Lesson Overview — Matrices (Ch 4 & 6)
Chapters
4 & 6
Days
9
Questions
80
Dates
8 Apr – 10 Jun
Chapter 4 — Matrices (Days 1–5)
DAY 1 · 8 APR
Types & Equality
DAY 2 · 9 APR
Ops & Transpose
DAY 3 · 10 APR
Sym & Skew
DAY 4 · 14 APR
Multiplication
DAY 5 · 15 APR
f(A) & MCQs
Chapter 6 — Adjoint & Inverse (Days 6–9)
DAY 6 · 22 APR
Cofactors & adj
DAY 7 · 23 APR
Inverse & C-H
DAY 8 · 9 JUN
ERO & ECO
DAY 9 · 10 JUN
AX=B & Word Probs
Common Misconceptions
✗ adj A is the TRANSPOSE of cofactor matrix — not cofactor matrix itself
✗ Matrix multiplication is NOT commutative: AB≠BA in general
✗ AB=O does NOT imply A=O or B=O (no cancellation law)
✗ NEVER mix row and column operations in the same ERO/ECO problem
✗ Reversal law: (AB)′=B′A′ and (AB)⁻¹=B⁻¹A⁻¹ — order always reverses
TUE 8 APR 2026
Day 1 — Types of Matrices & Equality
Concept Board — Day 1
Click to open · Draw diagrams & explain concepts
▶
Types of Matrices
Squarerows = cols; n×n matrix
Diagonalaᵢⱼ=0 for i≠j
Scalardiagonal with all equal entries
Identitydiagonal=1; denoted Iₙ
Zeroall entries=0; denoted O
EqualityA=B iff same order AND every aᵢⱼ=bᵢⱼ
Q1
Let A = 520-137614. State the order of A and find a₂₃ and a₃₁.
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Order of A = 3×3 (3 rows, 3 columns)
2
a₂₃ = entry in row 2, col 3 = 7
3
a₃₁ = entry in row 3, col 1 = 6
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Q2
If a matrix has 12 elements, list all possible orders. What if it has 7 elements?
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1
12 elements: factor pairs → 1×12, 2×6, 3×4, 4×3, 6×2, 12×1 → 6 possible orders
2
7 is prime → only 1×7 and 7×1 → 2 possible orders
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Q3
Construct a 2×2 matrix A=[aᵢⱼ] where aᵢⱼ=(i+2j)²/2.
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a₁₁=(1+2)²/2=9/2 ; a₁₂=(1+4)²/2=25/2 ; a₂₁=(2+2)²/2=8 ; a₂₂=(2+4)²/2=18
2
A = 9/225/2818
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Q4 (CBSE)
If a matrix has 8 elements, what are possible orders? What if 5 elements?
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8 = 1×8, 2×4, 4×2, 8×1 → 4 orders
2
5 is prime → 1×5, 5×1 → 2 orders
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Q5
Construct a 3×4 matrix: (i) aᵢⱼ=i−j (ii) aᵢⱼ=i·j (iii) aᵢⱼ=i/j.
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(i) R1=[0,−1,−2,−3]; R2=[1,0,−1,−2]; R3=[2,1,0,−1]
2
(ii) R1=[1,2,3,4]; R2=[2,4,6,8]; R3=[3,6,9,12]
3
(iii) R1=[1,½,⅓,¼]; R2=[2,1,⅔,½]; R3=[3,3/2,1,¾]
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Q6
For the matrix 5-217643-15 find: (i) order (ii) a₁₂, a₂₃ (iii) if aᵢⱼ=4, find i,j.
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(i) Order = 3×3
2
(ii) a₁₂ = −2 ; a₂₃ = 4
3
(iii) a₂₃=4 → i=2, j=3
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Q7 (Equality)
Find x,y,z,t if x+3x+2yz-14t-6 = 0-732t.
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x+3=0 → x=−3
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x+2y=−7 → −3+2y=−7 → y=−2
3
z−1=3 → z=4 ; 4t−6=2t → t=3
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Q8
Find x,y,z,w if x-y2x+z2x-y3z+w = -15013.
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x−y=−1 and 2x−y=0 → subtract: x=1, y=2
2
2x+z=5 → z=3 ; 3z+w=13 → w=4
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Q9
Find a,b,c,d if a-b2a+c2a-b3c+d = -15013.
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a−b=−1, 2a−b=0 → a=1, b=2
2
2a+c=5 → c=3 ; 3c+d=13 → d=4
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📎 Assignment Resources
A1
Homework
Construct a 3×4 matrix with aᵢⱼ = 2i − j.Day 1 — Video Resources
Exit Ticket
1Difference between scalar matrix and diagonal matrix?
2Can a 3×4 matrix equal a 4×3 matrix? Give reason.
Day Summary
▸Order: m×n → m rows, n cols
▸Equality: same order + every entry matches
▸Square: rows=cols
WED 9 APR 2026
Day 2 — Operations & Transpose
Concept Board — Day 2
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Matrix Operations
A+Bsame order; add entry-by-entry; commutative
kAmultiply every entry by scalar k
A−BA + (−1)B
TransposeA′: rows↔cols ; (AB)′=B′A′ reversal law
Q1
Compute 312214 + 102-130.
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1
Add entry by entry row by row
2
Result = 414144
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Q2
Find X: X + 13-12 = 4-162.
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X = 4-162 − 13-12
2
X = 3-470
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Q3
Find X: 3A − 2B + X = O, where A = 4213, B = -2132.
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X = 2B − 3A
2
3A=12639 , 2B=-4264
3
X = -16-43-5
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Q4
A=123-257, 2A−3B=45-9123. Find B.
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3B = 2A − 45-9123 = -2-115-5811
2
B = -2/3-1/35-5/38/311/3
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Q5 (ISC 1990)
Find non-zero k: 123-1-3-2 + k·102345 = 44104214.
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1
Entry (1,1): 1+k=4 → k=3
2
Verify: k=3 satisfies all entries
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Q6
A=235102345. Find B such that A+B−4I₃=O.
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B = 4I − A ; 4I₃=400040004
2
B = 2-3-5-14-2-3-4-1
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Q7
A=-2345-4-3729. Find ½(A−A′).
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A′=-2573-424-39
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A−A′=0-2-320-5350
3
½(A−A′)=0-1-3/210-5/23/25/20 — this is skew-symmetric!
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Q8
A=24-1-102, B=34-1221. Find (AB)′.
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AB=0131-2
2
(AB)′=0113-2 Verify: B′A′ gives same result ✓
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Q9 (CBSE)
A=cosα-sinαsinαcosα. Find α∈(0,π/2) such that A+A′=√2·I₂.
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A+A′=2cosα002cosα
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2cosα=√2 → cosα=1/√2 → α=π/4
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Q10
B=13-25, C=-2534. Find (BC)′.
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BC=7171910
2
(BC)′=7191710 ; Verify C′B′ gives same result ✓
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Q11
A=0ab-a0c-b-c0. Find ½(A+A′) and ½(A−A′).
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1
A′=0-a-ba0-cbc0 → A+A′=O
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½(A+A′)=O (zero matrix)
3
½(A−A′)=A itself — confirms A is skew-symmetric
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📎 Assignment Resources
A1
Homework
Ex 4B Q4 (A−B, 3A+2B), Q9. Ex 4E Q5 (transpose).Day 2 — Video Resources
Exit Ticket
1State the reversal law for (AB)′.
2If A is 2×3 and B is 3×2, is A+B defined? Is AB defined?
Day Summary
▸(A+B)′: = A′+B′
▸(AB)′: = B′A′ (order reverses)
▸(kA)′: = k·A′
FRI 10 APR 2026
Day 3 — Symmetric & Skew-Symmetric
Concept Board — Day 3
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Symmetric & Skew-Symmetric
SymmetricA′=A; symmetric about main diagonal
Skew-SymA′=−A; all diagonal entries must be 0
DecomposeA = ½(A+A′) + ½(A−A′) always
Sym partP=½(A+A′) is always symmetric
Skew partQ=½(A−A′) is always skew-symmetric
Q1
Show A=0ab-a0c-b-c0 is skew-symmetric.
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A′=0-a-ba0-cbc0
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A′=−A entry by entry ✓ ; all diagonal entries=0 ✓
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Q2
For A=1234, show (A+A′) is symmetric and (A−A′) is skew.
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A′=1324
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A+A′=2558 ; (A+A′)′=itself ✓ Symmetric
3
A−A′=0-110 ; transpose=−(A−A′) ✓ Skew
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Q3 (CBSE 2010)
Express A=213114-162 as sum of symmetric and skew-symmetric.
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P=½(A+A′), Q=½(A−A′), A=P+Q
2
A′=21-1116342 ; A+A′=42222102104
3
P=211115152, Q=00200-1-210 ; verify P+Q=A ✓
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Q4
If A,B symmetric of same order, prove A+B is symmetric.
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Given A′=A, B′=B
2
(A+B)′=A′+B′=A+B
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∴ A+B is symmetric ✓
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Q5 (ISC 2024)
If A,B symmetric, prove AB−BA is skew-symmetric.
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(AB−BA)′=(AB)′−(BA)′=B′A′−A′B′
2
=BA−AB=−(AB−BA)
3
∴ AB−BA is skew-symmetric ✓
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Q6
For any square A: (i) A+A′ is symmetric (ii) A−A′ is skew-symmetric.
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(i) (A+A′)′=A′+(A′)′=A′+A=A+A′ ✓
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(ii) (A−A′)′=A′−A=−(A−A′) ✓
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Q7
If A is both symmetric and skew-symmetric, show A=O.
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Symmetric: A′=A ; Skew: A′=−A
2
Both: A=−A → 2A=O → A=O
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Q8
Prove A′A is always symmetric for any matrix A.
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Let B=A′A
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B′=(A′A)′=A′(A′)′=A′A=B ✓
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Q9
Show ahghbfgfc is symmetric.
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M′ obtained by interchanging rows and columns
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M′=M (entry (i,j)=entry (j,i) throughout) ✓
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📎 Assignment Resources
A1
Homework
Prove A′A is always symmetric. Ex 4F Q3, Q7.Day 3 — Video Resources
Exit Ticket
1A = P + Q where P is ___ and Q is ___.
2If A is skew-symmetric, what must be true of all diagonal entries?
Day Summary
▸Symmetric: A′=A
▸Skew: A′=−A; diag=0
▸Decompose: A=½(A+A′)+½(A−A′)
TUE 14 APR 2026
Day 4 — Matrix Multiplication
Concept Board — Day 4
Click to open · Draw diagrams & explain concepts
▶
Matrix Multiplication
ConditionA(m×n) × B(n×p) — inner dims must match → result m×p
NOT commut.AB ≠ BA in general
AssociativeA(BC)=(AB)C always
DistributiveA(B+C)=AB+AC
AB=Odoes NOT imply A=O or B=O
Q1
State whether defined. If so, give order: (i) A₄ₓ₅·B₅ₓ₃ (ii) P₂ₓ₁·Q₂ₓ₃.
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(i) cols(A)=5=rows(B) ✓ → AB defined, order 4×3
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(ii) cols(P)=1 ≠ rows(Q)=2 → NOT defined
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Q2
A=12-13, B=2111. Find AB and BA. What do you observe?
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AB: R1=[4,3]; R2=[1,2] → AB=4312
2
BA: R1=[1,7]; R2=[0,5] → BA=1705
3
AB≠BA → Matrix multiplication is NOT commutative!
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Q3
A=312101, B=1-12131. Find AB.
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A is 2×3, B is 3×2 → AB is 2×2
2
R1: [3+2+6, −3+1+2]=[11,0] ; R2: [1+0+3, −1+0+1]=[4,0]
3
AB=11040
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Q4 (ISC 1992)
Find a,b: (3A−2B)·C where A=21-3142, B=1-202-13, C=20-1.
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3A−2B=47-9-1140
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Product = 17-2 = 17-2
3
a=17, b=−2
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Q5
Calculate 234345456 × 1-35024305.
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R1: [2+0+12, −6+6+0, 10+12+20]=[14,0,42]
2
R2: [3+0+15, −9+8+0, 15+16+25]=[18,−1,56]
3
R3: [4+0+18, −12+10+0, 20+20+30]=[22,−2,70]
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Q6
Find x: [x, 7] × 4x = [22].
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4x + 7x = 11x = 22
2
x = 2
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Q7
A=12-1-1, B=-1221, C=1-121. Find AB, BA. Verify (A+B)C=AC+BC.
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AB=34-1-3 , BA=-3-413 → AB≠BA ✓
2
AC=51-30 , BC=334-1
3
(A+B)C=841-1 = AC+BC=841-1 ✓
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📎 Assignment Resources
A1
Homework
Ex 4C Q5 (verify distributive), Q6 (verify associative).Day 4 — Video Resources
Exit Ticket
1Under what condition is AB defined?
2Give 2×2 matrices A≠O, B≠O such that AB=O.
Day Summary
▸Defined: cols(A)=rows(B)
▸NOT commut.: AB≠BA
▸Distrib.: A(B+C)=AB+AC
WED 15 APR 2026
Day 5 — f(A), Applications & MCQs
Concept Board — Day 5
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▶
Polynomial in A & Applications
f(A)Substitute matrix A for x in polynomial f(x)
Cayley-HamiltonEvery matrix satisfies its own characteristic equation
TrickIf A²=kA+mI → express higher powers without fresh multiplication
A²=A(I+A)³=I+7A → 7A−(I+A)³=−I
Q1 (CBSE 2003)
A=31-12. Show A²−5A+7I₂=O.
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A²=85-53
2
5A=155-510, 7I=7007
3
A²−5A+7I=0000 = O ✓
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Q2
A=3175. Find x,y such that A²+xI=yA.
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A²=1685632
2
Entry (1,2): 8=y → y=8
3
Entry (1,1): 16+x=24 → x=8
✓ Complete
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Q3 (NCERT)
A=3-24-2. Find k such that A²=kA−2I.
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A²=1-24-4
2
From entry (1,2): −2k=−2 → k=1. Verify all entries with k=1 ✓
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Q4
(A−2I)(A−3I)=O, A=42-1x. Find x.
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Compute (A−2I)(A−3I) and set equal to O
2
From entry (2,1)=0: gives x=1. Verify entry (1,2)=0 ✓
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Q5 (CBSE 2014)
A²=A. Find 7A−(I+A)³.
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(I+A)³=I+3A+3A²+A³. Since A²=A: A³=A
2
(I+A)³=I+3A+3A+A=I+7A
3
7A−(I+A)³=7A−I−7A=−I
✓ Complete
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Q6 (NCERT Exemplar)
A²=I. Find (A−I)³+(A+I)³−7A.
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A²=I → A³=A
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(A−I)³=4A−4I ; (A+I)³=4A+4I
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Sum=8A ; 8A−7A=A
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Q7 (ISC 2004)
A=3-41-1. Prove by induction Aⁿ=1+2n-4nn1-2n.
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Base n=1: A¹=3-41-1 = 3-41-1 ✓
2
Assume true for k. Compute Aᵏ⁺¹=Aᵏ·A using inductive hypothesis.
3
Result matches formula for n=k+1 ✓
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Q8 (CBSE App.)
Family A: 4 men, 6 women, 2 children. Family B: 2 men, 2 women, 4 children. Calories: man 2400, woman 1900, child 1800. Find total using matrix method.
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M=462224 , R=240019001800
2
MR=2460015800
3
Family A: 24800 cal. Family B: 17200 cal.
✓ Complete
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MCQ 1
P=004040400 is: (a) diagonal (b) symmetric (c) skew (d) none.
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P′=P (symmetric about diagonal) ✓
2
Answer: (b) Symmetric
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MCQ 2 (ISC 2024)
A,B symmetric same order. AB−BA is: (a) skew (b) symmetric (c) diagonal (d) identity.
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(AB−BA)′=B′A′−A′B′=BA−AB=−(AB−BA)
2
Answer: (a) Skew-symmetric
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MCQ 3 (JEE)
A=cosα-sinαsinαcosα, A³²=0-110. Find α.
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Aⁿ=rotation by nα ; A³²=R(32α)=R(π/2)
2
32α=π/2 → α=π/64. Answer: (c)
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MCQ 4 (A&R)
Assertion: [[0,k],[−2,0]] skew-sym → k=2. Reason: A−A′ is always skew-sym.
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For skew: A(1,2)=−A(2,1) → k=−(−2)=2. Assertion TRUE.
2
Reason: (A−A′)′=−(A−A′) ✓ TRUE and IS correct explanation.
3
Answer: (a)
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Q13
A=1002. Find A²−3A+2I.
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A²=1004 ; 3A=3006 ; 2I=2002
2
A²−3A+2I=0000 = O — A satisfies its characteristic equation ✓
✓ Complete
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📎 Assignment Resources
A1
Homework
Show A²−7A+10I=O. MCQ revision sheet.Day 5 — Video Resources
Exit Ticket
1If f(A)=O, how do you find A⁻¹ using the polynomial?
2AB=O. Can you cancel A to conclude B=O? Justify.
Day Summary
▸f(A): substitute A in polynomial
▸A²=A: 7A−(I+A)³=−I
▸A²=I: (A−I)³+(A+I)³−7A=A
WED 22 APR 2026
Day 6 — Cofactors & Adjoint
Concept Board — Day 6
Click to open · Draw diagrams & explain concepts
▶
Cofactors & Adjoint
CofactorAᵢⱼ=(−1)^(i+j)×Mᵢⱼ where Mᵢⱼ=minor (det of submatrix deleting row i, col j)
adj ATRANSPOSE of cofactor matrix ⚠ NOT cofactor matrix itself
TheoremA·(adj A)=(adj A)·A=|A|·Iₙ
|adj A|=|A|^(n−1) for n×n non-singular A
Q1
A=12457891012. Find cofactors A₂₂ and A₃₃.
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A₂₂=(−1)^4·det14912 = 12−36 = −24
2
A₃₃=(−1)^6·det1257 = 7−10 = −3
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Q2
Find adj A for A=1-12235-201.
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Compute all 9 cofactors: A₁₁=3, A₁₂=−12, A₁₃=6
2
A₂₁=1, A₂₂=5, A₂₃=2 ; A₃₁=−11, A₃₂=−1, A₃₃=5
3
adj A = TRANSPOSE of cofactor matrix
✓ Complete
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Q3 (ISC 2006)
A=10-13450-6-7. Find adj A. Verify A·(adj A)=|A|·I₃.
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|A|=1(−28+30)−0+(−1)(−18)=2+18=20
2
Compute all cofactors → transpose → adj A
3
A·(adj A)=20·I₃ ✓
✓ Complete
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Q4
A is 3×3, |A|=4. Find |2A|.
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|kA|=kⁿ·|A| for n×n matrix
2
|2A|=2³×4=8×4=32
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Q5
Find adj A for A=1234.
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2×2 shortcut: swap diagonal entries, negate off-diagonal entries
2
adj A=4-2-31
✓ Complete
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Q6 (AICBSE 2015)
A=-1-2-221-22-21. Find adj A. Verify A·(adj A)=|A|·I.
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|A|=−1(1−4)+2(2+4)+(−2)(−4−2)=3+12+12=27
2
Compute all cofactors → form adj A (transpose of cofactor matrix)
3
A·(adj A)=27·I₃ ✓
✓ Complete
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Q7 (NMOC 1994)
A=131210323. Find A·(adj A) WITHOUT finding adj A.
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1
Use: A·(adj A)=|A|·I — only need |A|
2
|A|=1(3−0)−3(6−0)+1(4−3)=3−18+1=−14
3
A·(adj A)=−14·I₃
✓ Complete
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Q8 (VSA)
|adj A|=64 for 3×3 matrix A. Find |A|.
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|adj A|=|A|^(n−1)=|A|² for n=3
2
|A|²=64 → |A|=±8
✓ Complete
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📎 Assignment Resources
A1
Homework
Prove A·adj A=O for singular A. Show adj(I)=I.Day 6 — Video Resources
Exit Ticket
1What exactly is adj A?
2State A·(adj A)=? and |adj A|=?
Day Summary
▸adj A: TRANSPOSE of cofactor matrix
▸Theorem: A·(adj A)=|A|·I
▸|adj A|: =|A|^(n−1)
THU 23 APR 2026
Day 7 — Inverse by Adjoint & Cayley-Hamilton
Concept Board — Day 7
Click to open · Draw diagrams & explain concepts
▶
Inverse of a Matrix
FormulaA⁻¹=(1/|A|)·adj A (only when |A|≠0)
2×2 shortcut[[a,b],[c,d]]⁻¹=(1/(ad−bc))·[[d,−b],[−c,a]]
Properties(AB)⁻¹=B⁻¹A⁻¹ ; (A′)⁻¹=(A⁻¹)′ ; det(A⁻¹)=1/det(A)
Cayley-Hamiltonf(A)=O → pre-multiply by A⁻¹ → express A⁻¹ in terms of A and I
Q1
Find A⁻¹ for A=-2534. Verify AA⁻¹=I.
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1
|A|=−8−15=−23≠0 → invertible
2
adj A=4-5-3-2
3
A⁻¹=(1/−23)·adj A. Verify AA⁻¹=I₂ ✓
✓ Complete
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Q2
Find A⁻¹ for A=1-1202-33-24.
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|A|=1(8−6)+1(0+9)+2(0−6)=2+9−12=−1≠0
2
Compute all 9 cofactors → form adj A
3
A⁻¹=(1/−1)·adj A. Verify AA⁻¹=I₃ ✓
✓ Complete
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Q3
A=53-1-2. Show A²−3A−7I=O. Hence find A⁻¹.
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A²=229-31. Verify A²−3A−7I=O ✓
2
A²=3A+7I → multiply by A⁻¹: A=3I+7A⁻¹
3
A⁻¹=(A−3I)/7=2/73/7-1/7-5/7
✓ Complete
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Q4 (NCERT)
A=11112-32-13. Show A³−6A²+5A+11I=O. Hence find A⁻¹.
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1
Compute A², then A³ step by step. Verify polynomial =O.
2
Multiply by A⁻¹: A²−6A+5I+11A⁻¹=O
3
A⁻¹=(1/11)(6A−A²−5I)
✓ Complete
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Q5
Find A⁻¹ for A=4523. Verify.
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|A|=12−10=2 ; adj A=3-5-24
2
A⁻¹=½·3-5-24. Verify AA⁻¹=I ✓
✓ Complete
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Q6
A=-1-12-2. Show A²+3A+4I=O. Find A⁻¹.
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A²=-13-62. A²+3A+4I=O ✓
2
A⁻¹=−¼(A+3I)=-1/21/4-1/2-1/4
✓ Complete
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Q7 (CBSE)
A=2-334. Show A²−6A+17I=O. Find A⁻¹.
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A²=-5-18187. Verify A²−6A+17I=O ✓
2
A⁻¹=(1/17)(6I−A)=(1/17)·43-34
✓ Complete
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Q8
A=4325. Find x,y: A²−xA+yI=O. Hence find A⁻¹.
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A²=22271831. From (1,2): 27−3x=0→x=9 ; from (1,1): y=14
2
A⁻¹=(1/14)(9I−A)=(1/14)·5-3-24
✓ Complete
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📎 Assignment Resources
A1
Homework
If A²−A+I=O show A⁻¹=I−A. Ex 6B Q16, Q20.Day 7 — Video Resources
Exit Ticket
1State the 2×2 shortcut for A⁻¹.
2First step to find A⁻¹ via Cayley-Hamilton?
Day Summary
▸A⁻¹: (1/|A|)·adj A — only if |A|≠0
▸2×2: swap diagonals, negate off-diagonal, divide by |A|
▸C-H: f(A)=O → pre-mult by A⁻¹
MON 9 JUN 2026
Day 8 — Inverse by ERO & ECO
Concept Board — Day 8
Click to open · Draw diagrams & explain concepts
▶
Elementary Operations
EROWrite A=IA; apply SAME row ops to both; LHS→I gives A⁻¹ on RHS
ECOWrite A=AI; apply SAME col ops to both; LHS→I gives A⁻¹ on RHS
⚠ WARNINGNEVER mix row and column operations in the same problem
SingularIf any row of LHS → all zeros → A is singular, A⁻¹ does not exist
Q1 (CBSE 2011)
Using ERO find A⁻¹: A=2513.
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[A|I]: R₁↔R₂ ; R₂→R₂−2R₁ → 130-1 | 011-2
2
R₂→−R₂ ; R₁→R₁−3R₂ → I | A⁻¹=3-5-12
✓ Complete
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Q2 (ISC 2018)
Using ERO find A⁻¹: A=-112123311.
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Set up [A|I₃] ; R₁→−R₁ ; R₂→R₂+R₁ ; R₃→R₃+3R₁
2
Continue eliminating col 2 and col 3 entries
3
When LHS→I, RHS=A⁻¹. Verify AA⁻¹=I₃ ✓
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Q3 (ISC 2018)
Using ECO find A⁻¹: A=20-1510013.
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1
Write A=AI. Apply COLUMN operations to both sides.
2
C₁↔C₂ ; C₁→C₁−5C₂ ; C₃→C₃+C₂ ; continue
3
When top block→I, bottom block=A⁻¹. Verify ✓
✓ Complete
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Q4 (NCERT)
Using ERO find A⁻¹: A=1-123.
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R₂→R₂−2R₁ : 1-105 | 10-21
2
R₂→R₂/5 ; R₁→R₁+R₂ → I | A⁻¹=3/51/5-2/51/5
✓ Complete
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Q5
Using ERO find A⁻¹: A=1-10253021.
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R₂→R₂−2R₁ ; continue eliminating to reduce LHS to I
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A⁻¹=11-3-2134-27. Verify ✓
✓ Complete
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Q6 (NCERT)
Using ERO find A⁻¹: A=13-2-30-5250.
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|A|=25≠0. Set up [A|I] ; R₂→R₂+3R₁ ; R₃→R₃−2R₁
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Complete reduction to I ; RHS=A⁻¹. Verify AA⁻¹=I₃ ✓
✓ Complete
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📎 Assignment Resources
A1
Homework
Ex 6C Q2, Q4, Q5. Never mix row and col ops.Day 8 — Video Resources
Exit Ticket
1If a row of LHS becomes all zeros, what does it mean?
2Can you use row ops then col ops in the same problem? Justify.
Day Summary
▸ERO: A=IA → same R-ops → LHS→I gives A⁻¹
▸ECO: A=AI → same C-ops → LHS→I gives A⁻¹
▸⚠: NEVER mix row and col ops
TUE 10 JUN 2026
Day 9 — Matrix Equations & Word Problems
Concept Board — Day 9
Click to open · Draw diagrams & explain concepts
▶
Solving AX = B
|A|≠0Unique solution: X=A⁻¹B (consistent)
|A|=0, (adj A)B≠ONo solution — INCONSISTENT
|A|=0, (adj A)B=OInfinitely many solutions — consistent dependent
HomogeneousAX=0: |A|≠0 → only trivial; |A|=0 → infinitely many
Q1 (ISC 2010)
Solve: 5x+3y+z=16, 2x+y+3z=19, x+2y+4z=25.
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A=531213124, B=161925; |A|=−22≠0
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X=A⁻¹B ; compute A⁻¹ via adj A
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x=1, y=2, z=3
✓ Complete
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Q2 (Consistency)
Test consistency: 2x+5y=7, 6x+15y=13.
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|A|=30−30=0 → singular
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adj A=15-5-62; (adj A)B=40-16
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(adj A)B≠O → INCONSISTENT — no solution
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Q3
Solve: x−y+z=2, 2x−y=0, 2y−z=1.
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A=1-112-1002-1, B=201; |A|=3≠0
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X=A⁻¹B → x=1, y=2, z=3
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Q4 (ISC Board)
A=4-5-111-3123-7. Find A⁻¹. Hence solve 4x−5y−11z=12, x−3y+z=1, 2x+3y−7z=2.
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|A|=4(21−3)+5(−7−2)+(−11)(3+6)=72−45−99=−72
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Compute A⁻¹=(1/−72)·adj A
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X=A⁻¹·1212; solve for x, y, z
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Q5 (Consistency)
Examine: x+y+z=6, x+2y+3z=14, x+4y+7z=30.
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|A|=1(14−12)−1(7−3)+1(4−2)=2−4+2=0
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Compute (adj A)B
3
(adj A)B=O → INFINITELY MANY solutions
✓ Complete
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Q6 (CBSE 2011)
A=1-202130-21, B=72-6-21-3-425. Find AB. Hence solve x−2y=10, 2x+y+3z=8, −2y+z=7.
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Compute AB — verify AB=7I₃
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A⁻¹=B/7 ; X=(1/7)·B·1087
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Solve to get x, y, z
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Q7 (Word Problem)
Gaurav: 3 pens, 2 bags, 1 box for ₹41. Dheeraj: 2 pens, 1 bag, 2 boxes for ₹29. Ankur: 2 pens, 2 bags, 2 boxes for ₹44. Find cost of each item.
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Let pen=x, bag=y, box=z. A=321212222, B=412944
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|A|=−4≠0 ; X=A⁻¹B
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pen=₹2, bag=₹10, box=₹5
✓ Complete
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Q8 (CBSE 2013)
School awards Honesty(h), Regularity(r), Hard Work(w). h+r+w=6000; h+3w=11000; h+w=2r. Find prize for each value.
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h−2r+w=0 ; A=1111031-21, B=6000110000
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|A|=6≠0 ; X=A⁻¹B
3
h=₹500, r=₹2000, w=₹3500
✓ Complete
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Q9 (ISC 2021)
Factory uses teakwood(T), rosewood(R), satinwood(S). Table needs [2,1,3], chair [3,1,2], cot [4,2,1] tonnes. Stock: 29T, 13R, 16S. Find number of each using AX=B.
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A=234112321, B=291316; |A|=5≠0
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X=A⁻¹B
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Solve to get tables, chairs, cots. Verify by substituting back ✓
✓ Complete
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📎 Assignment Resources
A1
Homework
Ex 6D Q3, Q15, Q19. Review ISC 2021 factory problem.Day 9 — Video Resources
Exit Ticket
1When does AX=B have no solution even when A is non-zero?
2For AX=O with |A|≠0, what is the only solution?
Day Summary
▸|A|≠0: unique solution X=A⁻¹B
▸|A|=0,(adj A)B≠O: no solution
▸|A|=0,(adj A)B=O: ∞ many solutions