MathsISC Class 12 · Determinants

Solving Linear Equations by the Matrix Method (Martin's Rule)

Updated 15 October 2026 · Fact-checked

Write the equations as AX = B, with A the coefficient matrix, X the unknowns and B the constants. If |A| ≠ 0, the system is consistent with the unique solution X = A⁻¹B = (adj A)B/|A|. If |A| = 0, find (adj A)B: if it is non-zero, the system has no solution.

First test |A|: non-zero gives the unique answer X = A⁻¹B; if zero, a non-zero (adj A)B means no solution.

Illustration drawn for QuestionBuddy.

Understand Solving Equations by Matrix Method (Martin's Rule)

A system of linear equations such as
$a_{1} x + b_{1} y + c_{1} z = d_{1}, a_{2} x + b_{2} y + c_{2} z = d_{2}, a_{3} x + b_{3} y + c_{3} z = d_{3}$
can be written as one matrix equation $A X = B$ , where
$A = a_{1} a_{2} a_{3} b_{1} b_{2} b_{3} c_{1} c_{2} c_{3}, X = x y z, B = d_{1} d_{2} d_{3} .$
The CISCE syllabus calls solving this with the inverse Martin's rule. Each row of $A$ holds the coefficients of one equation, in the same order of variables, and missing variables get a 0.

The method. If $∣ A ∣ \neq = 0$ , then $A^{- 1}$ exists and multiplying $A X = B$ on the left by $A^{- 1}$ gives
$X = A^{- 1} B = \frac{1}{∣ A ∣} (adj A) B .$
So you need three things: $∣ A ∣$ , the adjoint (transpose of the matrix of cofactors), and one matrix product. The answer is unique, because the inverse is unique.

Consistency. A system is consistent if it has at least one solution and inconsistent if it has none. The syllabus asks for the conditions using matrices:

$∣ A ∣ \neq = 0$ : consistent, with a unique solution.
$∣ A ∣ = 0$ and $(adj A) B \neq = O$ : inconsistent (no solution).
$∣ A ∣ = 0$ and $(adj A) B = O$ : the system may have infinitely many solutions or none; check by trying to solve the equations directly. For two variables (with $A$ not the zero matrix), it always has infinitely many.

Notice that whether a solution is unique depends only on $A$ , not on $B$ . In ISC 2025, an MCQ asked what the uniqueness of the solution of $x + y = λ$ , $5 x + k y = 2$ depends on; the answer is $k$ alone, since $∣ A ∣ = k - 5$ . The examiners noted that candidates did not see how the determinant decides existence and uniqueness.

Why the test works. If $∣ A ∣ = 0$ , then $A (adj A) = (adj A) A = ∣ A ∣ I = O$ . Multiplying $A X = B$ by $adj A$ gives $O = (adj A) B$ . So if $(adj A) B \neq = O$ , no $X$ can exist.

A shortcut from a given product. Some questions give a second matrix $B^{'}$ and ask you to find $A B^{'}$ first. If $A B^{'} = k I$ , then $A^{- 1} = \frac{1}{k} B^{'}$ , and you can solve the system without computing any cofactors. This is a favourite way of shortening a 5-mark question.

Word problems. Most real-life questions in the ISC paper (for example the 2024 furniture-factory question on three kinds of wood) need you to form the equations first. Choose clear variable names, write one equation per condition with the same order of variables, and only then build $A$ and $B$ . The 2024 examiners reported that candidates lost marks both in forming the system and in computing the inverse.

Solving Equations by Matrix Method (Martin's Rule): formulas to remember

Matrix form: $A X = B$; Rows of A: coefficients of each equation, in a fixed order of variables.
Martin's rule: $X = A^{- 1} B = \frac{1}{∣ A ∣} (adj A) B$; Valid when |A| ≠ 0; the solution is unique.
Inverse: $A^{- 1} = \frac{adj A}{∣ A ∣}, adj A = [C_{ij}]^{'}$; Adjoint = transpose of the matrix of cofactors.
Unique solution: $∣ A ∣ \neq = 0$; Depends only on the coefficients, not on B.
No solution: $∣ A ∣ = 0, (adj A) B \neq = O$; Inconsistent system.
Further check needed: $∣ A ∣ = 0, (adj A) B = O$; Infinitely many solutions or none; solve directly to decide.
Inverse from a product: $A B^{'} = k I \Rightarrow A^{- 1} = \frac{1}{k} B^{'}$; Use when the question supplies a second matrix.

How to solve Solving Equations by Matrix Method (Martin's Rule) questions

This order of work earns the method marks at each stage of a Martin's rule question.

1Arrange every equation with the variables in the same order ( $x$ , $y$ , $z$ ) and the constant on the right. Put 0 for a missing variable.
2Write $A$ , $X$ and $B$ , and state "the system is $A X = B$ ".
3Find $∣ A ∣$ . If it is non-zero, write " $A$ is non-singular, so the system has a unique solution".
4Find all nine cofactors $C_{ij} = (- 1)^{i + j} M_{ij}$ , then transpose to get $adj A$ . Check one entry of $A (adj A) = ∣ A ∣ I$ .
5Compute $X = \frac{1}{∣ A ∣} (adj A) B$ , keeping the scalar outside until the end.
6Substitute the answer into at least one original equation to check it.
7If $∣ A ∣ = 0$ , compute $(adj A) B$ : non-zero means no solution; zero means test further by solving the equations directly.

Solving Equations by Matrix Method (Martin's Rule) in real life

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.

Three-variable system by Martin's rule

Suppose three friends order at a café. Asha has 2 coffees, 1 sandwich and 1 muffin for Rs 250; Ben has 1 coffee, 2 sandwiches and 1 muffin for Rs 280; Charu has 1 coffee, 1 sandwich and 2 muffins for Rs 230.

Find the price of each item using the matrix method.

Try it first, then show the solutionSolution

Let the prices be $x, y, z$ . Then $A = 211121112$ , $B = 250280230$ .
$∣ A ∣ = 2 (4 - 1) - 1 (2 - 1) + 1 (1 - 2) = 6 - 1 - 1 = 4 \neq = 0$ .
$adj A = 3 - 1 - 1 - 1 3 - 1 - 1 - 1 3$ (the cofactor matrix is symmetric here).
$X = \frac{1}{4} 750 - 280 - 230 - 250 + 840 - 230 - 250 - 280 + 690 = \frac{1}{4} 240360160 = 609040$ .
Check: $2 (60) + 90 + 40 = 250$ .

Answer: Coffee Rs 60, sandwich Rs 90, muffin Rs 40.

Memory hook: Three bills, three prices: one inverse matrix solves them all at once.

Two-variable system and its matrix

Imagine your mobile bill is a fixed monthly charge plus a charge per GB of data. In one month you used 10 GB and paid Rs 350; the next month you used 16 GB and paid Rs 440.

Find the fixed charge and the charge per GB by the matrix method.

Try it first, then show the solutionSolution

Let the fixed charge be $a$ and the rate $b$ . Then $a + 10 b = 350$ , $a + 16 b = 440$ : $A = [111016]$ , $B = [350440]$ .
$∣ A ∣ = 16 - 10 = 6$ , $A^{- 1} = \frac{1}{6} [16 - 1 - 10 1]$ .
$X = \frac{1}{6} [5600 - 4400 - 350 + 440] = \frac{1}{6} [120090] = [20015]$ .

Answer: Fixed charge Rs 200; Rs 15 per GB.

Memory hook: Bill = fixed + rate × use: two months give two equations, and A⁻¹ splits the bill.

Tricks to remember Solving Equations by Matrix Method (Martin's Rule)

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

A-X-B line-up
Coefficients in A, unknowns in X, constants in B, all in the same order. "A times X is B, so X is A-inverse B."
Use it when: Setting up and the first line of every answer.
Determinant decides
|A| ≠ 0: one answer. |A| = 0: look at (adj A)B; non-zero means no answer.
Use it when: Consistency questions.
Transpose the cofactors
Adjoint = cofactors placed in transposed positions. Cofactor of row 1 goes into column 1.
Use it when: Avoiding the commonest error in finding A⁻¹.
Check with one equation
Always put your x, y, z back into the first equation; it takes ten seconds.
Use it when: Catching sign errors before the examiner does.

Solving Equations by Matrix Method (Martin's Rule): solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: An inconsistent pair of equations · 1 mark · MCQBoard level

The system $x + 3 y = 4$ , $2 x + 6 y = 9$ is:
(a) consistent with a unique solution
(b) consistent with infinitely many solutions
(c) inconsistent
(d) consistent only if $x = 0$

Try it first, then show the solutionSolution

$∣ A ∣ = 6 - 6 = 0$ . $(adj A) B = [6 - 2 - 3 1] [49] = [- 3 1] \neq = O$ .

Answer: (c) inconsistent. (The left side of the second equation is twice the first, but 9 is not twice 4.)

Example 2: Two variables by the matrix method · 3 marks · Short answerBoard level

Solve by the matrix method: $3 x - 2 y = 4$ , $5 x + 3 y = 13$ .

Try it first, then show the solutionSolution

$A = [35 - 2 3]$ , $B = [413]$ . $∣ A ∣ = 9 + 10 = 19 \neq = 0$ , so there is a unique solution. (1 mark)
$A^{- 1} = \frac{1}{19} [3 - 5 23]$ . (1 mark)
$X = \frac{1}{19} [12 + 26 - 20 + 39] = \frac{1}{19} [3819] = [21]$ . (1 mark)

Answer: $x = 2$ , $y = 1$ .

Example 3: Martin's rule in three variables · 5 marks · Long answerHarder board level

Solve by the matrix method: $2 x + y + z = 4$ , $x - y + 2 z = 5$ , $3 x + 2 y - z = 3$ .

Try it first, then show the solutionSolution

$A = 213 1 - 1 2 12 - 1$ , $B = 453$ .
$∣ A ∣ = 2 (1 - 4) - 1 (- 1 - 6) + 1 (2 + 3) = - 6 + 7 + 5 = 6 \neq = 0$ . (1 mark)
Cofactors: $C_{11} = - 3, C_{12} = 7, C_{13} = 5$ ; $C_{21} = 3, C_{22} = - 5, C_{23} = - 1$ ; $C_{31} = 3, C_{32} = - 3, C_{33} = - 3$ . (1½ marks)
$adj A = - 3 75 3 - 5 - 1 3 - 3 - 3$ . (½ mark)
$X = \frac{1}{6} - 12 + 15 + 9 28 - 25 - 9 20 - 5 - 9 = \frac{1}{6} 12 - 6 6 = 2 - 1 1$ . (1½ marks)
Check in the third equation: $6 - 2 - 1 = 3$ . (½ mark)

Answer: $x = 2$ , $y = - 1$ , $z = 1$ .

Example 4: Showing a 3-variable system is inconsistent · 3 marks · Short answerHarder board level

Show that the system $x + y + z = 1$ , $2 x + 2 y + 2 z = 3$ , $x - y + z = 0$ is inconsistent.

Try it first, then show the solutionSolution

$A = 121 12 - 1 121$ . Rows 1 and 2 are proportional, so $∣ A ∣ = 0$ . (1 mark)
$adj A = 40 - 4 - 2 02 000$ . (1 mark)
$(adj A) B = 4 - 6 + 0 0 - 4 + 6 + 0 = - 2 02 \neq = O$ , so the system is inconsistent. (1 mark)

Answer: Inconsistent: $∣ A ∣ = 0$ and $(adj A) B \neq = O$ .

Example 5: Using a given product to find the inverse · 5 marks · Long answerHarder board level

Let $A = 111124139$ and $C = 6 - 6 2 - 5 8 - 3 1 - 2 1$ . Find $A C$ , hence $A^{- 1}$ , and solve $x + y + z = 6$ , $x + 2 y + 3 z = 14$ , $x + 4 y + 9 z = 36$ .

Try it first, then show the solutionSolution

Row 1 of $A$ times the columns of $C$ : $6 - 6 + 2 = 2$ , $- 5 + 8 - 3 = 0$ , $1 - 2 + 1 = 0$ . Rows 2 and 3 give $(0, 2, 0)$ and $(0, 0, 2)$ . So $A C = 2 I$ . (2 marks)
Hence $A (\frac{1}{2} C) = I$ , so $A^{- 1} = \frac{1}{2} C$ . (1 mark)
The system is $A X = B$ with $B = 61436$ . $X = \frac{1}{2} C B = \frac{1}{2} 36 - 70 + 36 - 36 + 112 - 72 12 - 42 + 36 = \frac{1}{2} 246$ . (1½ marks)
Check: $1 + 4 + 9 = 14$ . (½ mark)

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

Example 6: Fruit packs: forming and solving the system · 5 marks · Case-basedHOTS / competency

Suppose a fruit seller sells packs of mangoes, bananas and apples at fixed prices. One customer buys 2 mango packs, 3 banana packs and 1 apple pack for Rs 190; a second buys 1, 2 and 3 packs for Rs 180; a third buys 3, 1 and 2 packs for Rs 230.
(i) Form the equations. (1 mark)
(ii) Find $∣ A ∣$ . (1 mark)
(iii) Find $adj A$ . (2 marks)
(iv) Find the price of each pack. (1 mark)

Try it first, then show the solutionSolution

(i) Prices $x, y, z$ : $2 x + 3 y + z = 190$ , $x + 2 y + 3 z = 180$ , $3 x + y + 2 z = 230$ .
(ii) $∣ A ∣ = 2 (4 - 3) - 3 (2 - 9) + 1 (1 - 6) = 2 + 21 - 5 = 18$ .
(iii) $adj A = 17 - 5 - 5 17 7 - 5 1$ .
(iv) $X = \frac{1}{18} 190 - 900 + 1610 1330 + 180 - 1150 - 950 + 1260 + 230 = \frac{1}{18} 900360540 = 502030$ .

Answer: Mango pack Rs 50, banana pack Rs 20, apple pack Rs 30.

Where students lose marks in Solving Equations by Matrix Method (Martin's Rule)

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

Mistake: Forming the wrong system from a word problem
The ISC 2024 examiners reported that candidates could not form the three equations in the furniture question.
Do this: Write a word equation for each condition ("teakwood used = 29") before writing symbols, and keep the variables in the same order in every row.
Usually costs: The whole 5- or 6-mark question.
Mistake: Forgetting to transpose the cofactor matrix
adj A is the transpose of the cofactor matrix; using the cofactors directly gives a wrong inverse unless the matrix is symmetric.
Wrong working
$A^{- 1} = \frac{1}{∣ A ∣} [C_{ij}]$ .
Right working
$A^{- 1} = \frac{1}{∣ A ∣} [C_{ij}]^{'}$ .
Do this: Write the cofactors of row 1 down column 1 of adj A.
Usually costs: 2 to 3 marks, as every unknown comes out wrong.
Mistake: Writing X = BA⁻¹
The inverse must multiply B from the left, because AX = B has A on the left.
Do this: Write $A^{- 1} A X = A^{- 1} B$ as a line of working.
Usually costs: Usually the product is undefined, so the rest is lost.
Mistake: Not checking |A| before inverting
If |A| = 0 the inverse does not exist and the consistency test is needed instead.
Do this: Compute |A| first and state the conclusion it gives.
Usually costs: 1 mark for the existence statement, more if the method is wrong.

Exam tips

State the conclusion from $∣ A ∣$ in words (" $∣ A ∣ \neq = 0$ , so the system is consistent with a unique solution"); it is usually worth a mark on its own.
List the nine cofactors before writing the adjoint, so the examiner can give part credit even if one is wrong.
Keep $\frac{1}{∣ A ∣}$ outside the matrices until the last line; it reduces fraction errors.
If the question gives a second matrix, multiply it with $A$ first: you will usually get a multiple of $I$ and save several minutes.
Finish with a check in one equation and a sentence answering the question in context (prices, amounts).

Practice questions on Solving Equations by Matrix Method (Martin's Rule)

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

Q1 · 1 mark · MCQBoard level
For which value of $k$ does the system $2 x + 3 y = 5$ , $4 x + k y = 10$ have infinitely many solutions?
(a) 3
(b) 6
(c) 0
(d) no value
Hint
Make $∣ A ∣ = 0$ , then check $(adj A) B$ .
Answer
(b) 6. $∣ A ∣ = 2 k - 12 = 0$ ; then $(adj A) B = [6 - 4 - 3 2] [510] = O$ , and the second equation is twice the first.
Q2 · 3 marks · Short answerBoard level
Solve by the matrix method: $2 x + y = 7$ , $3 x - 2 y = 0$ .
Hint
$∣ A ∣ = - 7$ .
Answer
$A^{- 1} = - \frac{1}{7} [- 2 - 3 - 1 2]$ ; $X = - \frac{1}{7} [- 14 - 21] = [23]$ . So $x = 2$ , $y = 3$ .
Q3 · 5 marks · Long answerHarder board level
Solve by the matrix method: $x + 2 y + 3 z = 5$ , $2 x - y + z = 5$ , $x + y - z = - 2$ .
Hint
$∣ A ∣ = 15$ .
Answer
$adj A = 033 5 - 4 1 55 - 5$ , so $X = \frac{1}{15} 0 + 25 - 10 15 - 20 - 10 15 + 5 + 10 = \frac{1}{15} 15 - 15 30$ : $x = 1$ , $y = - 1$ , $z = 2$ .
Q4 · 3 marks · Short answerHarder board level
Show that the system $x + y + z = 2$ , $2 x + y + 3 z = 5$ , $3 x + 2 y + 4 z = 8$ has no solution.
Hint
Row 3 of $A$ is the sum of rows 1 and 2.
Answer
$∣ A ∣ = 0$ (row 3 = row 1 + row 2). $(adj A) B = 2 - 1 - 1 \neq = O$ , so the system is inconsistent. (Adding the first two equations gives $3 x + 2 y + 4 z = 7$ , not 8.)
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If $∣ A ∣ = 0$ and $(adj A) B = O$ , then $A X = B$ has infinitely many solutions.
Reason (R): If $∣ A ∣ \neq = 0$ , then $A X = B$ has the unique solution $X = A^{- 1} B$ .
(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
Try $x + y + z = 1$ , $x + y + z = 2$ , $x + y + z = 3$ .
Answer
(d). Counter-example: $x + y + z = 1$ , $x + y + z = 2$ , $x + y + z = 3$ . Every $2 \times 2$ minor of $A$ is zero, so $∣ A ∣ = 0$ and $adj A = O$ , giving $(adj A) B = O$ ; yet the equations contradict each other, so there is no solution. A is false; R is true.
Q6 · 3 marks · Short answerHOTS / competency
For the system $x + y + z = 6$ , $x + 2 y + 3 z = 10$ , $x + 2 y + λ z = μ$ , find the conditions on $λ$ and $μ$ for (i) a unique solution, (ii) no solution, (iii) infinitely many solutions.
Hint
Subtract the second equation from the third.
Answer
$∣ A ∣ = λ - 3$ . (i) $λ \neq = 3$ . If $λ = 3$ , the third equation minus the second gives $0 = μ - 10$ . (ii) $λ = 3$ , $μ \neq = 10$ . (iii) $λ = 3$ , $μ = 10$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose Rs 50 000 is split among three savings schemes paying 5%, 6% and 8% simple interest a year. The total yearly interest is Rs 3300, and the amount in the 8% scheme equals the total of the other two.
(i) Form the equations, using $x, y, z$ for the amounts. (1 mark)
(ii) Show that $∣ A ∣ = - 2$ . (1 mark)
(iii) Find the amounts by the matrix method. (2 marks)
Hint
Clear the percentages: $5 x + 6 y + 8 z = 330000$ .
Answer
(i) $x + y + z = 50000$ , $5 x + 6 y + 8 z = 330000$ , $x + y - z = 0$ . (ii) $∣ A ∣ = 1 (- 14) - 1 (- 13) + 1 (- 1) = - 2$ . (iii) $adj A = - 14 13 - 1 2 - 2 0 2 - 3 1$ , so $X = - \frac{1}{2} - 40000 - 10000 - 50000$ : Rs 20 000 at 5%, Rs 5000 at 6%, Rs 25 000 at 8%.

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.Write the solution of $A X = B$ when $∣ A ∣ \neq = 0$ .Show answer
$X = A^{- 1} B = \frac{1}{∣ A ∣} (adj A) B$ .
2.What does $∣ A ∣ = 0$ and $(adj A) B \neq = O$ tell you?Show answer
The system is inconsistent (no solution).
3.Does the uniqueness of the solution depend on B?Show answer
No, only on whether $∣ A ∣ \neq = 0$ .
4.What is $A (adj A)$ ?Show answer
$∣ A ∣ I$ .
5.If $A C = 4 I$ , what is $A^{- 1}$ ?Show answer
$\frac{1}{4} C$ .
6.Solve $x + y = 5$ , $x - y = 1$ using $A^{- 1}$ .Show answer
$A^{- 1} = \frac{1}{2} [11 1 - 1]$ , so $x = 3$ , $y = 2$ .

Solving Equations by Matrix Method (Martin's Rule): frequently asked questions

What is Martin's rule in ISC Maths?

It is the CISCE syllabus name for solving a system of linear equations with matrices: write it as AX = B and, when |A| is not zero, find X = A⁻¹B using the adjoint.

How do I know if a system of equations has no solution?

Find |A|. If it is zero, compute (adj A)B. If that product is not the zero matrix, the system is inconsistent and has no solution.

What if |A| = 0 and (adj A)B = O?

Then the matrix test is inconclusive for three variables: the system may have infinitely many solutions or none. Solve the equations by elimination to decide. For two variables (with A not the zero matrix) it always has infinitely many solutions.

Is Cramer's rule needed for ISC?

No. The CISCE 2027 syllabus asks for the matrix method (Martin's rule) and consistency conditions using matrices. Cramer's rule is not listed.

How many marks does the matrix method carry in the ISC paper?

Algebra (matrices and determinants) carries 10 marks in the 2027 paper, and a full matrix-method question is usually one of the longer questions, often set in a real-life context such as the 2024 furniture-factory problem.