: inconsistent. (b) |A| = 0; (2) − (1): y + 2z = 1, (3) − (2): y + 2z = 1: consistent with infinitely many solutions.

12. Q12 · 5 marks · Long answerHarder board level

    (a) Find the inverse of $A=\left[\begin{array}{ccc}2& 1& 3\\ 1& 2& 1\\ 3& 1& 2\end{array}\right]$. [2]
    (b) Hence solve 2x + y + 3z = 520, x + 2y + z = 280, 3x + y + 2z = 560. [3]

    Hint

    |A| = −8.

    Answer

    (a) $A^{-1}=-\frac{1}{8}\left[\begin{array}{ccc}3& 1& -5\\ 1& -5& 1\\ -5& 1& 3\end{array}\right]$. (b) X = $A^{-1}$B = $-\frac{1}{8}$[−960, −320, −640]′ = [120, 40, 80]′: x = 120, y = 40, z = 80.

13. Q13 · 2 marks · Short answerHarder board level

    For $A=\left[\begin{array}{cc}2& 3\\ 1& 4\end{array}\right]$, find adj A and verify A(adj A) = |A|I.

    Hint

    Swap and flip.

    Answer

    adj A = $\left[\begin{array}{cc}4& -3\\ -1& 2\end{array}\right]$; A(adj A) = $\left[\begin{array}{cc}5& 0\\ 0& 5\end{array}\right]$ = 5I and |A| = 8 − 3 = 5.

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.Define the cofactor of $a_{ij}$.Show answer

   $C_{ij}=(-1{)}^{i+j}{M}_{ij}$, where $M_{ij}$ is the minor.

2. 2.What happens to |A| when two rows are interchanged?Show answer

   It changes sign.

3. 3.|kA| for a 3 × 3 matrix?Show answer

   $k^3|A|$.

4. 4.Condition for three points to be collinear?Show answer

   The determinant with rows $(x_i,y_i,1)$ is zero.

5. 5.Formula for $A^{-1}$?Show answer

   $\frac{1}{|A|}\text{adj}A$, |A| ≠ 0.

6. 6.|adj A| for an n × n matrix?Show answer

   $|A{|}^{n-1}$.

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

   When |A| ≠ 0.

8. 8.|A| = 0 and (adj A)B ≠ O means?Show answer

   The system is inconsistent (no solution).

Quick revision checklist

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

• ☐Can you expand a 3 × 3 determinant along the best row?
• ☐Can you write minors and cofactors with correct signs?
• ☐Can you show a determinant is 0 using properties?
• ☐Can you factorise a determinant into (a − b)(b − c)(c − a)?
• ☐Can you find area, unknown vertices and test collinearity?
• ☐Can you find adj A and $A^{-1}$ and verify?
• ☐Can you solve a 3-variable word problem by the matrix method?
• ☐Can you decide consistency when |A| = 0?

Determinants: frequently asked questions