Is it correct to say: Multiplying matrices as if it were commutative: AB = BA?

In general AB ≠ BA. Order matters. Even the ORDERS may not match if you swap.

Is it correct to say: Computing the adjoint by taking the cofactor matrix WITHOUT transposing?

adj(A) = (cofactor matrix)^T. Cofactor at position (i, j) goes to position (j, i) of the adjoint.

Is it correct to say: Writing A⁻¹ = adj(A)/|A| when |A| = 0?

|A| = 0 means A is SINGULAR — no inverse exists. Check |A| ≠ 0 before dividing.

Is it correct to say: Forgetting the (−1)^{i+j} sign pattern in cofactors?

Cofactor sign chart: + − + / − + − / + − + for a 3×3. A wrong sign flips the whole determinant.

Is it correct to say: Using (AB)^{-1} = A^{-1}B^{-1} instead of the reversed order?

(AB)^{-1} = B^{-1}A^{-1}. Same rule as transpose: order reverses when the operation crosses a product.

Is it correct to say: Declaring a system inconsistent whenever |A| = 0?

|A| = 0 alone is not enough. If (adj A)·B = 0 as well, the system has INFINITELY MANY solutions; if (adj A)·B ≠ 0, it is INCONSISTENT.

Matrices and Determinants

Matrix algebra, transpose, determinants, minors, cofactors, adjoint, inverse, system of linear equations — NCERT Class 12 Maths Ch 3 & 4

📐 14 formulas✏️ 3 examples🎯 6 practice⚖️ 8-10 marks🏫 CBSE📚 Class 12✓ 2025–26 syllabus

💡

Board Exam Tips

→Solving a 3×3 linear system by the matrix method (X = A⁻¹B) is a guaranteed 5-mark question.
→Verify (adj A)·A = |A|·I after computing the adjoint — a quick sanity check catches sign errors.
→Matrix multiplication is NOT commutative — never write AB = BA unless you have justified it.
→For 'find x such that |A| = 0' problems, expand along the row or column with the most zeros.
→Always state 'system is consistent' or 'inconsistent' based on |A| and (adj A)·B — CBSE gives 1 mark for the conclusion.

📐 Formulas(14)

Matrix Addition

(A + B)_{ij} = A_{ij} + B_{ij}

Scalar Multiplication

(k A)_{ij} = k \cdot A_{ij}

Matrix Multiplication★ Board fav

(A B)_{ij} = \sum_{k = 1}^{n} A_{ik} B_{k j}

Transpose★ Board fav

(A^{T})_{ij} = A_{j i}, (A B)^{T} = B^{T} A^{T}

Determinant of 2×2★ Board fav

det (a c b d) = a d - b c

Determinant of 3×3 (Cofactor Expansion)★ Board fav

∣ A ∣ = a_{11} C_{11} + a_{12} C_{12} + a_{13} C_{13}, C_{ij} = (- 1)^{i + j} M_{ij}

Product Rule for Determinants

∣ A B ∣ = ∣ A ∣ \cdot ∣ B ∣

Adjoint★ Board fav

adj (A) = (C_{ij})^{T}

Fundamental Identity

A \cdot adj (A) = adj (A) \cdot A = ∣ A ∣ I

Inverse via Adjoint★ Board fav

A^{- 1} = \frac{1}{∣ A ∣} adj (A), ∣ A ∣ \neq = 0

Matrix Method for Linear System★ Board fav

A X = B \Rightarrow X = A^{- 1} B, ∣ A ∣ \neq = 0

Consistency Criteria

∣ A ∣ \neq = 0 \Rightarrow unique; ∣ A ∣ = 0, (adj A) B = 0 \Rightarrow infinite; ∣ A ∣ = 0, (adj A) B \neq = 0 \Rightarrow inconsistent

Inverse of Transpose and Product

(A^{T})^{- 1} = (A^{- 1})^{T}, (A B)^{- 1} = B^{- 1} A^{- 1}

Symmetric and Skew-Symmetric

A^{T} = A \Rightarrow symmetric; A^{T} = - A \Rightarrow skew-symmetric

✏️ Solved Examples

1Solved Exampleeasy3 steps

If A = [[2, 3], [1, 4]], find |A| and A⁻¹.

Compute the determinant.

∣ A ∣ = 2 (4) - 3 (1) = 8 - 3 = 5

2Solved Exampleboard4 steps

Solve the system 2x + 3y = 8, x + 2y = 5 using the matrix method.

Write in matrix form AX = B.

A = (2132), X = (x y), B = (85)

3Solved ExampleHOTS5 steps

For A = [[1, −1, 2], [0, 2, −3], [3, −2, 4]], verify that A · (adj A) = |A|·I and hence find A⁻¹.

Expand |A| along row 1.

∣ A ∣ = 1 (2 \cdot 4 - (- 3) (- 2)) - (- 1) (0 \cdot 4 - (- 3) (3)) + 2 (0 \cdot (- 2) - 2 \cdot 3)

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Multiplying matrices as if it were commutative: AB = BA
✓In general AB ≠ BA. Order matters. Even the ORDERS may not match if you swap.
2
Computing the adjoint by taking the cofactor matrix WITHOUT transposing
✓adj(A) = (cofactor matrix)^T. Cofactor at position (i, j) goes to position (j, i) of the adjoint.
3
Writing A⁻¹ = adj(A)/|A| when |A| = 0
✓|A| = 0 means A is SINGULAR — no inverse exists. Check |A| ≠ 0 before dividing.
4
Forgetting the (−1)^{i+j} sign pattern in cofactors
✓Cofactor sign chart: + − + / − + − / + − + for a 3×3. A wrong sign flips the whole determinant.
5
Using (AB)^{-1} = A^{-1}B^{-1} instead of the reversed order
✓(AB)^{-1} = B^{-1}A^{-1}. Same rule as transpose: order reverses when the operation crosses a product.
6
Declaring a system inconsistent whenever |A| = 0
✓|A| = 0 alone is not enough. If (adj A)·B = 0 as well, the system has INFINITELY MANY solutions; if (adj A)·B ≠ 0, it is INCONSISTENT.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
If A = [[3, 1], [2, 4]], find A⁻¹.
Q2
Find the value of x for which the matrix [[x, 2], [3, x + 1]] is singular.
Q3
Solve x + y + z = 6, y + 3z = 11, x + z = 2 using the matrix method.
Q4
If A is a 3×3 matrix with |A| = 4, find |adj(A)| and |2A|.
Q5
Show that A = [[1, 2], [3, 4]] and B = [[5, 6], [7, 8]] satisfy (AB)^T = B^T A^T. Compute (AB)^T.
Q6
Find k so that [[k, 3], [2, k]] has determinant 2.

📝 Notes

Matrices and Determinants

Chapters 3 and 4 give the algebraic scaffolding for solving linear systems, coordinate transformations, and much of applied maths. A matrix is a table of numbers; a determinant is the signed volume scaling factor of the linear map that the matrix represents.

Reading a matrix problem

Small dimension (2×2) — direct formulas are quickest.
3×3 — cofactor expansion; choose row/column with zeros.
Bigger — row operations to create zeros first, then expand.

The invertibility ladder

For a square matrix A the following are equivalent:

A is invertible.
|A| ≠ 0.
The homogeneous system AX = 0 has only the trivial solution.
The rows (and columns) of A are linearly independent.

Any one of these implies all the others.

Solving AX = B

Write the system in the form AX = B.
Compute |A|.
If |A| ≠ 0 — invert A, then X = A⁻¹B. Unique solution.
If |A| = 0 — compute (adj A)·B. Zero ⇒ infinitely many solutions; non-zero ⇒ inconsistent.