NCERT Plus Two (Class 12) Mathematics • Chapter 4

Determinants

Q: Evaluate the determinant |cos θ −sin θ ; sin θ cos θ|.

Δ = cos θ · cos θ − (−sin θ) · sin θ = cos²θ + sin²θ = 1 .

Q: If A is a square matrix of order 3 with |A| = 5, find |2A| and |adj A|.

• |2A| = 2³ |A| = 8 × 5 = 40 . • |adj A| = |A|³⁻¹ = 5² = 25 .

Q: Find values of k if area of triangle is 4 sq. units and vertices are (k, 0), (4, 0), (0, 2).

½ |k 0 1 ; 4 0 1 ; 0 2 1| = ±4 Expanding along C₂ : ½ [−2(k − 4)] = ±4 ⇒ 4 − k = ±4 . • 4 − k = 4 ⇒ k = 0 . • 4 − k = −4 ⇒ k = 8 .

Q: Find the equation of the line joining (1, 2) and (3, 6) using determinants.

For collinear (x, y), (1, 2), (3, 6) : |x y 1 ; 1 2 1 ; 3 6 1| = 0 x(2 − 6) − y(1 − 3) + 1(6 − 6) = 0 ⇒ −4x + 2y = 0 ⇒ 2x − y = 0 .

Q: Using cofactors of elements of third column, evaluate Δ = |1 x yz ; 1 y zx ; 1 z xy| (or verify a₁₁A₁₁+a₁₂A₁₂+a₁₃A₁₃ = |A| for [[2,−3,5],[6,0,4],[1,5,−7]]).

For [[2,−3,5],[6,0,4],[1,5,−7]] along Row 1: • A₁₁ = +(0 − 20) = −20 • A₁₂ = −(−42 − 4) = +46 • A₁₃ = +(30 − 0) = 30 Δ = 2(−20) + (−3)(46) + 5(30) = −40 − 138 + 150 = −28 .

Q: For matrix A = [[3, 1], [−1, 2]], show A² − 5A + 7I = O and hence find A⁻¹.

A² = [[8, 5], [−5, 3]] , so A² − 5A + 7I = [[8−15+7, 5−5], [−5+5, 3−10+7]] = O . Multiply by A⁻¹ : A − 5I + 7A⁻¹ = O ⇒ 7A⁻¹ = 5I − A = [[2, −1], [1, 3]] . Hence A⁻¹ = 1 7 2 −1 1 3 .

Q: Solve the system of linear equations using matrix method: x − y + 2z = 7; 3x + 4y − 5z = −5; 2x − y + 3z = 12.

A = 1 −1 2 3 4 −5 2 −1 3 , B = 7 −5 12 . • |A| = 1(12−5) + 1(9+10) + 2(−3−8) = 7 + 19 − 22 = 4 ≠ 0 . • adj(A) = 7 1 −3 −19 −1 11 −11 −1 7 . • X = ¼ adj(A)·B = ¼ 49−5−36 −133+5+132 −77+5+84 = ¼ 8 4 12 = 2 1 3 . Solution: x = 2, y = 1, z = 3 .

Q: Use the product [[1, −1, 2], [0, 2, −3], [3, −2, 4]] [[−2, 0, 1], [9, 2, −3], [6, 1, −2]] to solve: x − y + 2z = 1; 2y − 3z = 1; 3x − 2y + 4z = 2.

Let first matrix be A and second be B . Multiplying gives AB = 1 0 0 0 1 0 0 0 1 = I₃ ! Therefore A⁻¹ = B . The linear system is AX = [1, 1, 2]' ⇒ X = A⁻¹ [1, 1, 2]' = B [1, 1, 2]' = −2+0+2 9+2−6 6+1−4 = 0 5 3 . Hence x = 0, y = 5, z = 3 !

2×2 & 3×3 determinants, triangle area, minors, cofactors, adjoint inverse & solving linear systems.

📄 Official NCERT Textbook PDF

Topic 4.2: Determinants of Orders 1, 2 & 3 (Expansion Rules & |kA| = kⁿ|A|)

A matrix is a whole grid of numbers, like a blueprint table. A Determinant (|A| or det A) boils any square matrix down into ONE magical single number — telling you the area-stretching factor of that matrix and whether linear equations can be solved uniquely!

ORDERS 1 & 2Order 1: |[a]| = a (NOT modulus!)
Order 2: |a b; c d| = ad − bc

ORDER 3 EXPANSIONExpand along ANY row/col using checkerboard signs + − + / − + − / + − +!

SUPER EXAM FORMULAFor an n × n matrix: |kA| = kⁿ |A|!
Also |AB| = |A||B| & |A'| = |A|

🧠 Memory Trick: Always expand along the row or column with the MOST ZEROS! Zero times anything vanishes instantly! And for |kA|, pull k out of ALL n rows → kⁿ |A|!

💡 Simple Kid-Check: For a 2 × 2 determinant, multiply the main diagonal going down-right (↘ ad) and subtract the cross diagonal going down-left (↙ bc)! For a 3 × 3, walk across one row with signs + − +, hiding each cell's row & column to get a tiny 2 × 2!

3×3 Sign Checkerboard: [+ − +] / [− + −] / [+ − +]
Along R₁:

|A| = +a₁₁(a₂₂a₃₃ − a₂₃a₃₂) − a₁₂(a₂₁a₃₃ − a₂₃a₃₁) + a₁₃(a₂₁a₃₂ − a₂₂a₃₁)

🧒 Smart Zero-Picking Example: Evaluate Δ = |1 2 4; −1 3 0; 4 1 0|.
Look at Column 3: it has TWO zeros! Expand along C₃ (signs + − +):
Δ = +4 |−1 3; 4 1| − 0 + 0 = 4(−1 − 12) = −52 in ONE line!

THREE huge 1-mark DHSE traps! (1) Writing |[−5]| = +5 — WRONG! Vertical bars mean determinant, NOT absolute value, so |[−5]| = −5! (2) Writing |2A| = 2|A| for a 3×3 matrix — WRONG! Pulling 2 out of all 3 rows gives |2A| = 2³|A| = 8|A|! (3) Solving |x 2; 18 x| = |6 2; 18 6|: don't equate entries like matrices! Compute determinants: x² − 36 = 0 ⇒ x = ±6 (BOTH signs!)

Topic 4.3: Area of a Triangle & Collinearity via Determinants

Picture three corner pillars of a paddy field in Palakkad at coordinates (x₁, y₁), (x₂, y₂), and (x₃, y₃). Want the exact triangle area in 15 seconds without finding base lengths or altitudes? Pack those 3 points with a third column of 1s into a 3×3 determinant and take half its magnitude! And if all 3 pillars sit on one straight road (Collinear), the triangle flattens to Area = 0!

TRIANGLE AREA (Δ)Δ = ½ | x₁ y₁ 1 ; x₂ y₂ 1 ; x₃ y₃ 1 |
(Always take positive magnitude |Δ| for area!)

COLLINEAR TEST3 points lie on one straight line IF AND ONLY IF Δ = 0!

LINE EQUATIONLine joining (x₁, y₁) & (x₂, y₂):
| x y 1 ; x₁ y₁ 1 ; x₂ y₂ 1 | = 0

🧠 Memory Trick: Area given backwards? USE BOTH ±Area! Because a determinant can come out positive OR negative depending on vertex order!

💡 Simple Kid-Check: Notice the third column is just 1, 1, 1! When three points lie on the same straight line, they enclose zero area — so setting det = 0 with general point (x, y) instantaneously gives the equation of the line through two points!

🧒 Classic 2-Mark Board Problem (The ± Trap!):
Find k if area of triangle with vertices (k, 0), (4, 0), (0, 2) is 4 sq. units.
Step 1: Set ½ |k 0 1 ; 4 0 1 ; 0 2 1| = ±4 (Notice ±!)
Step 2: Expand along Column 2 (two zeros!): ½ [ −2(k − 4) ] = ±4 ⇒ −k + 4 = ±4.
• Using +4: −k + 4 = 4 ⇒ k = 0.
• Using −4: −k + 4 = −4 ⇒ k = 8. So k = 0 or 8!

NEVER drop the ± when area is given in the question! If you write ½ det = +4 only, you get k = 0 and miss k = 8, losing half your marks! Also remember to multiply by ½ — forgetting ½ doubles the triangle area into a parallelogram area!

Topic 4.4: Minors (Mᵢⱼ) & Cofactors (Aᵢⱼ = (−1)ⁱ⁺ʲ Mᵢⱼ)

Imagine sitting at seat (i, j) inside a 3×3 cinema grid and firing two laser beams that cross out your entire Row i and Column j! The smaller 2×2 determinant left behind is your Minor Mᵢⱼ. Now multiply it by the checkerboard sign (−1)ⁱ⁺ʲ — and boom, you have your Cofactor Aᵢⱼ!

MINOR (Mᵢⱼ)Determinant left after deleting Row i & Column j (No sign change!)

COFACTOR (Aᵢⱼ)Aᵢⱼ = (−1)ⁱ⁺ʲ Mᵢⱼ
Even (i+j) → +Mᵢⱼ | Odd (i+j) → −Mᵢⱼ

ALIEN ROW THEOREM!Same row: a₁₁A₁₁ + a₁₂A₁₂ + a₁₃A₁₃ = |A|
Different row: a₁₁A₂₁ + a₁₂A₂₂ + a₁₃A₂₃ = 0!

🧠 Memory Trick: Odd-sum Diamond gets MINUS! Inside a 3×3 grid, ONLY the 4 middle-edge cells (1,2), (2,1), (2,3), (3,2) flip sign; corners & centre stay +!

💡 Simple Kid-Check: Minor = bare sub-determinant; Cofactor = Minor PLUS its checkerboard + / − badge! And remember: multiplying elements of ANY row by cofactors of the SAME row builds |A|, while pairing them with cofactors of a DIFFERENT row always crashes to 0!

🧒 Quick Example: For Δ = |2 −3 5; 6 0 4; 1 5 −7|, find M₂₃ and A₂₃.
• Cross out Row 2 and Column 3 → left with |2 −3; 1 5|.
• M₂₃ = 2(5) − (−3)(1) = 10 + 3 = 13.
• Since 2 + 3 = 5 (odd!), A₂₃ = (−1)⁵ M₂₃ = −13!

Watch TWO classic 1-mark exam traps! (1) Multiplying by the element aᵢⱼ when asked for Minor or Cofactor! NEVER multiply by aᵢⱼ inside Mᵢⱼ or Aᵢⱼ! (2) Classic NCERT Ex 4.4 MCQ: What is a₁₁A₃₁ + a₁₂A₃₂ + a₁₃A₃₃? Since Row 1 elements are paired with Row 3 cofactors (mismatch!), the answer is ALWAYS 0!

Topic 4.5: Adjoint of a Square Matrix & Fundamental Property A(adj A) = |A| I

Once you compute all cofactors Aᵢⱼ of a square matrix, don't stop yet — TRANSPOSE that cofactor matrix (flip rows ↔ columns)! That transposed cofactor matrix is the Adjoint adj(A)! Amazingly, multiplying A by adj(A) wipes every off-diagonal entry clean to 0 and stamps |A| down the diagonal: A(adj A) = |A| I!

ADJOINT DEFINITIONadj(A) = [Aᵢⱼ]'
(Transpose of the Cofactor Matrix!)

2×2 SUPER SHORTCUTFor abcd, swap a ↔ d, negate b, c:
adj = d−b−ca

MAGIC PROPERTIESA(adj A) = (adj A)A = |A| I
And |adj A| = |A|ⁿ⁻¹ (order n)!

🧠 Memory Trick: C → T = adj! (1. Find Cofactors, 2. Transpose!). For 2×2: Swap Main, Sign Change Off-diagonal!

💡 Simple Kid-Check: Why does A(adj A) magically produce |A| I? Remember the Alien Row Theorem! Diagonal entries multiply Row i of A by Row i's cofactors (giving |A|!), while off-diagonal entries multiply Row i by Row j's cofactors (giving 0!)!

🧒 Instant 2×2 Adjoint & Property Check: Let A = 2314 (so |A| = 8 − 3 = 5).
Swap 2 ↔ 4 and negate 3, 1 ⇒ adj(A) = 4−3−12.
Check: A · adj(A) = 8−3−6+64−4−3+8 = 5005 = 5 I₂ = |A| I₂ ✓!

Watch TWO classic board exam traps! (1) Forgetting to TRANSPOSE the cofactor matrix — leaving cofactors untransposed gives the wrong answer in every 3×3 problem! (2) Classic 1-mark MCQ: "If A is a 3×3 matrix with |A| = 5, find |adj A|." Using |A|ⁿ⁻¹ = 5³⁻¹ = 25 (NOT 5 and NOT 125)! Also, if A(adj A) = 10 I₃, then |A| = 10!

Topic 4.5: Matrix Inverse Formula A⁻¹ = adj(A)/|A| & Singular Criterion

Look at A(adj A) = |A| I from Topic 4 and divide both sides by the single number |A|! Instantly you uncover the secret UNDO formula: A · (adj A|A|) = I, meaning A⁻¹ = adj A|A|! There is just ONE unbreakable Maths rule — never divide by zero! So when |A| = 0 (Singular), no inverse exists!

SINGULAR vs NON-SINGULARSingular: |A| = 0 (NO inverse!)
Non-Singular: |A| ≠ 0 (Invertible!)

INVERSE FORMULAA⁻¹ = 1|A| adj(A)

MCQ SHORTCUTS|A⁻¹| = 1|A| | |AB| = |A||B|
(AB)⁻¹ = B⁻¹A⁻¹

🧠 Memory Trick: Step 1 ALWAYS check |A| first! If |A| = 0 → STOP, Singular, NO Inverse! If |A| ≠ 0 → Swap/Cofactors → Transpose → Divide by |A|!

💡 Simple Kid-Check: A square matrix is invertible if and only if it is Non-Singular (|A| ≠ 0)! And notice: if A and B are both non-singular, their product AB is also non-singular because |AB| = |A||B| ≠ 0!

1
Direct 4-Step Inverse Recipe: (1) Find |A| (stop if 0); (2) Find all cofactors Aᵢⱼ; (3) Transpose to get adj(A); (4) Multiply by 1 / |A|!
2
Polynomial Trick (Zero Cofactor Work!): If A² − 5A + 7I = O, multiply every term by A⁻¹: A − 5I + 7A⁻¹ = O ⇒ A⁻¹ = 17(5I − A)!

🧒 Quick Inverse Check: Find A⁻¹ for A = 31−12.
• Step 1: |A| = 3(2) − 1(−1) = 7 ≠ 0 (Non-singular!).
• Step 2: Swap & negate ⇒ adj(A) = 2−113.
• Step 3: A⁻¹ = 17 2−113!

Watch TWO classic 1-mark traps! (1) Computing all 9 cofactors BEFORE checking |A| — if |A| = 0, you wasted 10 minutes! Always check |A| ≠ 0 FIRST! (2) If |A| = 4, then |A⁻¹| = 1/4 (NOT −4!), and (2A)⁻¹ = ½ A⁻¹ (NOT 2 A⁻¹!).

Topic 4.6: Solving Linear System AX = B by Matrix Method & Consistency

Suppose 2 notebooks and 3 pens cost ₹110, while 3 notebooks and 1 pen cost ₹95. Pack the coefficients into matrix A, the unknowns (x, y) into column X, and the bills into column B so the whole puzzle becomes ONE sleek equation: AX = B! Left-multiply by A⁻¹ and the answer pops straight out: X = A⁻¹ B!

MATRIX METHODWrite system as AX = B. If |A| ≠ 0, unique solution is X = A⁻¹ B!

|A| = 0 CHECKLISTCompute (adj A) B:
• (adj A)B ≠ O → Inconsistent (No solution)
• (adj A)B = O → Consistent (∞ solutions) or No solution!

4/5-MARK VERIFYALWAYS plug (x, y, z) back into all 3 equations in the rough column — 100% guarantee!

🧠 Memory Trick: LEFT-multiply A⁻¹: X = A⁻¹ B (NEVER B A⁻¹)! And if |A| = 0, NEVER guess "no solution" without checking (adj A)B first!

💡 Simple Kid-Check: Geometrically for 2 equations, |A| ≠ 0 means two lines cross at ONE clean point (x, y)! When |A| = 0, the lines have equal slope: if (adj A)B ≠ O, they are parallel tracks that never meet (No Solution); if (adj A)B = O, they lie right on top of each other (Infinitely Many Solutions)!

🧒 Full 4-Mark DHSE Board Solution: Solve 2x + 5y = 1, 3x + 2y = 7.
• A = 2532, X = xy, B = 17.
• |A| = 4 − 15 = −11 ≠ 0 (Consistent & unique!).
• adj(A) = 2−5−32 ⇒ A⁻¹ = 1−112−5−32.
• X = A⁻¹ B = 1−112(1)−5(7)−3(1)+2(7) = 1−11−3311 = 3−1 ⇒ x = 3, y = −1!
(Quick check: 2(3)+5(−1)=1 ✓ and 3(3)+2(−1)=7 ✓!)

Watch THREE common 5-mark blunders! (1) Writing X = B A⁻¹ instead of X = A⁻¹ B — remember matrix multiplication is NOT commutative, and B A⁻¹ isn't even dimension-compatible! (2) Saying "No solution" immediately when |A| = 0 — you MUST compute (adj A)B to distinguish parallel vs coincident lines! (3) Arithmetic sign slips in 3×3 cofactors — ALWAYS check your final (x, y, z) back in all 3 equations!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 1 Marks): Evaluate the determinant |cos θ −sin θ ; sin θ cos θ|.

Δ = cos θ · cos θ − (−sin θ) · sin θ = cos²θ + sin²θ = 1.

Q2 (Kerala HSE Say • 1 Marks): If A is a square matrix of order 3 with |A| = 5, find |2A| and |adj A|.

• |2A| = 2³ |A| = 8 × 5 = 40.
• |adj A| = |A|³⁻¹ = 5² = 25.

Q3 (Kerala HSE March • 2 Marks): Find values of k if area of triangle is 4 sq. units and vertices are (k, 0), (4, 0), (0, 2).

½ |k 0 1 ; 4 0 1 ; 0 2 1| = ±4
Expanding along C₂: ½ [−2(k − 4)] = ±4 ⇒ 4 − k = ±4.
• 4 − k = 4 ⇒ k = 0.
• 4 − k = −4 ⇒ k = 8.

Q4 (NCERT / DHSE Model • 2 Marks): Find the equation of the line joining (1, 2) and (3, 6) using determinants.

For collinear (x, y), (1, 2), (3, 6):
|x y 1 ; 1 2 1 ; 3 6 1| = 0
x(2 − 6) − y(1 − 3) + 1(6 − 6) = 0 ⇒ −4x + 2y = 0 ⇒ 2x − y = 0.

Q5 (Kerala HSE March • 3 Marks): Using cofactors of elements of third column, evaluate Δ = |1 x yz ; 1 y zx ; 1 z xy| (or verify a₁₁A₁₁+a₁₂A₁₂+a₁₃A₁₃ = |A| for [[2,−3,5],[6,0,4],[1,5,−7]]).

For [[2,−3,5],[6,0,4],[1,5,−7]] along Row 1:
• A₁₁ = +(0 − 20) = −20
• A₁₂ = −(−42 − 4) = +46
• A₁₃ = +(30 − 0) = 30
Δ = 2(−20) + (−3)(46) + 5(30) = −40 − 138 + 150 = −28.

Q6 (Kerala HSE Say • 3 Marks): For matrix A = [[3, 1], [−1, 2]], show A² − 5A + 7I = O and hence find A⁻¹.

A² = [[8, 5], [−5, 3]], so A² − 5A + 7I = [[8−15+7, 5−5], [−5+5, 3−10+7]] = O.
Multiply by A⁻¹: A − 5I + 7A⁻¹ = O ⇒ 7A⁻¹ = 5I − A = [[2, −1], [1, 3]].
Hence A⁻¹ = 17 2−113.

Q7 (Kerala HSE March • 4 Marks): Solve the system of linear equations using matrix method: x − y + 2z = 7; 3x + 4y − 5z = −5; 2x − y + 3z = 12.

A = 1−1234−52−13, B = 7−512.
• |A| = 1(12−5) + 1(9+10) + 2(−3−8) = 7 + 19 − 22 = 4 ≠ 0.
• adj(A) = 71−3−19−111−11−17.
• X = ¼ adj(A)·B = ¼ 49−5−36−133+5+132−77+5+84 = ¼ 8412 = 213.
Solution: x = 2, y = 1, z = 3.

Q8 (Kerala HSE Model • 5 Marks): Use the product [[1, −1, 2], [0, 2, −3], [3, −2, 4]] [[−2, 0, 1], [9, 2, −3], [6, 1, −2]] to solve: x − y + 2z = 1; 2y − 3z = 1; 3x − 2y + 4z = 2.

Let first matrix be A and second be B. Multiplying gives AB = 100010001 = I₃!
Therefore A⁻¹ = B.
The linear system is AX = [1, 1, 2]' ⇒ X = A⁻¹ [1, 1, 2]' = B [1, 1, 2]' = −2+0+29+2−66+1−4 = 053.
Hence x = 0, y = 5, z = 3!