Visual walkthrough — Cramer's rule

4.5.24 · D2 · Maths › Linear Algebra (Full) › Cramer's rule

Hum ek ek idea build karenge. Pehle yeh agree karte hain ki raw objects ka matlab kya hai.

---

Step 1 — "$A\mathbf{x}=\mathbf{b}$" actually kya keh raha hai

KYA HAI. Hamare paas equations ka ek system hai. Compactly likha jaaye toh: $\underset{\text{the box of columns}}{\underbrace{A}}\underset{\text{the mix amounts}}{\underbrace{\mathbf{x}}}=\underset{\text{the target}}{\underbrace{\mathbf{b}}}.$ Yahan $A$ ek square grid of numbers hai jisme $n$ rows aur $n$ columns hain; $\mathbf{x}=(x_1,\dots ,x_n)$ unknowns ki list hai jo hum chahte hain; $\mathbf{b}$ ek known target vector hai.

Columns ke nazriye se kyun padhein. Neeche ki har cheez ke liye sabse important fact yeh hai: $A$ ko $\mathbf{x}$ se multiply karna same hai jaise $A$ ke columns ko mix karna, jahan har $x_i$ "column $i$ kitna lena hai" bataata hai: $A\mathbf{x}=x_1{\mathbf{a}}_1+x_2{\mathbf{a}}_2+\cdots +x_n{\mathbf{a}}_n,$ jahan ${\mathbf{a}}_k$ ka matlab hai "$A$ ka $k$-waan column." Toh sawaal "$A\mathbf{x}=\mathbf{b}$ solve karo" literally yeh hai: "har column kitna-kitna milaaun ki $\mathbf{b}$ mil jaaye?"

PICTURE. Neeche, ek $2\times 2$ matrix ke dono columns ko arrows ki tarah draw kiya gaya hai. Unhe $x_1$ aur $x_2$ se scale karke tip-to-tail add karne par $\mathbf{b}$ milta hai.

---

Step 2 — Determinant = columns ke box ka area (volume)

KYA HAI. Determinant $\det A$ ek single number hai jo ek square matrix se attached hota hai. Geometrically yeh us box (parallelogram) ka signed area (2D mein) ya signed volume (3D mein) hota hai jiske edges columns ${\mathbf{a}}_1,\dots ,{\mathbf{a}}_n$ hain.

Kyun zaroori hai. Cramer's rule do aisi areas ka ratio hai. Toh divide karne se pehle humein unhe picture karna hoga. Signed word important hai: agar columns ko swapped order mein likhein toh area minus sign ke saath aata hai. Yeh yaad rakhna — yahi "alternating" behaviour hai jis par hum baad mein rely karenge.

PICTURE. ${\mathbf{a}}_1$ aur ${\mathbf{a}}_2$ se bana parallelogram; iski area $\det A$ hai.

---

Step 3 — Clever helper matrix $X_i$ banao

KYA HAI. Identity matrix $I$ se shuru karo (diagonal par $1$s, baaki $0$s — iske columns standard arrows ${\mathbf{e}}_1,\dots ,{\mathbf{e}}_n$ hain jo har axis ki taraf point karte hain). Ab sirf $i$-waan column ko unknown vector $\mathbf{x}$ se overwrite karo. Result ko $X_i$ kaho. $n=3$, $i=2$ ke liye: $X_2=\left(\begin{array}{ccc}1& x_1& 0\\ 0& x_2& 0\\ 0& x_3& 1\end{array}\right).$ Har column ek plain axis-arrow ${\mathbf{e}}_k$ hai, siwaaye column $i$ ke, jo orange unknown $\mathbf{x}$ hai.

Yeh invent kyun karein? Do reasons hain jo agle steps mein kaam aayenge: (1) iska determinant bahut simple hai, aur (2) jab hum $A{X}_i$ multiply karenge toh axis-columns $A$ ke columns ko untouched "pick out" karenge, jabki orange column $A\mathbf{x}=\mathbf{b}$ ban jaayegi. Yeh exactly ek hi column ko $\mathbf{b}$ se swap karne ke liye engineer kiya gaya hai.

PICTURE. $X_2$ ka box: do edges abhi bhi unit axis-arrows hain; sirf middle edge $\mathbf{x}$ ki taraf tilted hai.

---

Step 4 — $\det X_i=x_i$ kyun hai (picture se read karo)

KYA HAI. $X_i$ ka determinant unknown ka ek single coordinate equal hota hai: $\det X_i=x_i.$

KYU. Step 3 wala box dekho. $i$-waan edge ke siwa har edge ek unit axis-arrow hai length $1$ ki, mutually perpendicular. Yeh ek unit "floor" banate hain. Orange column $\mathbf{x}$ akela slanted edge hai; us floor ke upar iski height exactly iski $i$-waan coordinate $x_i$ hai ($\mathbf{x}$ ke sideways parts floor ke andar flat lete hain aur koi volume add nahi karte). Volume $=$ floor-area $\times$ height $=1\times x_i=x_i$.

Term by term: $\det X_i=\underset{\text{unit floor}}{\underbrace{1}}\times \underset{\begin{array}{c}\text{height of the}\\ \text{orange column}\end{array}}{\underbrace{x_i}}=x_i.$

PICTURE. Orange edge decomposed: iska "flat" part floor ke along slide karta hai (koi volume nahi), sirf perpendicular part $x_i$ height set karta hai.

Recall

$\mathbf{x}$ ke baaki components $\det X_i$ ko affect kyun nahi karte? ::: Woh un doosre axis-columns se bane unit floor ke andar hote hain, isliye woh sirf box ko shear karte hain — shearing se volume kabhi nahi badalti; sirf perpendicular height $x_i$ counts.

---

Step 5 — Core idea: $A{X}_i=A_i$

KYA HAI. Apne helper $X_i$ ko $A$ se multiply karo. Rule "$A$ times a matrix = $A$ applied to each column separately" ka matlab hai hum $X_i$ ke har column par $A$ lagate hain:

• Ek pure axis-column ${\mathbf{e}}_k$ ke liye $A{\mathbf{e}}_k={\mathbf{a}}_k$ hota hai — yeh simply $A$ ka column $k$ select kar leta hai, unchanged.
• Orange column $\mathbf{x}$ se $A\mathbf{x}=\mathbf{b}$ milta hai (haara system!).

Toh product woh $A$ hai jisme har column waise ka waisa rehta hai siwaaye $i$-waan ke, jo $\mathbf{b}$ ban gaya hai. Yahi woh matrix hai jise parent note $A_i$ kehta hai: $A{X}_i=[{\mathbf{a}}_1\cdots \underset{i\text{-th column}}{\underbrace{\mathbf{b}}}\cdots {\mathbf{a}}_n]=A_i.$

Yeh trick kyun hai. Humein column haath se edit nahi karna pada. Matrix multiplication ne swap khud kar diya, aur usne ek column swap kiya (row nahi) — kyunki $A$ columns par act karta hai. Yahi exact reason hai ki common mistake "ek row replace karo" galat hota hai.

PICTURE. Teen column-slots: unchanged ${\mathbf{a}}_1,{\mathbf{a}}_3$ pass through ho jaate hain; middle slot ka $\mathbf{x}$ transform hokar $\mathbf{b}$ ban jaata hai.

---

Step 6 — Determinants multiply hote hain, toh cancel karke answer milta hai

KYA HAI. $A{X}_i=A_i$ ke dono sides ka determinant lo, yeh fact use karte hue ki determinants multiply karte hain, $\det (MN)=\det M\cdot \det N$: $\underset{\text{original box}}{\underbrace{\det A}}\cdot \underset{=x_i}{\underbrace{\det X_i}}=\underset{\text{swapped box}}{\underbrace{\det A_i}}.$ Step 4 se $\det X_i=x_i$ substitute karo: $\det A\cdot x_i=\det A_i.$

Divide kyun karein. Agar (aur sirf agar) $\det A\ne 0$ ho toh hum dono sides ko isse divide kar sakte hain, ek akele unknown ko isolate karte hue: $\boxed{x_i=\frac{\det A_i}{\det A}}$ Picture ko areas ki tarah padhein: $x_i$ woh factor hai jitna original box $\det A$ ko badhna hoga taaki swapped box $\det A_i$ ban sake. Woh growth factor hi unknown hai.

PICTURE. Do nested boxes: original (area $\det A$) aur swapped wala (area $\det A_i$); unka ratio height $x_i$ hai.

---

Step 7 — Degenerate case: $\det A=0$

KYA HAI. Upar sab kuch ek ek forbidden move par depend karta tha — $\det A$ se divide karna — jo legal ho. Jab $\det A=0$ ho toh woh step illegal hai, aur Cramer's rule apply nahi hota.

KYU. $\det A=0$ ka matlab hai column-box flat hai: columns ${\mathbf{a}}_k$ ek lower-dimensional space (ek line, ya ek plane) mein squash ho gaye hain. Do sub-cases hain:

• Agar $\mathbf{b}$ us flat span ke andar hai, toh infinitely many mixes usse reach kar sakte hain — infinitely many solutions.
• Agar $\mathbf{b}$ flat span se bahar hai, toh koi bhi mix usse reach nahi kar sakta — no solution.

Kisi bhi case mein koi unique $x_i$ read off karne ko nahi hota, aur indeed formula $\frac{\text{something}}{0}$ compute karne ki koshish karta hai. Iske bajaaye Gaussian Elimination use karo; yeh bhi dekho Invertible Matrix Theorem — kyun $\det A\ne 0$ exactly "unique solution" condition hai.

PICTURE. Ek collapsed (zero-area) column-box: dono column-arrows parallel hain, isliye line se bahar $\mathbf{b}$ unreachable hai.

---

Step 8 — Sanity check: $\mathbf{b}$ mein linearity

KYA HAI. Predict karo ki target $\mathbf{b}$ double karne se kya hoga, phir verify karo.

Yeh kyun hona chahiye. $A_i$ mein sirf $i$-waan column $\mathbf{b}$ hai; determinant us column mein linear hota hai (multilinearity, Step 2). Us ek column ko double karne se $\det A_i$ double ho jaata hai, jabki $\det A$ (jisme koi $\mathbf{b}$ nahi hai) unchanged rehta hai. Isliye: $x_i^{\text{new}}=\frac{\det (2\mathbf{b}\text{ swapped in})}{\det A}=\frac{2\det A_i}{\det A}=2x_i.$ Har unknown double ho jaata hai — exactly wahi jo ek linear system ko karna chahiye. Yeh ek free correctness check hai jo tum kisi bhi answer par run kar sakte ho.

PICTURE. Same box, lekin swapped edge $\mathbf{b}\to 2\mathbf{b}$ swapped area double kar deti hai, isliye height $x_i$ bhi double ho jaati hai.

---

Ek-picture summary

Poori derivation ek single canvas par: $X_i$ banao, $A$ se multiply karke ek column $\mathbf{b}$ se swap karo, determinants lo (jo multiply karte hain), aur dono boxes ka ratio hi unknown hai.

Recall Feynman retelling — ek dost ko batao

Tum jaanna chahte ho ki smoothie mein kitna banana jaata hai. Apne teen ingredients (banana, apple, milk) ko teen arrows ki tarah picture karo; jo "box" yeh banate hain uska ek certain size hai — woh hai $\det A$. Ab ek trick container banao: apple aur milk ko plain unit measuring-cups rakhho, lekin poori bani smoothie banana slot mein daalo. Jab tum machine ($A$) ko is trick container par run karte ho, plain cups real ingredients ko untouched wapas de dete hain, aur smoothie-slot smoothie $\mathbf{b}$ ban jaata hai — toh tumhe ek naya box milta hai jiska size $\det A_i$ hai. Jaadu (Step 4 se) yeh hai ki trick container ka apna size exactly banana amount hota hai, aur sizes multiply hote hain jab tum machine run karte ho. Toh (banana ka size) × (original box ka size) = (swapped box ka size), yaani $x_i\cdot \det A=\det A_i$. Divide karo, aur banana count nikal aata hai. Agar original box flat hai — ingredients secretly ek doosre ki copies the — toh koi single recipe nahi hai, aur tum zero se divide nahi kar sakte.

---

Connections

• Determinants — har figure mein har box ek determinant hai.
• Multilinear and Alternating Maps — Steps 2, 4, 8 ke "ek column scale karo / equal columns ⇒ 0" rules.
• Cofactor Expansion — har $\det A_i$ actually compute kaise karte hain.
• Matrix Inverse — wahi $A{X}_i=A_i$ idea $A^{-1}=\frac{1}{\det A}\mathrm{adj}A$ ke roop mein packaged.
• Gaussian Elimination — Step 7 degenerate case mein kya karna hai.
• Invertible Matrix Theorem — $\det A\ne 0⟺$ unique-solution wali duniya jahan Cramer rehta hai.