📌 Snapshot
- Defines the determinant of a square matrix of order 1, 2 and 3 and gives the expansion rule (sum of products of row/column elements with corresponding cofactors).
- Establishes the scalar property |kA| = k^n |A| for an n × n matrix.
- Gives the determinant formula for the area of a triangle with given vertices and its corollary for collinear points.
- Develops minors, cofactors, adjoint and proves A(adj A) = (adj A)A = |A| I, leading to A^(-1) = (1/|A|) adj A whenever |A| ≠ 0.
- Applies determinants and matrix inverse to solve a system of linear equations (matrix method AX = B → X = A^(-1)B) and to decide consistency when |A| = 0.
📖 Detailed Notes
2.1 Core concepts
- To every square matrix A = [a_ij] of order n, a unique number (real or complex) called its determinant |A| (also det A or Δ) is associated; it can be viewed as a function f : M → K from the set of square matrices to numbers (NCERT §4.2, p. 76–77).
- Only square matrices have determinants, and |A| is read as "determinant of A", not modulus of A (NCERT §4.2 Remarks, p. 77).
- For a 1 × 1 matrix A = [a], the determinant is defined to be a itself (NCERT §4.2.1, p. 77).
- For a 2 × 2 matrix with entries a_11, a_12, a_21, a_22, the determinant is a_11 a_22 − a_21 a_12 (NCERT §4.2.2, p. 77).
- A 3 × 3 determinant is evaluated by expansion along a row or column: multiply each element by (−1)^(i+j) and by the 2 × 2 determinant obtained on deleting its row and column, then add (NCERT §4.2.3, p. 77–79).
- Expansion along any of the three rows or three columns yields the same value — six equivalent ways exist (NCERT §4.2.3, p. 80).
- For easier calculation one should expand along the row or column with the maximum number of zeros (NCERT §4.2.3 Remark (i), p. 80).
- If A = kB where A and B are square matrices of the same order n, then |A| = k^n |B| (illustrated for n = 2 with |A| = 2^2 |B|) (NCERT §4.2.3 Remark (iii), p. 80).
- Area of a triangle with vertices (x_1, y_1), (x_2, y_2), (x_3, y_3) equals (1/2) of the determinant whose rows are (x_i, y_i, 1); absolute value is taken since area is positive (NCERT §4.3, p. 82).
- Three points are collinear iff this determinant is zero (NCERT §4.3 Remark (iii), p. 82).
- The minor M_ij of element a_ij is the determinant obtained by deleting the i-th row and j-th column; for an n × n determinant (n ≥ 2) the minor has order n − 1 (NCERT §4.4 Definition 1 and Remark, p. 84).
- The cofactor A_ij of a_ij is defined as A_ij = (−1)^(i+j) M_ij (NCERT §4.4 Definition 2, p. 84).
- |A| equals the sum of products of elements of any row (or column) with their corresponding cofactors (NCERT §4.4, p. 85).
- If elements of a row (or column) are multiplied with cofactors of a different row (or column), the sum is zero (NCERT §4.4 Note, p. 85).
- The adjoint adj A of A = [a_ij]_{n × n} is the transpose of the cofactor matrix [A_ij]_{n × n} (NCERT §4.5.1 Definition 3, p. 87).
- For a 2 × 2 matrix the adjoint can be written down directly by swapping the diagonal entries and changing the sign of the off-diagonal entries (NCERT §4.5.1 Remark, p. 88).
- Theorem 1: For any square matrix A of order n, A(adj A) = (adj A) A = |A| I, where I is the identity of order n (NCERT §4.5.1, p. 88).
- A square matrix A is singular if |A| = 0 and non-singular if |A| ≠ 0 (NCERT §4.5.1 Definitions 4 and 5, p. 89).
- Theorem 2: If A and B are non-singular matrices of the same order, then AB and BA are also non-singular of the same order (NCERT §4.5.1, p. 89).
- Theorem 3 (Product theorem): |AB| = |A| |B| for square matrices A, B of the same order (NCERT §4.5.1, p. 89).
- For an n × n matrix, |adj A| = |A|^(n − 1) (NCERT §4.5.1, p. 89–90).
- Theorem 4: A square matrix A is invertible iff A is non-singular; whenever |A| ≠ 0, A^(-1) = (1/|A|) adj A (NCERT §4.5.1, p. 90).
- A system of linear equations in three unknowns a_i x + b_i y + c_i z = d_i (i = 1, 2, 3) is written in matrix form as AX = B, where A is the coefficient matrix, X = [x y z]^T and B = [d_1 d_2 d_3]^T (NCERT §4.6.1, p. 94).
- Matrix Method (Case I): if |A| ≠ 0, A^(-1) exists and the unique solution is X = A^(-1) B (NCERT §4.6.1, p. 94).
- Case II (|A| = 0): compute (adj A) B. If (adj A) B ≠ O, no solution exists and the system is inconsistent; if (adj A) B = O, the system may have infinitely many solutions or none (NCERT §4.6.1, p. 94–95).
- A system is consistent if it has at least one solution and inconsistent if it has none (NCERT §4.6 definitions, p. 94).
2.2 Definitions to memorise
| Term | Definition | Page |
|---|---|---|
| Determinant of A | A number (real/complex) associated to every square matrix A = [a_ij], denoted | A |
| Determinant of order 1 | If A = [a], then | A |
| Determinant of order 2 | a_11 a_22 − a_21 a_12 | 77 |
| Determinant of order 3 | Sum, with signs (−1)^(i+j), of each element of a chosen row/column multiplied by its 2 × 2 minor | 77–79 |
| Minor M_ij | Determinant obtained by deleting i-th row and j-th column of the determinant of order n; itself of order n − 1 | 84 |
| Cofactor A_ij | A_ij = (−1)^(i+j) M_ij | 84 |
| Adjoint adj A | Transpose of the matrix of cofactors [A_ij] | 87 |
| Singular matrix | Square matrix A with | A |
| Non-singular matrix | Square matrix A with | A |
| Inverse A^(-1) | A^(-1) = (1/ | A |
| Area of triangle | (1/2) · | determinant of rows (x_i, y_i, 1) |
| Consistent system | System of equations that has at least one solution | 94 |
| Inconsistent system | System of equations whose solution does not exist | 94 |
2.3 Diagrams / processes to remember
- The four-step expansion of a 3 × 3 determinant along R_1 — multiply a_11 by (−1)^(1+1) and its minor, a_12 by (−1)^(1+2) and its minor, a_13 by (−1)^(1+3) and its minor, then sum (NCERT §4.2.3 Steps 1–4, p. 77–78).
- The schematic showing six equivalent expansions of a 3 × 3 determinant along R_1, R_2, R_3, C_1, C_2, C_3 yielding the same value (NCERT §4.2.3, p. 78–80).
- The 2 × 2 adjoint shortcut: swap a_11 ↔ a_22 and negate a_12 and a_21 (NCERT §4.5.1 Remark, p. 88).
- The verification array A(adj A) = diag(|A|, |A|, |A|) = |A| I used to prove Theorem 1 (NCERT §4.5.1, p. 88).
- Flow for solving AX = B: compute |A|; if non-zero, find adj A → A^(-1) = (1/|A|) adj A → X = A^(-1) B (NCERT §4.6.1 Case I, p. 94).
2.5 Key formulas & theorems
| Formula | Statement | NCERT page |
|---|---|---|
| 2 × 2 determinant | a₁₁a₂₂ − a₁₂a₂₁ | 77 |
| 3 × 3 expansion | Sum of element × cofactor along a row/column | 78 |
| Scalar property | kA | |
| Area of triangle | (1/2) | det of (xᵢ, yᵢ, 1) |
| Collinearity | det = 0 | 82 |
| Minor M_ij | Delete row i and column j | 84 |
| Cofactor A_ij | (−1)^(i+j) M_ij | 84 |
| Expansion formula | A | |
| Cross-row/column sum | 0 | 85 |
| Adjoint | Transpose of cofactor matrix | 87 |
| A(adj A) | A | |
| Singular | A | |
| Non-singular | A | |
| Product theorem | AB | |
| adj A | ||
| Invertibility | A invertible iff non-singular | 90 |
| Inverse formula | A⁻¹ = (1/ | A |
| AX = B form | A is coefficient, X is unknowns, B is constants | 94 |
| Unique solution | A | |
| Inconsistent | A | |
| Indeterminate | A | |
| det(A⁻¹) | 1/ | A |
| 2 × 2 adjoint shortcut | Swap diagonal, negate off-diagonal | 88 |
| Identity matrix | I_n with diagonal 1, off-diagonal 0 | 88 |
2.6 Solved examples (NCERT-grounded)
Example A (NCERT §4.2.2 Example 1, p. 77). det([[2, 4], [−1, 2]]).
Step 1 — apply formula: 2·2 − 4·(−1) = 4 + 4. Step 2 — sum: 8. Step 3 — answer: 8.
Example B (NCERT §4.2.3, Example 4, p. 81). Compute |3A| for |A| = 4, order 3.
Step 1 — apply scalar rule: |3A| = 3³ |A|. Step 2 — compute: 27 · 4. Step 3 — answer: 108.
Example C (NCERT §4.3 Example 6, p. 83). Area of triangle with vertices (3, 8), (−4, 2), (5, 1).
Step 1 — set up determinant: (1/2)|det([[3, 8, 1], [−4, 2, 1], [5, 1, 1]])|. Step 2 — expand R₁: 3(2 − 1) − 8(−4 − 5) + 1(−4 − 10) = 3 + 72 − 14 = 61. Step 3 — divide by 2: area = 61/2 sq units.
Example D (NCERT §4.5.1 Example 12, p. 88). Find adj A for A = [[2, 3], [1, 4]].
Step 1 — swap diagonal: (4, 2). Step 2 — negate off-diagonal: −3, −1. Step 3 — assemble: adj A = [[4, −3], [−1, 2]].
Example E (NCERT §4.6.1 Example 16, p. 95). Solve 2x + 5y = 1, 3x + 2y = 7 by matrix method.
Step 1 — |A| = 4 − 15 = −11: non-singular, unique solution exists. Step 2 — A⁻¹ = (−1/11)[[2, −5], [−3, 2]]: compute X = A⁻¹B. Step 3 — multiply: X = (−1/11)[2(1) − 5(7); −3(1) + 2(7)] = (−1/11)[−33; 11] = [3; −1], so x = 3, y = −1.
2.4 Common confusions / NTA trap points
- Students often read |A| as a modulus — but |A| means "determinant of A" and is signed, not absolute (NCERT §4.2 Remark (i), p. 77).
- |kA| = k|A| is a common error; the correct rule for an n × n matrix is |kA| = k^n |A| (NCERT §4.2.3 Remark (iii), p. 80).
- Cofactor and minor are confused: the cofactor carries the sign (−1)^(i+j), the minor does not (NCERT §4.4 Definitions 1–2, p. 84).
- Multiplying elements of one row with cofactors of the same row gives |A|; with cofactors of a different row, the sum is 0 — a frequent trap in MCQs (NCERT §4.4 Note, p. 85).
- When |A| = 0 students often jump to "no solution"; instead, check (adj A) B: zero → may be consistent or inconsistent, non-zero → inconsistent (NCERT §4.6.1 Case II, p. 94–95).
- Area being a positive quantity is forgotten; the absolute value of the determinant must be taken, but if area is given, both signs must be solved (NCERT §4.3 Remarks (i)–(ii), p. 82).
- Sign of the cofactor (−1)^(i+j) is checkerboard: + − +, − + −, + − +. Forgetting this sign is the single biggest error in 3×3 expansions.
🎯 Practice MCQs
First 3 questions free · create a free account to unlock the rest — answers & explanations included, no payment needed
Q1. Evaluate the determinant of the matrix [[2, 4], [−1, 2]].
▸ Show answer & explanation
Answer: C
For a 2 × 2 determinant a_11 a_22 − a_21 a_12 = 2(2) − 4(−1) = 4 + 4 = 8. Option B is the trap that forgets the −a_21 a_12 sign.
Q2. For a square matrix A of order n, which of the following statements is correct?
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Answer: C
|A| = k^n |B| when A = kB for square matrices of order n. Only square matrices have determinants (so A is wrong); |A| is the signed determinant, not modulus (B wrong); the 1 × 1 determinant is a itself, not |a| (D wrong).
Q3. Evaluate det([[3, −4, 5], [1, 1, −2], [2, 3, 1]]).
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Answer: C
Expanding along R_1: 3(1·1 − (−2)·3) − (−4)(1·1 − (−2)·2) + 5(1·3 − 1·2) = 3(1 + 6) + 4(1 + 4) + 5(3 − 2) = 21 + 20 + 5 = 40. Option A drops the sign on (−4); option D forgets the last (+5).
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Q4. If A is a square matrix of order 3 and |A| = 4, then |3A| equals:
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Answer: C
For an n × n matrix, |kA| = k^n |A|. With n = 3 and k = 3, |3A| = 3^3 · 4 = 27 · 4 = 108. Option A is the (wrong) "|kA| = k|A|" trap and option B uses k^2 instead of k^3.
Q5. The area of the triangle with vertices (3, 8), (−4, 2) and (5, 1) is:
▸ Show answer & explanation
Answer: B
Δ = (1/2)|det([[3, 8, 1], [−4, 2, 1], [5, 1, 1]])| = (1/2)|3(2 − 1) − 8(−4 − 5) + 1(−4 − 10)| = (1/2)|3 + 72 − 14| = 61/2. Option C forgets the 1/2 factor.
Q6. Three points A(a, b + c), B(b, c + a), C(c, a + b) in the plane are:
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Answer: C
Three points are collinear iff the determinant det([[a, b + c, 1], [b, c + a, 1], [c, a + b, 1]]) = 0. Adding C_2 to C_1 makes column 1 = a + b + c in every row, so two columns become proportional and the determinant vanishes. Hence the points are collinear.
Q7. In the determinant [[1, 2, 3], [4, 5, 6], [7, 8, 9]], the minor M_23 of the element 6 is:
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Answer: B
Delete R_2 and C_3 to get [[1, 2], [7, 8]] whose value is 1·8 − 2·7 = 8 − 14 = −6. The cofactor here would be A_23 = (−1)^{2+3}·(−6) = 6; option A confuses cofactor with minor.
Q8. For the determinant ∆ = det([[a_11, a_12, a_13], [a_21, a_22, a_23], [a_31, a_32, a_33]]) with cofactors A_ij, ∆ equals:
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Answer: D
∆ is the sum of products of elements of any one row (or column) with their corresponding cofactors. Option D is expansion along C_1: a_11 A_11 + a_21 A_21 + a_31 A_31. Options A, B and C mix elements of one row with cofactors of another row, giving 0, not ∆.
Q9. If A = [[2, 3], [1, 4]], then adj A is:
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Answer: A
For a 2 × 2 matrix, swap the diagonal entries (2 ↔ 4) and change the signs of the off-diagonal entries (3 → −3, 1 → −1) to get [[4, −3], [−1, 2]]. Option B forgets the sign change; option D negates everything wrongly.
Q10. Let A be a non-singular square matrix of order 3 × 3. Then |adj A| equals:
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Answer: B
|adj A| = |A|^(n − 1). With n = 3, |adj A| = |A|^2. Option C confuses with |kA| = k^n |A|, and option D is dimensionally nonsensical.
Q11. If A is an invertible matrix of order 2, then det(A^(-1)) equals:
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Answer: B
Since A A^(-1) = I, by the product theorem |A| · |A^(-1)| = |I| = 1, so |A^(-1)| = 1/|A| (with |A| ≠ 0 since A is invertible). Option D would require A^(-1) singular, contradicting invertibility.
Q12. For the system 2x + 5y = 1, 3x + 2y = 7 written as AX = B with A = [[2, 5], [3, 2]], the solution by the matrix method X = A^(-1) B is:
▸ Show answer & explanation
Answer: B
|A| = 4 − 15 = −11 ≠ 0, so A is non-singular and the system has a unique solution. A^(-1) = (−1/11)[[2, −5], [−3, 2]], and X = A^(-1) B = (−1/11)[[2·1 − 5·7], [−3·1 + 2·7]] = (−1/11)[−33, 11]^T = [3, −1]^T. Option A is the sign trap in y.
Q13. For the system of equations AX = B, which option lists the correct conclusions for a square matrix A?
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Answer: B
This restates the matrix-method algorithm — non-singular A guarantees a unique solution via X = A^(-1) B; if |A| = 0, (adj A) B determines inconsistency vs the indeterminate "may or may not be consistent" case. Options A and C invert or oversimplify the criterion; D is wrong because (adj A) B = O leaves both possibilities open.
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