4.5.6 · D3 · Maths › Linear Algebra (Full) › Matrices — review, operations, types
Intuition Ye page kis liye hai
Parent note ne bataya tha ki operations kaise define hote hain. Ye page ensure karta hai ki tum koi bhi matrix problem dekho jo tumne pehle na dekhi ho. Hum pehle ek scenario matrix banate hain — har case type ki checklist — phir har cell ke liye ek example karte hain, aur compute karne se pehle answer forecast karte hain.
Har matrix exercise inme se kisi ek box mein aati hai. Agar tum har box se ek example kar sako, toh exam mein kuch bhi naya nahi hoga.
Cell
Scenario
Kya mushkil hai
A
Compatible product A B , dono nonzero
row·column pairing sahi karna
B
Non-commutativity: A B aur B A compute karo
ye dikhana ki dono alag hain
C
Degenerate : zero matrix / identity se multiply karna
"kuch nahi karta" aur "collapse" wale cases
D
Zero divisors : A B = 0 with A = 0 , B = 0
woh surprising box jo number-intuition tod deta hai
E
Transpose reversal ( A B ) ⊤ = B ⊤ A ⊤
order flip ho jaata hai
F
Symmetric + skew split
2 1 trick, zero diagonal
G
Shape mismatch : jab product exist hi nahi karta
dimensions padhna
H
Limiting / repeated : powers A 2 , A 3 of a matrix
khud ke saath composition
I
Word problem : matrices se kuch real cheez model karna
words ko grid mein translate karna
J
Exam twist : orthogonal matrix, verify A ⊤ A = I
length-preserving check
Neeche har symbol scratch se banaya gaya hai. Ek matrix bas ek rectangular grid of numbers hai; ek entry a ij row i , column j mein hoti hai (rows upar se neeche, columns left se right). Ek row aur ek column ka dot product matlab: matching numbers ko multiply karo aur add karo — yahi ek idea is page ke har product ko power karta hai. Ise ek horizontal strip ko ek vertical strip pe slide karte hue socho:
A B compute karo
A = [ 2 0 − 1 3 ] , B = [ 1 2 4 − 2 ]
Forecast: aage padhne se pehle, top-left entry guess karo. Ye A ki row 1 = ( 2 , − 1 ) aur B ki column 1 = ( 1 , 2 ) ka dot product hai. Apna guess yaad rakho.
Step 1 — shapes check karo.
Ye step kyun? Product tab hi exist karta hai jab inner dimensions match karein: A is 2 × 2 , B is 2 × 2 , inner numbers dono 2 hain. ✓ Result 2 × 2 hoga.
Step 2 — entry ( 1 , 1 ) : row 1 · col 1.
2 ⋅ 1 + ( − 1 ) ⋅ 2 = 2 − 2 = 0
Ye step kyun? Har output entry mein ek row aur ek column milte hain; kuch aur combine nahi hota.
Step 3 — baaki entries.
( 1 , 2 ) : 2 ⋅ 4 + ( − 1 ) ( − 2 ) = 8 + 2 = 10
( 2 , 1 ) : 0 ⋅ 1 + 3 ⋅ 2 = 6
( 2 , 2 ) : 0 ⋅ 4 + 3 ⋅ ( − 2 ) = − 6
A B = [ 0 6 10 − 6 ]
Verify: ( 1 , 1 ) entry 0 aayi — yahan ek coincidence hai, koi rule nahi. Sanity check: A ki row 2 mein 0 se shuru hota hai, isliye A B ki row 2 ki dono entries sirf 3 se aati hain: 3 ⋅ ( 2 , − 2 ) = ( 6 , − 6 ) . ✓ Match karta hai.
A , B ke liye B A compute karo
Forecast: kya B A , Cell A ke A B ke barabar hoga? "Nahi" guess karo — lekin kitna farq hoga?
Step 1 — entry ( 1 , 1 ) : B ki row 1 · A ki col 1.
1 ⋅ 2 + 4 ⋅ 0 = 2
Ye step kyun? Ab B likhne mein doosre lekin kaam mein pehle act karta hai — roles swap ho gaye, isliye hum B ki rows ko A ki columns se dot karte hain.
Step 2 — finish karo.
( 1 , 2 ) : 1 ⋅ ( − 1 ) + 4 ⋅ 3 = 11 , ( 2 , 1 ) : 2 ⋅ 2 + ( − 2 ) ⋅ 0 = 4 , ( 2 , 2 ) : 2 ⋅ ( − 1 ) + ( − 2 ) ⋅ 3 = − 8
B A = [ 2 4 11 − 8 ]
Verify: A B = [ 0 6 10 − 6 ] = [ 2 4 11 − 8 ] = B A . Har ek entry alag hai — concrete proof ki "pehle B phir A " = "pehle A phir B ." Jaise pehle moje phir joote.
Worked example Identity aur zero matrix se multiply karo
Maano I = [ 1 0 0 1 ] (identity : diagonal pe 1's, baaki jagah 0's) aur 0 = [ 0 0 0 0 ] .
Forecast: I A = ? aur 0 ⋅ A = ?
Step 1 — I A . I ki row 1 hai ( 1 , 0 ) ; kisi bhi column ke saath dot karne pe sirf us column ki pehli entry milti hai. Toh I har column ko waise ka waisa return karta hai.
I A = [ 2 0 − 1 3 ] = A
Ye step kyun? I ek "kuch nahi karta" transformation hai; multiply karne se kuch nahi badalta.
Step 2 — 0 ⋅ A . 0 ki har row ( 0 , 0 ) hai; har dot product 0 hai.
0 ⋅ A = [ 0 0 0 0 ] = 0
Verify: I A = A ✓ aur 0 A = 0 ✓ — dono degenerate limiting cases, parent ke "I A = A I = A " se match karte hain.
Worked example Do nonzero matrices jinका product zero hai
P = [ 1 0 0 0 ] , Q = [ 0 0 0 1 ]
Forecast: ordinary numbers mein, ab = 0 ka matlab hai a = 0 ya b = 0 . Kya tum yahan bhi yehi expect karte ho?
Step 1 — P Q compute karo.
( 1 , 1 ) : 1 ⋅ 0 + 0 ⋅ 0 = 0 , ( 1 , 2 ) : 1 ⋅ 0 + 0 ⋅ 1 = 0 , ( 2 , ⋅ ) : 0.
P Q = [ 0 0 0 0 ] = 0
Ye step kyun? P sirf x -axis ko rakhta hai; Q sirf y -axis ko. Pehle Q karo (y pe survive karo), phir P karo (y ko kill karo) — kuch bhi nahi bachta.
Verify: P = 0 , Q = 0 , phir bhi P Q = 0 . Toh rule "A B = 0 ⇒ A = 0 or B = 0 " matrices ke liye fail ho jaata hai. Ye zero divisors hain. Isliye tum matrices ko cancel nahi kar sakte jab tak wo invertible na hon — dekho Determinant and Inverse of a Matrix .
Worked example Reversal law verify karo
Cell A ke A , B use karo. Transpose A ⊤ grid ko uski diagonal ke across flip karta hai: ( A ⊤ ) ij = a j i .
Forecast: kaun sa order — A ⊤ B ⊤ ya B ⊤ A ⊤ — ( A B ) ⊤ se match karta hai?
Step 1 — answer ko transpose karo. A B = [ 0 6 10 − 6 ] ⇒ ( A B ) ⊤ = [ 0 10 6 − 6 ] .
Step 2 — B ⊤ A ⊤ banao.
B ⊤ = [ 1 4 2 − 2 ] , A ⊤ = [ 2 − 1 0 3 ]
B ⊤ A ⊤ = [ 1 ⋅ 2 + 2 ( − 1 ) 4 ⋅ 2 + ( − 2 ) ( − 1 ) 1 ⋅ 0 + 2 ⋅ 3 4 ⋅ 0 + ( − 2 ) 3 ] = [ 0 10 6 − 6 ]
Ye step kyun? Transpose roles aur order dono reverse karta hai — "moje aur joote": ulti order mein undo karo.
Verify: B ⊤ A ⊤ = ( A B ) ⊤ ✓. Agar A ⊤ B ⊤ try karo toh [ 4 − 5 − 6 9 ] milega — galat hai.
M ko symmetric aur skew-symmetric parts mein split karo
M = [ 5 − 4 2 1 ]
Ek symmetric matrix satisfy karta hai S ⊤ = S (diagonal ke across mirror); ek skew-symmetric satisfy karta hai K ⊤ = − K (diagonal pe zeros force hote hain, kyunki k ii = − k ii ).
Forecast: skew part ka diagonal sab zeros hona chahiye. Dhyan rakhna.
Step 1 — transpose. M ⊤ = [ 5 2 − 4 1 ] .
Step 2 — symmetric part 2 1 ( M + M ⊤ ) .
2 1 [ 10 − 2 − 2 2 ] = [ 5 − 1 − 1 1 ]
Ye step kyun? M ko uske mirror ke saath average karne se s ij = s j i force hota hai.
Step 3 — skew part 2 1 ( M − M ⊤ ) .
2 1 [ 0 − 6 6 0 ] = [ 0 − 3 3 0 ]
Verify: sum = [ 5 − 4 2 1 ] = M ✓; skew diagonal 0 hai ✓. Ye trick hamesha kaam karti hai kyunki tumne same 2 1 M ⊤ add aur subtract kiya.
Worked example Product kab illegal hai?
C = [ 1 4 2 5 3 6 ] ( 2 × 3 ) , D = [ 7 9 8 0 ] ( 2 × 2 )
Forecast: kya C D exist karta hai? Kya D C exist karta hai?
Step 1 — C D test karo. Inner dimensions: C ke 3 columns hain, D ke 2 rows hain. 3 = 2 → koi product nahi .
Ye step kyun? Dot product ke liye equal-length strips chahiye; C ki row mein 3 entries hain, D ki column mein 2. Pair nahi ho sakta.
Step 2 — D C test karo. D is 2 × 2 , C is 2 × 3 : inner 2 = 2 ✓, result 2 × 3 .
D C = [ 7 ⋅ 1 + 8 ⋅ 4 9 ⋅ 1 + 0 ⋅ 4 7 ⋅ 2 + 8 ⋅ 5 9 ⋅ 2 + 0 ⋅ 5 7 ⋅ 3 + 8 ⋅ 6 9 ⋅ 3 + 0 ⋅ 6 ] = [ 39 9 54 18 69 27 ]
Verify: D C is 2 × 3 jaise predict kiya ✓. D ki row 2 hai ( 9 , 0 ) , toh D C ki row 2 hai 9 ⋅ ( 1 , 2 , 3 ) = ( 9 , 18 , 27 ) ✓.
Worked example Ek shear ke liye
A 2 aur A 3 compute karo
A = [ 1 0 1 1 ]
Ye ek shear hai: ye points ko unki height ke barabar amount se right slide karta hai (dekho Linear Transformations ).
Forecast: A 2 guess karo. Shear do baar lagane se "double shear" honi chahiye.
Step 1 — A 2 = A ⋅ A .
[ 1 ⋅ 1 + 1 ⋅ 0 0 1 ⋅ 1 + 1 ⋅ 1 1 ] = [ 1 0 2 1 ]
Ye step kyun? A 2 matlab hai "pehle shear karo, phir dobara shear karo" — composition.
Step 2 — A 3 = A 2 ⋅ A .
[ 1 0 2 1 ] [ 1 0 1 1 ] = [ 1 0 3 1 ]
Verify: top-right entry count karta hai kitne shears hue — A n = [ 1 0 n 1 ] . Toh A 3 mein 3 hai ✓. Limiting picture: jaise n → ∞ , shear bina ruke badhti rehti hai.
Worked example Do shops, do products
Shop prices (₹ mein) rakhti hai pen = 10 , book = 50 , stored as a row P = [ 10 50 ] . Monday–Tuesday quantities sold hain columns of
Q = [ 3 1 5 2 ] ← pens ← books
Forecast: P Q ka kya matlab hai, aur uski kya shape hogi?
Step 1 — shapes. P is 1 × 2 , Q is 2 × 2 , inner 2 = 2 ✓ → result 1 × 2 : har din ke liye ek revenue number.
Ye step kyun? Inner dimensions match karna ensure karta hai ki har din ke price·quantity dot product mein sahi number of terms hain — ye ek dot product chhupa hua hai.
Step 2 — compute karo.
P Q = [ 10 ⋅ 3 + 50 ⋅ 1 10 ⋅ 5 + 50 ⋅ 2 ] = [ 80 150 ]
Verify (units): ₹/pen × pens + ₹/book × books = ₹. Monday: 30 + 50 = ₹80 ; Tuesday: 50 + 100 = ₹150 ✓. Matrix ne poora revenue calculation ek product mein pack kar diya.
Worked example Kya ye rotation matrix orthogonal hai?
R = [ cos θ sin θ − sin θ cos θ ]
Ek matrix orthogonal hoti hai jab R ⊤ R = I ; aisi maps lengths preserve karti hain — pure rotations/reflections (Eigenvalues and Eigenvectors mein central hain).
Forecast: expand karne se pehle R ⊤ R guess karo — kya wo I hona chahiye?
Step 1 — R ⊤ R banao. R ⊤ = [ cos θ − sin θ sin θ cos θ ] .
( 1 , 1 ) : cos 2 θ + sin 2 θ , ( 1 , 2 ) : − cos θ sin θ + sin θ cos θ , …
Ye step kyun? Har entry R ke do columns ka dot product hai; orthogonality ka matlab hai columns unit-length hain aur perpendicular hain.
Step 2 — cos 2 θ + sin 2 θ = 1 use karke simplify karo.
R ⊤ R = [ 1 0 0 1 ] = I
Verify (numeric, θ = 3 0 ∘ ): cos 3 0 ∘ = 2 3 , sin 3 0 ∘ = 2 1 . Toh ( 1 , 1 ) = 4 3 + 4 1 = 1 ✓ aur off-diagonal − 2 3 ⋅ 2 1 + 2 1 ⋅ 2 3 = 0 ✓. Rotation length preserve karta hai jaise expected tha.
Recall Kaun sa cell kaun sa tha?
Har product tab exist karta hai jab ::: inner dimensions match karein (pehle matrix ke columns = doosre matrix ke rows).
Woh box jo number-intuition tod deta hai ::: zero divisors, P Q = 0 with P , Q = 0 (Cell D).
Ek product ka transpose ::: order reverse karta hai: ( A B ) ⊤ = B ⊤ A ⊤ (Cell E).
Skew part mein hamesha hoti hai ::: zero diagonal (Cell F).
Shear [ 1 0 1 1 ] ke liye A n ::: [ 1 0 n 1 ] (Cell H).
Orthogonal test ::: R ⊤ R = I (Cell J).
"Pehle shapes, phir dot, order kabhi assume mat karo." Dimensions check karo, row·column dot products lo, aur kabhi mat mano ki A B = B A hai ya A B = 0 ka matlab koi factor zero hai.