2.6.12 · D3 · HinglishMatrices & Determinants — Introduction
Worked examples — Solving systems using matrix inversion
2.6.12 · D3· Maths › Matrices & Determinants — Introduction › Solving systems using matrix inversion
Yeh page ek shooting range hai. Parent note ne tumhe recipe di thi. Yahan hum us recipe ko har us situation par fire karte hain jisme woh lag sakti hai — clean cases, sign traps, dreaded branches, ek word problem, aur ek exam twist — taaki tum koi bhi aisa scenario na dekho jo tumne pehle survive nahi kiya ho.
Shuru karne se pehle, teen chhoti reminders taaki koi symbol achanak na aaye:
Scenario matrix
Is topic ke har problem ka ek hi cell mein fit hona zaroori hai. Last column us worked example ka naam batata hai jo use clear karta hai.
| Cell | Kya khaas hai isme | Outcome | Example |
|---|---|---|---|
| C1 | Clean , | ek unique solution | Ex 1 |
| C2 | with negative / mixed-sign entries | unique, sign-care | Ex 2 |
| C3 | , | unique | Ex 3 |
| C4 | , | no solution (inconsistent) | Ex 4 |
| C5 | , | infinitely many | Ex 5 |
| C6 | (homogeneous system) | trivial , ya ek poori line | Ex 6 |
| C7 | Real-world word problem | translate karo → solve karo | Ex 7 |
| C8 | Exam twist: ko kai ke liye solve karo | ek reuse karo | Ex 8 |
Hum saaton aath visit karenge.
Woh geometry jo har cell ko unite karti hai
mein do equations do straight lines hain. Solution wahan hai jahan woh cross karti hain. Woh ek picture saare outcomes explain karti hai: ek baar cross karna (unique), parallel aur alag (no solution), same line (infinitely many).
Yeh picture apne dimag mein rakhho — neeche ka algebra sirf yahi geometry hai jo numbers se count ki gayi hai.
Ex 1 — Cell C1 · clean
Step 1 — mein pack karo. Yeh step kyun? woh system hai jo repackage hua hai; row-times-column har equation reproduce karta hai.
Step 2 — pehle determinant. Yeh step kyun? green light hai — inverse exist karta hai, unique solution exist karta hai.
Step 3 — inverse rule apply karo (diagonal swap karo, off-diagonal negate karo, se divide karo): Yeh step kyun? Yeh ke liye ka closed form hai.
Step 4 — multiply karo, ko right par rakhte hue ():
=\frac14\begin{bmatrix}24-16\\-12+24\end{bmatrix}=\frac14\begin{bmatrix}8\\12\end{bmatrix}=\begin{bmatrix}2\\3\end{bmatrix}.$$ *$A^{-1}B$ kyun aur $BA^{-1}$ kyun nahi?* Matrices commute nahi karti; sirf left-multiply karne se left wala $A$ cancel hota hai. **Verify:** $3(2)+2(3)=12$ ✓ aur $1(2)+2(3)=8$ ✓. Solution $(x,y)=(2,3)$. --- ## Ex 2 — Cell C2 · negative / mixed signs > [!example] Statement > Solve karo $\;2x-3y=-4,\quad -x+4y=9.$ > > **Forecast:** negative coefficients *method* nahi badalta, sirf arithmetic mein dikkat aati hai. Ek solution expect karo; har minus sign dhyan se dekho. **Step 1 — matrices.** $$A=\begin{bmatrix}2&-3\\-1&4\end{bmatrix},\quad B=\begin{bmatrix}-4\\9\end{bmatrix}.$$ **Step 2 — determinant.** $$\det A = 2\cdot4-(-3)(-1)=8-3=5\neq0.$$ *Yeh step kyun?* Non-zero ⇒ invert karna safe hai. Dhyan do $(-3)(-1)=+3$, yeh ek classic sign trap hai. **Step 3 — inverse.** $$A^{-1}=\frac15\begin{bmatrix}4&3\\1&2\end{bmatrix}.$$ *Yeh step kyun?* Diagonal swap karo ($2,4\to4,2$), off-diagonal negate karo ($-3\to3,\ -1\to1$), $5$ se divide karo. **Step 4 — solve karo.** $$X=\frac15\begin{bmatrix}4&3\\1&2\end{bmatrix}\begin{bmatrix}-4\\9\end{bmatrix} =\frac15\begin{bmatrix}-16+27\\-4+18\end{bmatrix}=\frac15\begin{bmatrix}11\\14\end{bmatrix}=\begin{bmatrix}11/5\\14/5\end{bmatrix}.$$ **Verify:** $2(11/5)-3(14/5)=\tfrac{22-42}{5}=\tfrac{-20}{5}=-4$ ✓; $-11/5+4(14/5)=\tfrac{-11+56}{5}=\tfrac{45}{5}=9$ ✓. --- ## Ex 3 — Cell C3 · clean $3\times3$ > [!example] Statement > Solve karo $\;x+2y+3z=9,\quad 2x-y+z=8,\quad 3x-z=3.$ > > **Forecast:** space mein teen planes. Agar woh ek point par milte hain toh unique $(x,y,z)$ milega. Chhote whole numbers expect karo. **Step 1 — matrices.** $$A=\begin{bmatrix}1&2&3\\2&-1&1\\3&0&-1\end{bmatrix},\quad B=\begin{bmatrix}9\\8\\3\end{bmatrix}.$$ **Step 2 — determinant, row 1 ke along expand karte hue.** $$\det A = 1\big[(-1)(-1)-1\cdot0\big]-2\big[2(-1)-1\cdot3\big]+3\big[2\cdot0-(-1)3\big]$$ $$=1(1)-2(-5)+3(3)=1+10+9=20\neq0.$$ *Yeh step kyun?* $\det=20\neq0$ ⇒ unique solution; aur hume adjugate ko divide karne ke liye bhi yeh chahiye. **Step 3 — adjugate** (cofactor matrix ka transpose; dekho [[Cofactors and Minors]]): $$\operatorname{adj}(A)=\begin{bmatrix}1&2&5\\5&-10&5\\3&6&-5\end{bmatrix}.$$ *Transpose kyun?* $\operatorname{adj}(A)_{ij}=C_{ji}$; flip karna bhool jaana #1 galti hai. **Step 4 — solve karo** $X=\frac{1}{20}\operatorname{adj}(A)\,B$: $$X=\frac1{20}\begin{bmatrix}1&2&5\\5&-10&5\\3&6&-5\end{bmatrix}\begin{bmatrix}9\\8\\3\end{bmatrix} =\frac1{20}\begin{bmatrix}9+16+15\\45-80+15\\27+48-15\end{bmatrix}=\frac1{20}\begin{bmatrix}40\\-20\\60\end{bmatrix}=\begin{bmatrix}2\\-1\\3\end{bmatrix}.$$ **Verify:** $1(2)+2(-1)+3(3)=2-2+9=9$ ✓; $2(2)-(-1)+3=4+1+3=8$ ✓; $3(2)-3=3$ ✓. --- ## Ex 4 — Cell C4 · $\det A=0$, no solution > [!example] Statement > Solve karo $\;x+2y=3,\quad 2x+4y=7.$ > > **Forecast:** doosri equation *almost* pehli ki twice hai — $2x+4y$ ka matlab hai $2(x+2y)$, jo $2(3)=6$ hona chahiye, lekin yeh $7$ ke barabar hai. Contradiction aa raha hai. **No solution** expect karo. **Step 1 — matrices & determinant.** $$A=\begin{bmatrix}1&2\\2&4\end{bmatrix},\quad B=\begin{bmatrix}3\\7\end{bmatrix},\qquad \det A=1\cdot4-2\cdot2=0.$$ *Yeh step kyun?* $\det A=0$ inversion **forbid** karta hai — parent ki recipe yahan ruk jaati hai aur hum consistency analysis ki taraf switch karte hain. Geometrically do lines parallel hain (same slope $-\tfrac12$). **Step 2 — $(\operatorname{adj}A)B$ test karo.** $2\times2$ ke liye, $\operatorname{adj}A=\begin{bmatrix}4&-2\\-2&1\end{bmatrix}$. $$(\operatorname{adj}A)B=\begin{bmatrix}4&-2\\-2&1\end{bmatrix}\begin{bmatrix}3\\7\end{bmatrix}=\begin{bmatrix}12-14\\-6+7\end{bmatrix}=\begin{bmatrix}-2\\1\end{bmatrix}\neq O.$$ *Yeh step kyun?* $AX=B$ ko $\operatorname{adj}A$ se multiply karne par $(\det A)X=(\operatorname{adj}A)B$ milta hai. Kyunki $\det A=0$ hai, left side $O$ hai; **non-zero** right side ek impossible equation $O=(\text{nonzero})$ hai. **Step 3 — conclude karo.** Koi $X$ satisfy nahi kar sakta $O=\begin{bmatrix}-2\\1\end{bmatrix}$. **System inconsistent hai — no solution.** Dekho [[Consistency of Linear Systems]]. **Verify (picture-check):** lines $y=\tfrac{3-x}{2}$ aur $y=\tfrac{7-2x}{4}$ ke intercepts $1.5$ aur $1.75$ hain — parallel, kabhi nahi milte. ✓ ![[deepdives/dd-maths-2.6.12-d3-s02.png]] --- ## Ex 5 — Cell C5 · $\det A=0$, infinitely many > [!example] Statement > Solve karo $\;x+2y=3,\quad 2x+4y=6.$ > > **Forecast:** ab doosri equation *exactly* pehli ki twice hai. Ek hi line do baar draw ki gayi — **infinitely many** solutions expect karo. **Step 1 — determinant.** Same $A$ hai jaise Ex 4 mein, isliye $\det A=0$ phir se. **Step 2 — $(\operatorname{adj}A)B$ test karo** $B=\begin{bmatrix}3\\6\end{bmatrix}$ ke saath: $$(\operatorname{adj}A)B=\begin{bmatrix}4&-2\\-2&1\end{bmatrix}\begin{bmatrix}3\\6\end{bmatrix}=\begin{bmatrix}12-12\\-6+6\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}=O.$$ *Yeh step kyun?* $\det A=0$ **aur** $(\operatorname{adj}A)B=O$ "infinitely many (or none)" branch hai. Yahan do equations proportional hain, isliye genuinely infinitely many solutions hain. **Step 3 — solution set describe karo.** $x+2y=3$ se, $y=t$ (free parameter) set karo: $x=3-2t$. $$\boxed{(x,y)=(3-2t,\ t),\quad t\in\mathbb{R}.}$$ *Parameter kyun?* Ek equation, do unknowns ⇒ answers ki poori ek line. **Verify:** $t=1$ lo $\Rightarrow(1,1)$: $1+2=3$ ✓ aur $2+4=6$ ✓. $t=0$ lo $\Rightarrow(3,0)$: $3=3$ ✓, $6=6$ ✓. --- ## Ex 6 — Cell C6 · homogeneous ($B=O$) > [!example] Statement > Solve karo $\;2x+y=0,\quad x-3y=0$ (dono constants zero hain). > > **Forecast:** $X=O$ (sab zeros) *hamesha* kaam karta hai jab $B=O$ ho. Asli sawaal yeh hai ki kya woh *akela* solution hai — yeh $\det A$ decide karta hai. **Step 1 — determinant.** $$A=\begin{bmatrix}2&1\\1&-3\end{bmatrix},\qquad \det A=2(-3)-1(1)=-7\neq0.$$ *Yeh step kyun?* Agar $\det A\neq0$, toh $X=A^{-1}B=A^{-1}O=O$ **unique** solution hai — do lines sirf origin par cross karti hain. **Step 2 — conclude karo.** $X=A^{-1}\begin{bmatrix}0\\0\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$. Sirf **trivial solution** $(0,0)$. **Contrast (degenerate homogeneous):** agar instead $2x+y=0,\ 4x+2y=0$ ho, toh $\det A=2\cdot2-1\cdot4=0$; doosri pehli ki twice hai, jo line $(x,y)=(t,-2t)$ deti hai — **infinitely many** solutions origin se guzarti hain. **Verify:** pehle system ke liye $(0,0)$: $0=0,\ 0=0$ ✓. Degenerate wale ke liye, $t=1\Rightarrow(1,-2)$: $2-2=0$ ✓, $4-4=0$ ✓. --- ## Ex 7 — Cell C7 · word problem > [!example] Statement > Ek shop pens aur notebooks bechti hai. **3 pens + 2 notebooks ₹80 mein aate hain**; **2 pens + 5 notebooks ₹110 mein aate hain**. Har cheez ki price pata karo. > > **Forecast:** do prices, do facts ⇒ positive rupee amounts ka ek unique pair expect karo. **Step 1 — translate karo.** Maano $p$ = pen ki price, $n$ = notebook ki price. $$A=\begin{bmatrix}3&2\\2&5\end{bmatrix},\quad X=\begin{bmatrix}p\\n\end{bmatrix},\quad B=\begin{bmatrix}80\\110\end{bmatrix}.$$ *Yeh step kyun?* Har real sentence $A$ ki ek row aur $B$ ki ek entry ban jaati hai. **Step 2 — determinant.** $\det A=3\cdot5-2\cdot2=15-4=11\neq0$. Unique prices exist karti hain. **Step 3 — inverse & solve karo.** $A^{-1}=\frac1{11}\begin{bmatrix}5&-2\\-2&3\end{bmatrix}$. $$X=\frac1{11}\begin{bmatrix}5&-2\\-2&3\end{bmatrix}\begin{bmatrix}80\\110\end{bmatrix} =\frac1{11}\begin{bmatrix}400-220\\-160+330\end{bmatrix}=\frac1{11}\begin{bmatrix}180\\170\end{bmatrix}.$$ Hmm — yeh non-integers de raha hai; chalte hain inhe rakhte hain: $p=\tfrac{180}{11},\ n=\tfrac{170}{11}$… chalte hain **arithmetic dobara verify karte hain**, kyunki prices sensible honi chahiye. Recompute karo: $5(80)-2(110)=400-220=180$; $-2(80)+3(110)=-160+330=170$. Toh $p=180/11\approx16.36$, $n=170/11\approx15.45$. **Verify (units = rupees):** $3(180/11)+2(170/11)=\tfrac{540+340}{11}=\tfrac{880}{11}=80$ ✓; $2(180/11)+5(170/11)=\tfrac{360+850}{11}=\tfrac{1210}{11}=110$ ✓. Dono totals given ₹ amounts se match karte hain, isliye (non-round lekin valid) prices hain $p=\tfrac{180}{11},\ n=\tfrac{170}{11}$. > [!intuition] > Real data mein round answers rarely milte hain. *Method* bilkul unfazed hai — $\det A\neq0$ exactly ek price pair guarantee karta hai, integer ho ya nahi. --- ## Ex 8 — Cell C8 · exam twist: ek $A^{-1}$, do right-hand sides > [!example] Statement > $A=\begin{bmatrix}2&1\\1&1\end{bmatrix}$ ke liye, $AX=B_1$ solve karo jahan $B_1=\begin{bmatrix}5\\3\end{bmatrix}$ **aur** $AX=B_2$ solve karo jahan $B_2=\begin{bmatrix}0\\4\end{bmatrix}$. > > **Forecast:** machine $A$ same hai; sirf output badal raha hai. Toh $A^{-1}$ **ek baar** compute karo aur do baar multiply karo — yahi reason hai ki inversion elimination dobara karne se better hai. **Step 1 — determinant & inverse.** $\det A=2\cdot1-1\cdot1=1\neq0$, toh $$A^{-1}=\frac11\begin{bmatrix}1&-1\\-1&2\end{bmatrix}=\begin{bmatrix}1&-1\\-1&2\end{bmatrix}.$$ *Yeh step kyun?* Jab $A^{-1}$ haath mein ho, toh har naye $B$ ke liye sirf ek matrix-vector product karna padta hai. **Step 2 — pehla solution.** $$X_1=A^{-1}B_1=\begin{bmatrix}1&-1\\-1&2\end{bmatrix}\begin{bmatrix}5\\3\end{bmatrix}=\begin{bmatrix}5-3\\-5+6\end{bmatrix}=\begin{bmatrix}2\\1\end{bmatrix}.$$ **Step 3 — doosra solution ($A^{-1}$ reuse karo).** $$X_2=A^{-1}B_2=\begin{bmatrix}1&-1\\-1&2\end{bmatrix}\begin{bmatrix}0\\4\end{bmatrix}=\begin{bmatrix}0-4\\0+8\end{bmatrix}=\begin{bmatrix}-4\\8\end{bmatrix}.$$ *Yeh step kyun?* Payoff demonstrate karta hai — koi re-elimination nahi. Compare karo [[Cramer's Rule]] se, jisko har naye $B$ ke liye determinants recompute karne padte hain. **Verify:** $X_1$: $2(2)+1=5$ ✓, $2+1=3$ ✓. $X_2$: $2(-4)+8=0$ ✓, $-4+8=4$ ✓. --- > [!recall]- Scenario matrix par quick self-test > Given $\det A=0$, main kis branch mein hoon? ::: $(\operatorname{adj}A)B$ test karo: $\neq O$ ⇒ no solution; $=O$ ⇒ infinitely many (ya none — substitute karo). > $B=O$ hone par hamesha kam se kam ek solution kyun hota hai? ::: $X=O$ satisfy karta hai $A\cdot O=O$; sirf sawaal uniqueness ka hai, jo $\det A$ decide karta hai. > Kai right-hand sides ke liye inversion Cramer se better kyun hai? ::: $A^{-1}$ ek baar compute karo, phir har $B$ ke liye ek multiply; Cramer har baar determinants recompute karta hai. > [!mnemonic] > **"Pehle Det, phir branch decide karo; tabhi invert karo."** > $\det\neq0\Rightarrow$ unique; $\det=0\Rightarrow$ *nothing* vs *everything* ke liye $(\operatorname{adj}A)B$ test karo. --- ## Connections - [[Solving systems using matrix inversion]] — parent recipe jinhe yeh examples exercise karte hain. - [[Determinants]] — $\det A$ gate jo har example mein branch choose karta hai. - [[Adjoint and Inverse of a Matrix]] — upar har $A^{-1}$ kaise build hua. - [[Cofactors and Minors]] — Ex 3–5 mein $\operatorname{adj}(A)$ ke andar ke entries. - [[Consistency of Linear Systems]] — Ex 4 aur Ex 5 yahan rehte hain. - [[Cramer's Rule]] — Ex 8 mein contrast kiya gaya rival. - [[Gaussian Elimination]] — large-system ka alternative. ## 🖼️ Scenario Map ```mermaid flowchart TD START["Compute det A"] -->|not zero| UNIQUE["Unique X = A inv B"] START -->|zero| TEST["Test adj A times B"] TEST -->|not zero| NONE["No solution Ex4"] TEST -->|equals zero| INF["Infinitely many Ex5"] UNIQUE --> C1["2x2 clean Ex1 Ex2"] UNIQUE --> C3["3x3 clean Ex3"] UNIQUE --> HOM["B is zero gives only origin Ex6"] UNIQUE --> WORD["Word problem Ex7"] UNIQUE --> MULTI["Reuse A inv for many B Ex8"] ```