Solving 2×2 systems using Cramer's rule — Hinglish notes

2.6.11 · Maths › Matrices & Determinants — Introduction

WHAT are we solving?

HOW we derive Cramer's rule (scratch se — kuch yaad nahi karna)

Hum chahte hain $x$ akela ho. Elimination ka trick: $y$ ko khatam karne ke liye, equations ko un numbers se multiply karo jo $y$ -terms ko cancel kar den.

Step 1 — $y$ ko Eliminate karo. Eq.(1) ko $b_{2}$ se aur eq.(2) ko $b_{1}$ se multiply karo: $a_{1} b_{2} x + b_{1} b_{2} y = c_{1} b_{2}$ $a_{2} b_{1} x + b_{1} b_{2} y = c_{2} b_{1}$ Yeh step kyun? Humne multipliers precisely isliye choose kiye taaki dono $y$ -coefficients $b_{1} b_{2}$ ban jayein — ab woh cleanly subtract ho jaate hain.

Step 2 — Subtract karo. $(a_{1} b_{2} - a_{2} b_{1}) x = c_{1} b_{2} - c_{2} b_{1}$ Yeh step kyun? $y$ chala gaya; $x$ akela khada hai.

Step 3 — $x$ ke liye solve karo. $x = \frac{c _{1} b _{2} - c _{2} b _{1}}{a _{1} b _{2} - a _{2} b _{1}}$

Ab har piece dekho — yeh determinants hain:

c_1b_2 - c_2b_1 = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} \equiv D_x$$ Notice karo ki $D_x$ bas $A$ hai jisme uska **pehla column $\mathbf{c}$ se replace** ho gaya hai! **Step 4 — $y$ ke liye bhi same karo** (eq.(1) ko $a_2$ se, eq.(2) ko $a_1$ se multiply karo, subtract karo): $$(a_1b_2 - a_2b_1)\,y = a_1c_2 - a_2c_1 \;\Rightarrow\; y = \frac{a_1c_2 - a_2c_1}{a_1b_2 - a_2b_1}$$ aur $a_1c_2 - a_2c_1 = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} \equiv D_y$ — column 2 ko $\mathbf{c}$ se replace kiya. > [!formula] Cramer's Rule (2×2) > $$\boxed{\,x = \dfrac{D_x}{D}, \qquad y = \dfrac{D_y}{D}\,}$$ > $$D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix},\quad > D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix},\quad > D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix}$$ > **Valid tabhi hai jab $D \neq 0$.** ![[2.6.11-Solving-2×2-systems-using-Cramer's-rule.png]] --- ## WHY does $D=0$ break everything? > [!intuition] $D=0$ ki Geometry > Har equation ek **line** hai. Solution wahan hai jahan woh cross karti hain. > - $D \neq 0$: lines exactly **ek point** par milti hain → unique solution. > - $D = 0, D_x=D_y=0$: lines **identical** hain → infinitely many solutions. > - $D = 0$, lekin $D_x$ ya $D_y \neq 0$: lines **parallel, alag** hain → no solution. > > $D$ do rows ke beech ka "spread" measure karta hai (woh parallelogram ka area jo woh banaate hain). Agar rows same direction mein point kar rahe hain, toh woh area $0$ hai — system degenerate hai. --- ## Worked examples > [!example] Example 1 — ek clean unique solution > $$2x + 3y = 8, \qquad 4x - y = 6$$ > **Step 1: $D$.** $\;D=\begin{vmatrix}2&3\\4&-1\end{vmatrix}=(2)(-1)-(3)(4)=-2-12=-14.$ > *Pehle kyun?* Agar $D=0$ hota toh hum ruk jaate — Cramer apply nahi hota. > **Step 2: $D_x$** (col 1 ko $\begin{bmatrix}8\\6\end{bmatrix}$ se replace karo): > $D_x=\begin{vmatrix}8&3\\6&-1\end{vmatrix}=(8)(-1)-(3)(6)=-8-18=-26.$ > **Step 3: $D_y$** (col 2 replace karo): > $D_y=\begin{vmatrix}2&8\\4&6\end{vmatrix}=(2)(6)-(8)(4)=12-32=-20.$ > **Step 4:** $x=\dfrac{-26}{-14}=\dfrac{13}{7},\quad y=\dfrac{-20}{-14}=\dfrac{10}{7}.$ > **Check:** $2(\tfrac{13}{7})+3(\tfrac{10}{7})=\tfrac{26+30}{7}=\tfrac{56}{7}=8.$ ✓ > [!example] Example 2 — no solution ($D=0$) > $$x + 2y = 3, \qquad 2x + 4y = 10$$ > $D=\begin{vmatrix}1&2\\2&4\end{vmatrix}=4-4=0.$ Cramer koi value nahi de sakta. > $D_x=\begin{vmatrix}3&2\\10&4\end{vmatrix}=12-20=-8\neq0$ test karo. > Kyunki $D=0$ lekin $D_x\neq0$: **no solution** (parallel distinct lines: doosri line $x+2y=5$ hai, jo $x+2y=3$ se contradict karti hai). > [!example] Example 3 — infinitely many ($D=0$) > $$x + 2y = 3, \qquad 2x + 4y = 6$$ > $D=0$ phir se. $D_x=\begin{vmatrix}3&2\\6&4\end{vmatrix}=12-12=0,\;D_y=\begin{vmatrix}1&3\\2&6\end{vmatrix}=6-6=0.$ > Sab zero → **infinitely many solutions** (same line). Ek solve karo: $x=3-2y$, $y$ free. --- ## Forecast-then-Verify drill > [!recall] Compute karne se *pehle* predict karo > $3x - y = 5,\; x + 2y = 4$ ke liye: guess karo ki $D$ positive hai ya negative, phir compute karo. > *Verify:* $D=(3)(2)-(-1)(1)=6+1=7>0$. Unique solution: $D_x=\begin{vmatrix}5&-1\\4&2\end{vmatrix}=10+4=14$, $D_y=\begin{vmatrix}3&5\\1&4\end{vmatrix}=12-5=7$. Toh $x=14/7=2,\ y=7/7=1$. --- ## Common mistakes (Steel-manned) > [!mistake] "Main column ki jagah row replace kar raha hoon." > **Kyun sahi lagta hai:** equations *rows* mein likhi hoti hain, isliye row replace karna natural lagta hai. > **Sach yeh hai:** har *variable ek column* of coefficients ka malik hai. $D_x$ column 1 replace karta hai kyunki $x$ ke coefficients column 1 mein hain. **Fix:** dhire se bolo "variable = column." > [!mistake] "$D=0$ ka matlab $x=0$ hai." > **Kyun sahi lagta hai:** divide karne par usually chhote numbers aate hain, toh $0$ ek value jaisi lagti hai. > **Sach yeh hai:** $D=0$ se $D_x/D$ *undefined* ho jaata hai, zero nahi. Yeh signal hai ki koi unique solution nahi hai — $D_x,D_y$ investigate karo. **Fix:** hamesha pehle $D$ compute karo aur agar $0$ hai toh ruko. > [!mistake] Determinant mein sign slip. > **Kyun sahi lagta hai:** $a_1b_2 - a_2b_1$ mein ek minus hai jo easily drop ho jaata hai. > **Fix:** hamesha "main-diagonal product **minus** anti-diagonal product," usi order mein. --- ## #flashcards/maths Cramer's rule ka formula $x$ aur $y$ ke liye ::: $x=D_x/D,\ y=D_y/D$ jahan $D$ coefficient determinant hai. $D_x$ kaise banate hain? ::: $A$ ke column 1 ($x$-coefficients) ko constant column $\mathbf{c}$ se replace karo. $D_y$ kaise banate hain? ::: $A$ ke column 2 ($y$-coefficients) ko constant column $\mathbf{c}$ se replace karo. $a_1x+b_1y=c_1,\ a_2x+b_2y=c_2$ ke liye $D$ kya hai? ::: $a_1b_2-a_2b_1$. Cramer's rule kab valid hai? ::: Sirf jab $D\neq0$ ho (unique solution / lines ek baar intersect karti hain). $D=0$ aur $D_x\neq0$ ka matlab? ::: No solution — parallel distinct lines. $D=D_x=D_y=0$ ka matlab? ::: Infinitely many solutions — dono equations same line hain. $D$ ka geometric matlab? ::: Coefficient rows se bana signed area; $0$ matlab rows parallel/degenerate hain. Column replace karne se numerator kyun milta hai? ::: Elimination se $c_1b_2-c_2b_1$ milta hai, jo exactly $A$ hai jisme target variable ka column $\mathbf{c}$ se swap hua ho. > [!recall]- Feynman: ek 12-saal ke bachche ko samjhao > Socho do dost hain jo tumhe do secret numbers ke baare mein rules bata rahe hain. Pehla secret number find karne ke liye, tum uski "jagah" cover karo aur answer numbers wahan slide karte ho, phir dono covered board aur original board par thoda criss-cross math karte ho ($\times$ phir subtract). Dono results divide karo — woh tumhara pehla number hai! Doosre number ke liye bhi same karo. Agar "original board" ka math zero deta hai, toh do doston ke rules secretly same line hain ya kabhi milte hi nahi, isliye koi single answer nahi hai. > [!mnemonic] Yaad rakho > **"D neeche, column swap upar."** > $x$ → **1st** column swap karo, $y$ → **2nd** swap karo. Determinant = **diagonal down minus diagonal up.** --- ## Connections - [[Determinant of a 2×2 matrix]] — woh engine $D$ jo yahan use hota hai. - [[Matrix form of linear equations Ax=b]] — Cramer isko $A$ invert kiye bina solve karta hai. - [[Inverse of a 2×2 matrix]] — $x = A^{-1}c$ mein wahi $1/D$ factor chhupa hai. - [[Consistency of linear systems]] — $D=0$ ke cases yahan classify kiye gaye hain. - [[Cramer's rule for 3×3 systems]] — same idea, teen column swaps. ## 🖼️ Concept Map ```mermaid flowchart TD SYS[2x2 linear system Ax=c] -->|written as| COEF[Coefficient matrix A] SYS -->|solved via| ELIM[Eliminate y by subtraction] ELIM -->|yields| XFORM[x = ratio of two determinants] XFORM -->|generalized to| CRAMER[Cramer's Rule] COEF -->|determinant of A| D[D main determinant] D -->|replace col 1 with c| DX[Dx] D -->|replace col 2 with c| DY[Dy] CRAMER -->|x equals| DX CRAMER -->|y equals| DY CRAMER -->|requires| VALID[D not equal 0] D -->|measures| GEOM[Area between rows / line spread] GEOM -->|D=0 same lines| INF[Infinite solutions] GEOM -->|D=0 parallel| NONE[No solution] GEOM -->|D not 0| UNIQUE[Unique solution] ```