NCERT Plus Two (Class 12) Mathematics • Chapter 3

Matrices

Q: Construct a 2 × 2 matrix A = [aᵢⱼ] whose elements are given by aᵢⱼ = |i − 2j| / 2.

• a₁₁ = |1−2|/2 = 1/2 , a₁₂ = |1−4|/2 = 3/2 • a₂₁ = |2−2|/2 = 0 , a₂₂ = |2−4|/2 = 1 A = 1/2 3/2 0 1

Q: Prove that all diagonal elements of a skew-symmetric matrix are zero.

If A = [aᵢⱼ] is skew-symmetric, A' = −A ⇒ aⱼᵢ = −aᵢⱼ . For diagonal elements, set j = i ⇒ aᵢᵢ = −aᵢᵢ ⇒ 2aᵢᵢ = 0 ⇒ aᵢᵢ = 0 .

Q: Find x, y, z, w given 3[[x, y], [z, w]] = [[x, 6], [−1, 2w]] + [[4, x+y], [z+w, 3]].

• 3x = x + 4 ⇒ 2x = 4 ⇒ x = 2 • 3y = 6 + x + y ⇒ 2y = 8 ⇒ y = 4 • 3w = 2w + 3 ⇒ w = 3 • 3z = −1 + z + w ⇒ 2z = 2 ⇒ z = 1

Q: If A = [[3, 1], [−1, 2]], show that A² − 5A + 7I = O.

A² = [[3,1],[−1,2]][[3,1],[−1,2]] = [[8, 5], [−5, 3]] . 5A = [[15, 5], [−5, 10]] , 7I = [[7, 0], [0, 7]] . A² − 5A + 7I = [[8−15+7, 5−5+0], [−5+5+0, 3−10+7]] = [[0, 0], [0, 0]] = O !

Q: For A = [[1], [−4], [3]] and B = [−1, 2, 1], verify that (AB)' = B'A'.

AB = [[−1, 2, 1], [4, −8, −4], [−3, 6, 3]] ⇒ (AB)' = [[−1, 4, −3], [2, −8, 6], [1, −4, 3]] . B' = [[−1], [2], [1]] and A' = [1, −4, 3] . B'A' = [[−1, 4, −3], [2, −8, 6], [1, −4, 3]] = (AB)' . Verified!

Q: Express A = [[6, −2, 2], [−2, 3, −1], [2, −1, 3]] as the sum of a symmetric and a skew-symmetric matrix (or verify A = P + Q for B = [[1, 5], [−1, 2]]).

Notice A' = A , so P = ½(A + A') = A (Symmetric) and Q = ½(A − A') = O (Skew-Symmetric)! For B = [[1, 5], [−1, 2]] : P = ½(B+B') = 1 2 2 2 and Q = ½(B−B') = 0 3 −3 0 ; check P + Q = B !

Q: Find X and Y if 2X + 3Y = [[2, 3], [4, 0]] and 3X + 2Y = [[2, −2], [−1, 5]].

Multiply first by 3 and second by 2, then subtract: 6X + 9Y = [[6, 9], [12, 0]] 6X + 4Y = [[4, −4], [−2, 10]] 5Y = [[2, 13], [14, −10]] ⇒ Y = 2/5 13/5 14/5 −2 . Similarly eliminate Y to get 5X = [[2, −12], [−11, 15]] ⇒ X = 2/5 −12/5 −11/5 3 .

Q: If A = [[1, 0, 2], [0, 2, 1], [2, 0, 3]], prove that A³ − 6A² + 7A + 2I = O.

Compute A² = [[5, 0, 8], [2, 4, 5], [8, 0, 13]] . Compute A³ = A² · A = [[21, 0, 34], [12, 8, 23], [34, 0, 55]] . Substitute: A³ − 6A² + 7A + 2I = [[21−30+7+2, 0, 34−48+14+0], [12−12+0+0, 8−24+14+2, 23−30+7+0], [34−48+14+0, 0, 55−78+21+2]] = O !

Matrix order, special types, addition/multiplication, transpose, symmetric/skew-symmetric & inverse.

📄 Official NCERT Textbook PDF

Topic 3.2 – 3.3: Matrix Order, Element Indexing (aᵢⱼ) & Special Matrix Types

Walk into a Kerala spice shop and look at the jar rack: 2 horizontal shelves (rows) and 3 vertical columns of jars! Need the cardamom jar on Shelf 2, Column 3? Its secret address is a₂₃! A neat rectangular number table like that is a Matrix of order 2 × 3 — the exact grid powering digital photos, game screens, and spreadsheets!

ORDER (m × n)m Rows × n Columns (Total elements = m · n)

ELEMENT ADDRESSaᵢⱼ lives in Row i (top→down) and Column j (left→right)

EQUALITY (A = B)Same order AND every matching cell equal: aᵢⱼ = bᵢⱼ!

🧠 Memory Trick: Think RC (Row × Column)! Just like reading a book — go across a Row first, then pick the Column!

💡 Simple Kid-Check: Every number inside a matrix has a two-digit GPS coordinate (i, j): first number = which floor (row), second number = which room (column). For A = 471905, order is 2 × 3 and a₂₁ = 9!

Matrix Type	Quick Rule	Visual Example
Row / Column Matrix	Only 1 row (`1×n`) or 1 col (`m×1`)	`[2 5 8]` (`1×3`)
Square Matrix	Rows = Columns (`m = n`)	`1342` (order 2)
Diagonal Matrix	Square; all non-diagonal entries = `0`	`500−2`
Scalar Matrix	Diagonal with ALL diagonal entries equal (`k`)	`3003`
Identity Matrix (I)	Diagonal with ALL diagonal entries = `1`	`1001`
Zero Matrix (O)	Every entry = `0` (any order)	`0000`

🧒 Classic 1-Mark Construction: Build a 2 × 2 matrix A = [aᵢⱼ] with aᵢⱼ = (i + j)²2.
• a₁₁ = 4/2 = 2, a₁₂ = 9/2, a₂₁ = 9/2, a₂₂ = 16/2 = 8 ⇒ A = 29/29/28!

Watch TWO 1-mark traps! (1) When asked "If a matrix has 8 elements, what are the possible orders?", students forget 1 × 8 and 8 × 1 — always list ALL factor pairs: 1×8, 8×1, 2×4, 4×2 (4 orders)! (2) "How many 3×3 matrices exist with entries 0 or 1?" Each of the 9 cells has 2 choices ⇒ 2⁹ = 512 (NOT 9 × 2 = 18)!

Topic 3.4.1 – 3.4.3: Matrix Addition, Scalar Multiplication & Properties

Suppose Kozhikode Bakery Branch A and Branch B each record Monday–Tuesday sales of Chips and Halwa inside matching 2 × 2 tables. Want combined sales? Just add matching cells: A + B! Expecting triple crowds for Onam? Multiply EVERY single cell by 3 — that's Scalar Multiplication 3A!

ADDITION (A ± B)Possible ONLY if A & B have the exact same order! Add cell-by-cell: (A+B)ᵢⱼ = aᵢⱼ + bᵢⱼ.

SCALAR MULTIPLY (kA)Multiply EVERY element of A by scalar k: (kA)ᵢⱼ = k · aᵢⱼ!

ALGEBRA LAWSA + B = B + A (Commutative) | A + O = A | A + (−A) = O

🧠 Memory Trick: Twin Shapes for +/−; Everyone Pays Tax for kA! You CANNOT add a 2×3 box to a 3×2 box, and kA multiplies ALL m×n entries!

💡 Simple Kid-Check: Matrix addition and subtraction work just like adding two identical scoreboards side-by-side — top-left adds to top-left, bottom-right adds to bottom-right! Scalar multiplication kA acts like zooming every number on the board by k!

🧒 Super-Easy 2-Mark Equation Solver:
Find X and Y if X + Y = 7025 and X − Y = 3003.
• Add both equations: 2X = 10028 ⇒ X = 5014!
• Subtract to get Y = X − 3003 = 2011!

Massive Matrix vs Determinant Trap! When scaling a MATRIX A by k, you multiply EVERY element of A by k! (Don't confuse this with Chapter 4 Determinants, where factoring k pulls out k from ONE row at a time, giving |kA| = kⁿ|A|!). Also remember: A + B is NOT defined if orders differ (e.g. 2×3 + 3×2)!

Topic 3.4.4 – 3.4.5: Matrix Multiplication (Row-by-Column) & Non-Commutativity

Suppose you buy 2 Puttu (₹30 each) and 3 Kadala (₹40 each). Your brain sweeps across the Row of Quantities [2, 3] and dives down the Column of Prices [30; 40] to compute one bill: 2×30 + 3×40 = ₹180! That "Row-runs-into-Column" dot product is how Matrix Multiplication works!

COMPATIBILITY RULE(m × n) · (n × p) = (m × p)
Inner numbers MUST match! Outer numbers give product size!

ROW × COLUMN DOTCell cᵢⱼ = (Row i of A) · (Column j of B)

NON-COMMUTATIVE!In general, AB ≠ BA! Also AB = O does NOT force A = O or B = O!

🧠 Memory Trick: The L-Dive! Left hand runs RIGHT across Row i (→), right hand dives DOWN Column j (↓), multiply pairs and add!

💡 Simple Kid-Check: NEVER multiply matrices entry-by-entry like addition! Think of each cell cᵢⱼ as a handshake between Row i of the first matrix and Column j of the second matrix! Also: A·I = I·A = A (Identity acts like the number 1!).

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Dimension Check: If A is 2 × 3 and B is 3 × 4, inner 3 = 3 matches ⇒ AB is 2 × 4! What about BA? (3×4)·(2×3) → 4 ≠ 2 ⇒ BA doesn't even exist!
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Zero-Divisor Surprise: Non-zero matrices can multiply to zero! Check 0−102 · 3500 = 0000!

🧒 Step-by-Step AB vs BA Check: Let A = 1234, B = 2012.
• AB = 1(2)+2(1)1(0)+2(2)3(2)+4(1)3(0)+4(2) = 44108
• BA = 2(1)+0(3)2(2)+0(4)1(1)+2(3)1(2)+2(4) = 24710 ⇒ AB ≠ BA!

THREE classic board exam traps! (1) Multiplying entry-by-entry (a₁₁b₁₁) — ALWAYS use Row × Column! (2) Expanding (A + B)² = A² + 2AB + B² — WRONG in general! Since AB ≠ BA, (A + B)² = A² + AB + BA + B²! (3) Assuming AB = O ⇒ A = O or B = O — FALSE for matrices!

Topic 3.5: Transpose of a Matrix (A') & Reversal Property (AB)' = B'A'

Imagine holding a 2 × 3 table of 2 students × 3 subjects and flipping it across its diagonal so rows become columns (now 3 × 2)! Nothing inside changes value — entry a₁₃ simply slides to (3, 1)! That clean diagonal flip is the Transpose A' (or Aᵀ)!

TRANSPOSE (A')Interchange Rows ↔ Columns: if A is m×n, then A' is n×m with (A')ᵢⱼ = aⱼᵢ!

BASIC LAWS(A')' = A | (kA)' = kA' | (A + B)' = A' + B'

REVERSAL LAW(AB)' = B' A' (Order FLIPS when transpose breaks brackets!)

🧠 Memory Trick: Product Transpose Flips Order: (AB)' = B' A'! Just check dimensions: if A is 2×3 & B is 3×4, then (AB)' is 4×2 — only B'A' (4×3 · 3×2) matches 4×2!

💡 Simple Kid-Check: To write A' in one second, read Row 1 of A left-to-right and write it straight DOWN as Column 1 of A'; then Row 2 becomes Column 2!

🧒 Classic 3-Mark NCERT Verification ((AB)' = B'A'):
Let A = 1−43 (3×1) and B = [−1 2 1] (1×3).
• AB = −1214−8−4−363 ⇒ (AB)' = −14−32−861−43
• B'A' = −121 · [1 −4 3] = −14−32−861−43 = (AB)' ✓!

Orthogonal Rotation Bonus: If A = cos αsin α−sin αcos α, then A'A = 1001 = I!

NEVER write (AB)' = A'B'! That's a guaranteed zero-mark blunder — always REVERSE the product order: (AB)' = B'A'! Also watch scalar transpose: (kA)' = kA' (scalars don't transpose or disappear!).

Topic 3.6: Symmetric vs Skew-Symmetric Matrices & Unique Sum Decomposition

Hold a square matrix up to its main diagonal mirror! If the top-right triangle mirrors the bottom-left identically (A' = A), it is Symmetric! If the bottom-left is the exact negative of the top-right (A' = −A), it is Skew-Symmetric! Best of all, EVERY square matrix splits uniquely into P + Q — one Symmetric piece plus one Skew-Symmetric piece!

SYMMETRIC (A' = A)Mirror twins across diagonal: aᵢⱼ = aⱼᵢ (e.g. A + A' is ALWAYS symmetric!)

SKEW-SYMMETRIC (A' = −A)Opposite twins: aᵢⱼ = −aⱼᵢ. Main diagonal entries MUST be ALL ZEROS!

UNIQUE SPLIT (A = P + Q)P = ½(A + A') (Symmetric)
Q = ½(A − A') (Skew-Symmetric)

🧠 Memory Trick: Diagonal of Skew-Symmetric = 0, 0, 0! Why? Because aᵢᵢ = −aᵢᵢ ⇒ 2aᵢᵢ = 0 ⇒ aᵢᵢ = 0! And don't forget the ½ in P + Q!

💡 Simple Kid-Check: Any number splits into two parts: take half the forward + backward sum and half the difference! For ANY square matrix A, A = 12(A + A') + 12(A − A') works like magic!

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Why Diagonal = 0 in Skew-Symmetric: Put i = j in aᵢⱼ = −aⱼᵢ ⇒ aᵢᵢ = −aᵢᵢ ⇒ 2aᵢᵢ = 0 ⇒ aᵢᵢ = 0!
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Classic MCQ Result: If A & B are symmetric, then AB − BA is ALWAYS Skew-Symmetric ((AB − BA)' = B'A' − A'B' = BA − AB = −(AB − BA))!

🧒 Worked 3-Mark Decomposition: Express A = 351−1 as Symmetric + Skew-Symmetric.
• A' = 315−1
• P = ½(A + A') = ½666−2 = 333−1 (P' = P ✓)
• Q = ½(A − A') = ½04−40 = 02−20 (Q' = −Q ✓)
• Check: P + Q = 351−1 = A ✓!

Watch TWO guaranteed board exam traps! (1) Forgetting the factor ½ when writing P = ½(A + A') and Q = ½(A − A') — without ½, P + Q = 2A ≠ A! (2) Putting non-zero numbers on the diagonal of a Skew-Symmetric matrix — IMPOSSIBLE, every diagonal element of a skew-symmetric matrix MUST be 0! (Only the Zero matrix is BOTH symmetric and skew-symmetric!)

Topic 3.7: Invertible Matrices, Uniqueness of Inverse & (AB)⁻¹ = B⁻¹A⁻¹

In regular numbers, multiplying 5 by its reciprocal 1/5 brings you back to 1. For a square matrix A, its exact UNDO partner is another square matrix B satisfying AB = BA = I! We call B the Inverse A⁻¹ — and just like taking off shoes before socks, inverting a product reverses order: (AB)⁻¹ = B⁻¹ A⁻¹!

DEFINITIONSquare matrix A is invertible if ∃ B of same order with AB = BA = I. Then B = A⁻¹!

UNIQUENESS THEOREMIf an inverse exists, it is UNIQUE! (No matrix has two different inverses!)

SHOE-SOCKS LAW(AB)⁻¹ = B⁻¹ A⁻¹ | (A⁻¹)⁻¹ = A | (A')⁻¹ = (A⁻¹)'

🧠 Memory Trick: Only SQUARE matrices can be invertible! (A rectangular matrix NEVER has an inverse!) And product inverse ALWAYS flips: (AB)⁻¹ = B⁻¹ A⁻¹!

💡 Simple Kid-Check: Why can't a rectangular 2 × 3 matrix have an inverse? Because for AB = I and BA = I to both produce square Identity matrices of the same size, A and B MUST be square! (And in Chapter 4 you'll see the exact formula using |A| ≠ 0!).

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Proof of Uniqueness (3 Lines!): Suppose B and C are both inverses of A. Then B = BI = B(AC) = (BA)C = IC = C! Done!
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Proof of (AB)⁻¹ = B⁻¹ A⁻¹: Multiply (AB)(B⁻¹ A⁻¹) = A(BB⁻¹)A⁻¹ = AIA⁻¹ = AA⁻¹ = I! Similarly (B⁻¹ A⁻¹)(AB) = I!

🧒 Quick Verification Check: Verify B = 2−3−12 is the inverse of A = 2312.
AB = 4−3−6+62−2−3+4 = 1001 = I₂, and BA = I₂! Thus A⁻¹ = B!

Watch TWO critical exam traps! (1) Writing (AB)⁻¹ = A⁻¹ B⁻¹ — WRONG! Like transpose, product inversion MUST reverse order: (AB)⁻¹ = B⁻¹ A⁻¹! (2) Thinking every non-zero square matrix is invertible — FALSE! For example, 1224 has NO inverse because no matrix can multiply it to give I₂!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 1 Marks): Construct a 2 × 2 matrix A = [aᵢⱼ] whose elements are given by aᵢⱼ = |i − 2j| / 2.

• a₁₁ = |1−2|/2 = 1/2, a₁₂ = |1−4|/2 = 3/2
• a₂₁ = |2−2|/2 = 0, a₂₂ = |2−4|/2 = 1
A = 1/23/201

Q2 (Kerala HSE Say • 1 Marks): Prove that all diagonal elements of a skew-symmetric matrix are zero.

If A = [aᵢⱼ] is skew-symmetric, A' = −A ⇒ aⱼᵢ = −aᵢⱼ.
For diagonal elements, set j = i ⇒ aᵢᵢ = −aᵢᵢ ⇒ 2aᵢᵢ = 0 ⇒ aᵢᵢ = 0.

Q3 (Kerala HSE March • 2 Marks): Find x, y, z, w given 3[[x, y], [z, w]] = [[x, 6], [−1, 2w]] + [[4, x+y], [z+w, 3]].

• 3x = x + 4 ⇒ 2x = 4 ⇒ x = 2
• 3y = 6 + x + y ⇒ 2y = 8 ⇒ y = 4
• 3w = 2w + 3 ⇒ w = 3
• 3z = −1 + z + w ⇒ 2z = 2 ⇒ z = 1

Q4 (NCERT / DHSE Model • 2 Marks): If A = [[3, 1], [−1, 2]], show that A² − 5A + 7I = O.

A² = [[3,1],[−1,2]][[3,1],[−1,2]] = [[8, 5], [−5, 3]].
5A = [[15, 5], [−5, 10]], 7I = [[7, 0], [0, 7]].
A² − 5A + 7I = [[8−15+7, 5−5+0], [−5+5+0, 3−10+7]] = [[0, 0], [0, 0]] = O!

Q5 (Kerala HSE March • 3 Marks): For A = [[1], [−4], [3]] and B = [−1, 2, 1], verify that (AB)' = B'A'.

AB = [[−1, 2, 1], [4, −8, −4], [−3, 6, 3]] ⇒ (AB)' = [[−1, 4, −3], [2, −8, 6], [1, −4, 3]].
B' = [[−1], [2], [1]] and A' = [1, −4, 3].
B'A' = [[−1, 4, −3], [2, −8, 6], [1, −4, 3]] = (AB)'. Verified!

Q6 (Kerala HSE March • 3 Marks): Express A = [[6, −2, 2], [−2, 3, −1], [2, −1, 3]] as the sum of a symmetric and a skew-symmetric matrix (or verify A = P + Q for B = [[1, 5], [−1, 2]]).

Notice A' = A, so P = ½(A + A') = A (Symmetric) and Q = ½(A − A') = O (Skew-Symmetric)!
For B = [[1, 5], [−1, 2]]: P = ½(B+B') = 1222 and Q = ½(B−B') = 03−30; check P + Q = B!

Q7 (Kerala HSE Say • 4 Marks): Find X and Y if 2X + 3Y = [[2, 3], [4, 0]] and 3X + 2Y = [[2, −2], [−1, 5]].

Multiply first by 3 and second by 2, then subtract:
6X + 9Y = [[6, 9], [12, 0]]
6X + 4Y = [[4, −4], [−2, 10]]
5Y = [[2, 13], [14, −10]] ⇒ Y = 2/513/514/5−2.
Similarly eliminate Y to get 5X = [[2, −12], [−11, 15]] ⇒ X = 2/5−12/5−11/53.

Q8 (Kerala HSE Model • 4 Marks): If A = [[1, 0, 2], [0, 2, 1], [2, 0, 3]], prove that A³ − 6A² + 7A + 2I = O.

Compute A² = [[5, 0, 8], [2, 4, 5], [8, 0, 13]].
Compute A³ = A² · A = [[21, 0, 34], [12, 8, 23], [34, 0, 55]].
Substitute:

A³ − 6A² + 7A + 2I = [[21−30+7+2, 0, 34−48+14+0], [12−12+0+0, 8−24+14+2, 23−30+7+0], [34−48+14+0, 0, 55−78+21+2]] = O