MathsCBSE Class 12 · Determinants

Solving Linear Equations by the Matrix Method (X = A⁻¹B), Class 12

Updated 15 October 2026 · Fact-checked

Write the system as AX = B, with A the coefficient matrix. If |A| ≠ 0, the system has the unique solution X = A⁻¹B = (1/|A|)(adj A)B. If |A| = 0, compute (adj A)B: if it is not zero the system is inconsistent; if it is zero, the system has no solution or infinitely many.

Always find |A| first: non-zero means one answer X = A⁻¹B; zero means check (adj A)B to decide between no solution and infinitely many.

Illustration drawn for QuestionBuddy.

Understand Solving Linear Equations by Matrix Method

A system such as
$x + y + z = 6, x - y + 2 z = 5, 2 x + y - z = 1$
can be written as one matrix equation $A X = B$ , where $A = 112 1 - 1 1 12 - 1$ holds the coefficients (each row is one equation, each column one variable), $X = x y z$ and $B = 651$ .

Unique solution. If $∣ A ∣ \neq = 0$ , $A^{- 1}$ exists. Multiply $A X = B$ on the left by $A^{- 1}$ : $A^{- 1} A X = A^{- 1} B$ , so
$X = A^{- 1} B = \frac{1}{∣ A ∣} (adj A) B$ .
The system is then consistent with exactly one solution. This is the 5-mark question that appears in almost every board paper: find $∣ A ∣$ , the nine cofactors, $adj A$ , $A^{- 1}$ , and then $X$ .

**Consistency when $∣ A ∣ = 0$ ** (NCERT Section 4.6). A system is consistent if it has at least one solution and inconsistent if it has none. When $∣ A ∣ = 0$ , compute $(adj A) B$ :

If $(adj A) B \neq = O$ , the system is inconsistent (no solution).
If $(adj A) B = O$ , the system is either consistent with infinitely many solutions or inconsistent; you decide by looking at the equations.

For two variables you can see why: $∣ A ∣ = 0$ means the two lines are parallel; they either never meet or are the same line. The 2026-27 syllabus asks you to solve only systems with a unique solution, but consistency questions are asked 'by examples'.

The product trick. Some questions give a second matrix $C$ and ask you to find $A C$ first. If $A C = k I$ for a number $k \neq = 0$ , then $A^{- 1} = \frac{1}{k} C$ with no cofactor work. Read carefully whether the system's coefficient matrix is $A$ , $C$ or $A^{'}$ : if the equations' coefficients appear as the columns of a given matrix, the coefficient matrix is its transpose.

Word problems. Choose variables for the unknown prices, amounts or rates, write one equation per sentence of information, and then use the matrix method. Finish with a sentence answering the question (for example, 'A pen costs Rs 10').

Keep the order: $X = A^{- 1} B$ , never $B A^{- 1}$ (which is not even defined for a column $B$ ). Cramer's rule is not in the syllabus, so use the inverse.

Solving Linear Equations by Matrix Method: formulas to remember

Matrix form: $A X = B$; Row i of A holds the coefficients of equation i; B holds the constants.
Unique solution: $∣ A ∣ \neq = 0 \Rightarrow X = A^{- 1} B = \frac{1}{∣ A ∣} (adj A) B$; Multiply by $A^{- 1}$ on the left.
Inconsistent: $∣ A ∣ = 0 and (adj A) B \neq = O$; No solution.
Further check needed: $∣ A ∣ = 0 and (adj A) B = O$; Infinitely many solutions or none; inspect the equations.
Product trick: $A C = k I (k \neq = 0) \Rightarrow A^{- 1} = \frac{1}{k} C$; Saves finding cofactors.

How to solve Solving Linear Equations by Matrix Method questions

The standard 5-mark layout for solving a system by the matrix method.

1Arrange each equation with the variables in the same order ( $x, y, z$ ) and constants on the right; put 0 for a missing variable.
2Write $A$ , $X$ and $B$ , and the statement $A X = B$ .
3Find $∣ A ∣$ . If it is non-zero, write 'A is non-singular, so $A^{- 1}$ exists and the system has a unique solution'.
4Find all nine cofactors with signs and write $adj A$ (transpose of the cofactor matrix).
5Write $A^{- 1} = \frac{1}{∣ A ∣} adj A$ , then $X = A^{- 1} B$ , multiplying $adj A$ by $B$ row by row before dividing by $∣ A ∣$ .
6Read off $x$ , $y$ , $z$ and substitute in one original equation to check.
7If $∣ A ∣ = 0$ , compute $(adj A) B$ and state the conclusion about consistency.

Solving Linear Equations by Matrix Method 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.

Unique solution of a 2 × 2 system

Suppose a small cinema keeps no price list but remembers two days' sales: on Monday it sold 3 adult and 2 child tickets for Rs 630, and on Tuesday 2 adult and 5 child tickets for Rs 750.

Use the matrix method to find the price of an adult ticket and a child ticket.

Try it first, then show the solutionSolution

Let the prices be $a$ and $c$ : $3 a + 2 c = 630$ , $2 a + 5 c = 750$ . So $A = [3225]$ , $B = [630750]$ .
$∣ A ∣ = 15 - 4 = 11 \neq = 0$ ; $adj A = [5 - 2 - 2 3]$ .
$X = \frac{1}{11} [3150 - 1500 - 1260 + 2250] = \frac{1}{11} [1650990] = [15090]$ .
Check: $3 (150) + 2 (90) = 630$ and $2 (150) + 5 (90) = 750$ .

Answer: Adult ticket Rs 150, child ticket Rs 90.

Memory hook: Two days of sales, one matrix inverse, two prices.

Inconsistent system

Imagine two shop assistants reporting the same sale. One says 'a pen and a notebook together cost Rs 50'; the other says 'two pens and two notebooks cost Rs 110'.

Use the matrix method to decide whether both statements can be true.

Try it first, then show the solutionSolution

$x + y = 50$ , $2 x + 2 y = 110$ : $A = [1212]$ , $∣ A ∣ = 2 - 2 = 0$ .
$adj A = [2 - 2 - 1 1]$ , so $(adj A) B = [100 - 110 - 100 + 110] = [- 10 10] \neq = O$ .
So the system is inconsistent: no prices satisfy both statements.
Indeed, the first statement implies two pens and two notebooks cost Rs 100, not Rs 110.

Answer: The system is inconsistent, so one assistant has made a mistake.

Memory hook: |A| = 0 and (adj A)B ≠ O: the stories contradict each other.

Tricks to remember Solving Linear Equations by Matrix Method

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

Inverse on the left
$A X = B$ becomes $X = A^{- 1} B$ : the inverse always stands in front of B.
Use it when: Avoiding the illegal product $B A^{- 1}$ .
Zero determinant, check adj A times B
$∣ A ∣ = 0$ : look at $(adj A) B$ . Non-zero means no solution; zero means look closer.
Use it when: Consistency questions.
Multiply by adj first, divide last
Compute $(adj A) B$ with whole numbers, then divide by $∣ A ∣$ at the end; fractions appear only once.
Use it when: Keeping arithmetic clean in 5-mark questions.
If AC = kI, C is almost the inverse
A product equal to $k I$ hands you $A^{- 1} = \frac{1}{k} C$ for free.
Use it when: 'Use the product to solve the system' questions.

Solving Linear Equations by Matrix Method: solved examples

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

Example 1: Two equations by the matrix method · 3 marks · Short answerBoard level

Solve by the matrix method: $3 x - 2 y = 4$ , $5 x + y = 11$ .

Try it first, then show the solutionSolution

$A = [35 - 2 1]$ , $X = [x y]$ , $B = [411]$ ; $∣ A ∣ = 3 + 10 = 13 \neq = 0$ . (1 mark)
$adj A = [1 - 5 23]$ , so $A^{- 1} = \frac{1}{13} [1 - 5 23]$ . (1 mark)
$X = \frac{1}{13} [4 + 22 - 20 + 33] = [21]$ . (1 mark)

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

Example 2: An inconsistent pair · 2 marks · Very short answerBoard level

Examine the consistency of the system $x + 2 y = 3$ , $2 x + 4 y = 7$ .

Try it first, then show the solutionSolution

$A = [1224]$ , $∣ A ∣ = 4 - 4 = 0$ . (½ mark)
$adj A = [4 - 2 - 2 1]$ . (½ mark)
$(adj A) B = [12 - 14 - 6 + 7] = [- 2 1] \neq = O$ , so the system is inconsistent. (1 mark)

Answer: Inconsistent (no solution).

Example 3: Three equations in three unknowns · 5 marks · Long answerHarder board level

Solve by the matrix method: $x + y + z = 6$ , $x - y + 2 z = 5$ , $2 x + y - z = 1$ .

Try it first, then show the solutionSolution

$A = 112 1 - 1 1 12 - 1$ , $B = 651$ . $∣ A ∣ = 1 (1 - 2) - 1 (- 1 - 4) + 1 (1 + 2) = - 1 + 5 + 3 = 7 \neq = 0$ . (1 mark)
Cofactors: $A_{11} = - 1$ , $A_{12} = 5$ , $A_{13} = 3$ ; $A_{21} = 2$ , $A_{22} = - 3$ , $A_{23} = 1$ ; $A_{31} = 3$ , $A_{32} = - 1$ , $A_{33} = - 2$ . (1½ marks)
$adj A = - 1 53 2 - 3 1 3 - 1 - 2$ , $A^{- 1} = \frac{1}{7} adj A$ . (1 mark)
$X = \frac{1}{7} - 6 + 10 + 3 30 - 15 - 1 18 + 5 - 2 = \frac{1}{7} 71421 = 123$ . (1 mark)
Check in the third equation: $2 + 2 - 3 = 1$ . (½ mark)

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

Example 4: Using a given product to find the inverse · 5 marks · Long answerHarder board level

Let $A = 101210012$ and $C = 21 - 1 - 4 22 2 - 1 1$ . Find $A C$ and use it to solve $x + 2 y = 5$ , $y + z = 5$ , $x + 2 z = 7$ .

Try it first, then show the solutionSolution

Row 1 of A times the columns of C: $(2 + 2 + 0, - 4 + 4 + 0, 2 - 2 + 0) = (4, 0, 0)$ . (1 mark)
Rows 2 and 3 similarly give $(0, 4, 0)$ and $(0, 0, 4)$ , so $A C = 4 I$ . (1 mark)
Hence $A (\frac{1}{4} C) = I$ and $A^{- 1} = \frac{1}{4} C$ . (1 mark)
The system has coefficient matrix A (rows $(1, 2, 0)$ , $(0, 1, 1)$ , $(1, 0, 2)$ ) and $B = 557$ . (½ mark)
$X = \frac{1}{4} 10 - 20 + 14 5 + 10 - 7 - 5 + 10 + 7 = \frac{1}{4} 4812 = 123$ . (1½ marks)

Answer: $A C = 4 I$ ; $x = 1$ , $y = 2$ , $z = 3$

Example 5: Consistency depends on a parameter · 3 marks · Short answerHOTS / competency

For the system $k x + 2 y = 4$ , $3 x + (k - 1) y = 6$ , find the values of $k$ for which there is a unique solution, and show that for every other value the system is inconsistent.

Try it first, then show the solutionSolution

$∣ A ∣ = k (k - 1) - 6 = k^{2} - k - 6 = (k - 3) (k + 2)$ . Unique solution when $k \neq = 3$ and $k \neq = - 2$ . (1 mark)
$k = 3$ : $A = [3322]$ , $adj A = [2 - 3 - 2 3]$ , $(adj A) B = [8 - 12 - 12 + 18] = [- 4 6] \neq = O$ : inconsistent. (1 mark)
$k = - 2$ : $A = [- 2 3 2 - 3]$ , $adj A = [- 3 - 3 - 2 - 2]$ , $(adj A) B = [- 24 - 24] \neq = O$ : inconsistent. (1 mark)

Answer: Unique solution for $k \neq = 3, - 2$ ; inconsistent for $k = 3$ and $k = - 2$ .

Example 6: Case study: three customers' bills · 4 marks · Case-basedHOTS / competency

Suppose a stationery shop sells pens, notebooks and school bags. Customer 1 buys 2 pens, 3 notebooks and 1 bag for Rs 340; customer 2 buys 5 pens, 2 notebooks and 1 bag for Rs 330; customer 3 buys 1 pen, 4 notebooks and 2 bags for Rs 570.
(i) Write the information as $A X = B$ . (1 mark)
(ii) Find $∣ A ∣$ . (1 mark)
(iii) Find the price of each item using $A^{- 1}$ . (2 marks)

Try it first, then show the solutionSolution

(i) $251324112 p n b = 340330570$ . (1 mark)
(ii) $∣ A ∣ = 2 (4 - 4) - 3 (10 - 1) + 1 (20 - 2) = 0 - 27 + 18 = - 9$ . (1 mark)
(iii) $adj A = 0 - 9 18 - 2 3 - 5 13 - 11$ ; $(adj A) B = - 660 + 570 - 3060 + 990 + 1710 6120 - 1650 - 6270 = - 90 - 360 - 1800$ . (1 mark)
$X = - \frac{1}{9} - 90 - 360 - 1800 = 1040200$ . (1 mark)

Answer: Pen Rs 10, notebook Rs 40, bag Rs 200.

Where students lose marks in Solving Linear Equations by Matrix Method

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

Mistake: Writing X = BA⁻¹
B is a column matrix; $B A^{- 1}$ is not defined, and order matters in matrix products.
Wrong working
$X = B A^{- 1}$ .
Right working
$X = A^{- 1} B$ .
Do this: Pre-multiply both sides of $A X = B$ by $A^{- 1}$ and write that step.
Usually costs: The method mark.
Mistake: Placing coefficients in columns instead of rows
Each equation must be a row of A.
Wrong working
$x + y + z = 6$ , $x - y + 2 z = 5$ , $2 x + y - z = 1$ with $A = 111 1 - 1 2 21 - 1$ .
Right working
$A = 112 1 - 1 1 12 - 1$ .
Do this: Read the first equation across the first row.
Usually costs: The whole question.
Mistake: Concluding 'infinitely many solutions' from |A| = 0 alone
A zero determinant also allows inconsistency.
Wrong working
$∣ A ∣ = 0$ , so infinitely many solutions.
Right working
$∣ A ∣ = 0$ ; $(adj A) B \neq = O$ , so the system is inconsistent.
Do this: Always compute $(adj A) B$ when $∣ A ∣ = 0$ .
Usually costs: 1 to 2 marks.
Mistake: Forgetting the transpose in adj A
Using the cofactor matrix instead of the adjoint gives a wrong $A^{- 1}$ and wrong solutions.
Wrong working
$A^{- 1} = \frac{1}{∣ A ∣} [A_{ij}]$ .
Right working
$A^{- 1} = \frac{1}{∣ A ∣} [A_{ij}]^{'}$ .
Do this: Substitute your answer back into one equation; a wrong adjoint usually fails the check.
Usually costs: 2 to 3 marks.

Exam tips

A 5-mark 'solve by the matrix method' question appeared in the 2023, 2024, 2025 and 2026 board papers; practise the full layout until it takes under 10 minutes.
Always state ' $∣ A ∣ \neq = 0$ , so $A^{- 1}$ exists' before using the inverse; it is a marked step.
If a question gives a product of matrices, compute it first: it usually equals $k I$ and gives $A^{- 1}$ directly.
Watch whether the system's coefficient matrix is the given matrix or its transpose.
End with a clear statement, such as ' $x = 1$ , $y = 2$ , $z = 3$ ', or a sentence for word problems.

Practice questions on Solving Linear Equations by Matrix Method

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
If $∣ A ∣ \neq = 0$ , the system $A X = B$ has:
(a) no solution
(b) a unique solution
(c) infinitely many solutions
(d) either no solution or infinitely many
Hint
Does $A^{- 1}$ exist?
Answer
(b) a unique solution, $X = A^{- 1} B$ .
Q2 · 1 mark · MCQBoard level
If $∣ A ∣ = 0$ and $(adj A) B \neq = O$ , the system $A X = B$ is:
(a) consistent with a unique solution
(b) consistent with infinitely many solutions
(c) inconsistent
(d) cannot be decided
Hint
Recall the consistency test.
Answer
(c) inconsistent.
Q3 · 3 marks · Short answerBoard level
Solve by the matrix method: $x - y = 1$ , $2 x + 3 y = 12$ .
Hint
$∣ A ∣ = 5$ .
Answer
$A = [12 - 1 3]$ , $∣ A ∣ = 5$ , $A^{- 1} = \frac{1}{5} [3 - 2 11]$ . $X = \frac{1}{5} [3 + 12 - 2 + 12] = [32]$ : $x = 3$ , $y = 2$ .
Q4 · 2 marks · Very short answerHarder board level
Examine the consistency of $2 x - y = 5$ , $4 x - 2 y = 10$ .
Hint
$∣ A ∣ = 0$ ; compute $(adj A) B$ , then look at the equations.
Answer
$∣ A ∣ = - 4 + 4 = 0$ ; $adj A = [- 2 - 4 12]$ , $(adj A) B = [- 10 + 10 - 20 + 20] = O$ . The second equation is twice the first, so the system is consistent with infinitely many solutions ( $y = 2 x - 5$ ).
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The system $x + y = 2$ , $2 x + 2 y = 5$ has no solution.
Reason (R): The determinant of its coefficient matrix is zero.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
A zero determinant alone allows infinitely many solutions too.
Answer
(b). Both are true, but $∣ A ∣ = 0$ alone does not decide between no solution and infinitely many; A follows from $(adj A) B = [- 1 1] \neq = O$ .
Q6 · 5 marks · Long answerHarder board level
Solve by the matrix method: $2 x - y + z = 3$ , $x + y - z = 0$ , $3 x + 2 y + z = 10$ .
Hint
$∣ A ∣ = 9$ .
Answer
$∣ A ∣ = 2 (1 + 2) + 1 (1 + 3) + 1 (2 - 3) = 9$ . $adj A = 3 - 4 - 1 3 - 1 - 7 033$ . $X = \frac{1}{9} 9 + 0 + 0 - 12 + 0 + 30 - 3 + 0 + 30 = 123$ : $x = 1$ , $y = 2$ , $z = 3$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a family invests Rs 50,000 in three schemes paying 5%, 6% and 8% simple interest a year. The yearly interest is Rs 3300, and the amount in the 8% scheme is Rs 10,000 more than that in the 5% scheme.
(i) Form the equations. (1 mark)
(ii) Solve them by the matrix method. (3 marks)
Hint
Multiply the interest equation by 100 to clear decimals.
Answer
(i) $x + y + z = 50000$ , $5 x + 6 y + 8 z = 330000$ , $- x + z = 10000$ . (ii) $A = 15 - 1 160181$ , $∣ A ∣ = - 1$ , $adj A = 6 - 13 6 - 1 2 - 1 2 - 3 1$ . $X = - 300000 - 330000 + 20000 - 650000 + 660000 - 30000 300000 - 330000 + 10000 = 100002000020000$ . Rs 10,000 at 5%, Rs 20,000 at 6% and Rs 20,000 at 8% (interest $500 + 1200 + 1600 = 3300$ ).

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.When is a system called consistent?Show answer
When it has at least one solution.
3. $∣ A ∣ = 0$ and $(adj A) B \neq = O$ : conclusion?Show answer
The system is inconsistent.
4.If $A C = 6 I$ , what is $A^{- 1}$ ?Show answer
$\frac{1}{6} C$ .
5.Is Cramer's rule in the 2026-27 syllabus?Show answer
No; use the matrix (inverse) method.
6.What does each row of A represent?Show answer
The coefficients of one equation.

Solving Linear Equations by Matrix Method: frequently asked questions

When can I use the matrix method?

When the coefficient matrix A is square and non-singular (|A| ≠ 0). Then the system has a unique solution X = A⁻¹B. The 2026-27 syllabus asks you to solve only such systems in two or three variables.

How do I check consistency when |A| = 0?

Compute (adj A)B. If it is not the zero matrix, the system is inconsistent. If it is the zero matrix, the system has either infinitely many solutions or none, and you decide by examining the equations.

Is Cramer's rule allowed?

Cramer's rule is not part of the CBSE 2026-27 syllabus. Board questions ask for the matrix method, so use X = A⁻¹B.

Which NCERT section covers the matrix method?

Section 4.6 (Applications of Determinants and Matrices) of Chapter 4, Determinants, in NCERT Mathematics Part I for Class 12.

How many marks is a matrix-method question worth?

Usually 5 marks in Section D, sometimes as a case study of 4 marks. Marks are given for |A|, cofactors, adj A, A⁻¹ and the final solution.