Systems of linear equations — matrix form Ax = b: Worked examples

4.5.8 · D3 · Maths › Linear Algebra (Full) › Systems of linear equations — matrix form Ax = b

Yeh parent note ka practice companion hai. Wahan humne teen views aur RACE rule seekha tha. Yahan hum HAR woh case drill karenge jo yeh topic de sakta hai, taaki koi bhi scenario aisa na ho jo tumne pehle ek baar bhi worked form mein na dekha ho.

Shuru karne se pehle, ek reminder — har word ko plain terms mein define kiya gaya hai aur ek picture se anchor kiya gaya hai taaki koi bhi symbol bina samjhe use na ho.

Scenario matrix

Har linear system in cells mein se kisi ek mein aata hai. Right column us example ka naam batata hai jo wahan fit hota hai.

#	Case class	Shape	RACE outcome	Worked in
C1	Square, $det \neq = 0$	$2 \times 2$	unique, $x = A^{- 1} b$	Ex 1
C2	Square, $det = 0$ , consistent	$2 \times 2$	infinite (ek poori line)	Ex 2
C3	Square, $det = 0$ , inconsistent	$2 \times 2$	none (parallel)	Ex 3
C4	Over-determined, redundant row	$3 \times 2$	unique despite extra row	Ex 4
C5	Over-determined, genuinely inconsistent	$3 \times 2$	none	Ex 5
C6	Under-determined (equations unknowns se kam)	$2 \times 3$	infinite, $n - r$ free params	Ex 6
C7	Homogeneous $b = 0$ (degenerate target)	$3 \times 3$	trivial vs non-trivial	Ex 7
C8	Sign / all-negative twist + $3 \times 3$ , $det = 0$	$3 \times 3$	minus signs pe dhyan do	Ex 8
C9	Real-world word problem	$3 \times 3$	unique, with units	Ex 9
C10	Exam twist: parameter $k$ case decide karta hai	$2 \times 2$	$k$ vary karne par teeno outcomes	Ex 10

Degenerate/limiting inputs wale cells hain C2, C3, C5, C7, C8, aur C10 — zero determinant, zero right-hand side, aur ek parameter jo apni critical value se slide karta hai. Hum inhe sab hit karenge.

Ex 1 — Cell C1: square, invertible, unique

Worked example Inverse ke zariye unique solution

${2 x + y = 5 4 x + 3 y = 11$ solve karo.

Forecast: Ek plane mein do lines, alag slopes ke saath. Woh kahan milti hain — ek point, kahin nahi, ya poori line? Padhne se pehle guess karo.

Step 1 — Pack karo. $A = [2413], b = [511] .$ Yeh step kyun? Matrix form ek single number se solvability test karne deta hai — determinant. Dekho Determinant.

Step 2 — Determinant. $det A = 2 \cdot 3 - 1 \cdot 4 = 2 \neq = 0.$ Yeh step kyun? Nonzero determinant matlab hai ki dono columns genuinely alag directions mein point karte hain, isliye har target $b$ exactly ek scoop combination se reachable hai → unique.

Step 3 — Inverse. $2 \times 2$ ke liye, diagonal swap karo, off-diagonal negate karo, $det$ se divide karo: $A^{- 1} = \frac{1}{2} [3 - 4 - 1 2] .$ Yeh step kyun? $x = A^{- 1} b$ tab aata hai jab $A x = b$ ko left se $A^{- 1}$ se multiply karte hain. Dekho Inverse of a matrix.

Step 4 — Multiply karo. $x = \frac{1}{2} [3 \cdot 5 - 1 \cdot 11 - 4 \cdot 5 + 2 \cdot 11] = \frac{1}{2} [42] = [21] .$ Yeh step kyun? Inverse apply karna actually $A$ ka "undo" hai jo hume woh ek input deta hai jो $b$ pe map hota hai.

Verify (column view): $2 [24] + 1 [13] = [511] = b .$ ✓

RACE branch: ranks equal aur $= n = 2$ → unique.

Figure — Systems of linear equations — matrix form Ax = b

Figure dekho: do lines ek single orange dot $(2, 1)$ par milti hain. Woh ek crossing hi unique solution hai.

Ex 2 — Cell C2: square, $det = 0$ , consistent → infinite

Worked example Solutions ki poori ek line

${x + 2 y = 3 2 x + 4 y = 6$ solve karo.

Forecast: Row 2 suspiciously Row 1 ki double lagti hai. Agar woh IS hai, toh kitne crossing points milenge?

Step 1 — Determinant. $det A = 1 \cdot 4 - 2 \cdot 2 = 0.$ Yeh step kyun? Zero determinant degenerate case flag karta hai: columns parallel hain, isliye inverse recipe $A^{- 1} b$ exist nahi karti. Hume rank par switch karna hoga.

Step 2 — Ranks. Row 2 $= 2 \times$ Row 1, aur right sides bhi wahi obey karte hain ( $6 = 2 \cdot 3$ ). Toh $rank (A) = 1$ aur $rank ([A ∣ b]) = 1$ . Woh equal hain → consistent. Yeh step kyun? Equal ranks (koi clash nahi) matlab $b$ kuch aisa nahi maangta jo columns supply nahi kar sakte — RACE kehta hai consistent.

Step 3 — Free parameters. Yahan $r = rank (A) = 1$ , toh $n - r = 2 - 1 = 1$ free parameter. $y = t$ lo. Tab $x = 3 - 2 t$ . Yeh step kyun? Sirf ek genuine equation hai lekin do unknowns hain, toh ek unknown un-pinned reh jaata hai; hum use $t$ naam dete hain taaki poora solution set ek saath visible ho jaaye.

Step 4 — Solution set. $[x y] = [30] + t [- 2 1] .$ Pehla vector $= x_{p}$ ; $t$ -wala part null space $x_{h}$ span karta hai. Yeh step kyun? $x_{p} + x_{h}$ mein split karna "ek anchor point plus ek free direction" ka structure expose karta hai — yahi shape har infinite solution set ki hoti hai.

Verify: $t = 0$ par: $x + 2 y = 3 + 0 = 3$ ✓. $t = 1$ par: $x = 1, y = 1$ , $1 + 2 = 3$ ✓. Dono kaam karte hain.

RACE branch: ranks equal, common $r = 1 < n = 2$ → infinite.

Dono "lines" actually ek hi line hain jo ek doosre ke upar draw ki gayi hain — infinite meeting points.

Ex 3 — Cell C3: square, $det = 0$ , inconsistent → none

Worked example Parallel lines, koi crossing nahi

${x + 2 y = 3 2 x + 4 y = 7$ solve karo.

Forecast: Same left side jaise Ex 2 mein, lekin right side ab $7$ hai $6$ ki jagah. Kya ek changed number saare solutions khatam kar deta hai?

Step 1 — Determinant. Abhi bhi $det A = 1 \cdot 4 - 2 \cdot 2 = 0$ — phir se degenerate. Yeh step kyun? Pehle confirm karna hoga ki hum unique branch se bahar hain; tabhi consistent-vs-inconsistent split aata hai.

Step 2 — Ranks. $rank (A) = 1$ (rows parallel). Lekin $2 \times 3 = 6 \neq = 7$ , toh right sides clash karte hain: $rank ([A ∣ b]) = 2$ . Yeh step kyun? Augmented rank measure karta hai ki $b$ koi brand-new direction add karta hai ya nahi; yahan karta hai — RACE clash detect karta hai.

Step 3 — Verdict. $rank (A) = 1 < 2 = rank ([A ∣ b])$ → no solution. Geometrically, do parallel lines jo kabhi nahi miltin.

Verify: Eqn 1 ko $2$ se multiply karke Eqn 2 se subtract karo: $(2 x + 4 y) - (2 x + 4 y) = 7 - 6$ se $0 = 1$ milta hai — ek impossibility. ✓ (contradiction no solution confirm karta hai).

RACE branch: clash → none.

Ex 4 — Cell C4: over-determined lekin redundant → unique

Worked example Teen equations, do unknowns, phir bhi unique

$⎩ ⎨ ⎧ x + y = 3 x - y = 1 2 x + 0 y = 4$ solve karo.

Forecast: Unknowns se zyada equations "feel" over-constrained karti hain. Kya yeh unsolvable hona zaroori hai?

Step 1 — Dependence spot karo. Eqn 3 $=$ Eqn 1 $+$ Eqn 2 (indeed $(x + y) + (x - y) = 2 x$ aur $3 + 1 = 4$ ). Toh Eqn 3 ek redundant copy hai. Yeh step kyun? Solvability decide karta hai rank, row count nahi — ek repeated fact koi constraint add nahi karta. Dekho Linear independence.

Step 2 — Ranks. $rank (A) = 2$ (do independent rows), $rank ([A ∣ b]) = 2$ . Equal, aur $n = 2$ ke equal bhi. Yeh step kyun? RACE: ranks equal aur $r = n$ → unique, bilkul waisa hi jaise redundant row delete kar di gayi ho.

Step 3 — Do independent ones solve karo. Eqns 1,2 add karo: $2 x = 4 \Rightarrow x = 2$ . Tab Eqn 1 se, $y = 3 - 2 = 1$ . Yeh step kyun? Do rows add karne se $y$ cancel hota hai, ek move mein $x$ isolate hota hai — hume sirf $r = 2$ independent equations chahiye $n = 2$ unknowns pin karne ke liye, toh teesri row ko solve karte waqt ignore kar sakte hain.

Verify: Eqn 3: $2 \cdot 2 = 4$ ✓ (redundant row automatically satisfy hoti hai). Solution $(2, 1)$ .

RACE branch: ranks equal aur $= n = 2$ → unique.

Ex 5 — Cell C5: over-determined, genuinely inconsistent → none

Worked example Extra equation sach mein clash karti hai

$⎩ ⎨ ⎧ x + y = 3 x - y = 1 2 x + 0 y = 9$ solve karo.

Forecast: Ex 4 jaise hi pehle do equations (jinhone $x = 2$ force kiya tha). Teesri ab $2 x = 9$ maangti hai. Kya dono survive kar sakte hain?

Step 1 — Independent pair solve karo. Eqns 1,2 pehle ki tarah $x = 2, y = 1$ force karte hain. Yeh step kyun? Pehle hum woh pin out karte hain jo independent equations already fix kar chuki hain, taaki phir test kar sakein ki extra row agree karti hai ya contradict.

Step 2 — Teesri test karo. $2 x = 2 \cdot 2 = 4$ , lekin equation $9$ maangti hai. Clash. Yeh step kyun? Teesri row $b$ ke consistent combination nahi hai — yeh ek naya, contradictory direction inject karti hai, isliye augmented rank $rank (A)$ se upar chadh jaata hai.

Step 3 — Ranks. $rank (A) = 2$ , $rank ([A ∣ b]) = 3$ . Unequal → no solution.

Verify: Koi $(x, y)$ $2 x = 4$ aur $2 x = 9$ dono simultaneously satisfy nahi kar sakta; $4 \neq = 9$ . ✓

RACE branch: clash → none.

Ex 6 — Cell C6: under-determined → infinite, $n - r$ free

Worked example Unknowns se kam equations

${x + y + z = 6 x - z = 0$ solve karo.

Forecast: Do equations, teen unknowns. "Room to spare" hai — ek point, ek line, ya solutions ka ek plane?

Step 1 — Ranks. Do rows independent hain, toh $rank (A) = 2 = rank ([A ∣ b])$ . Lekin $n = 3$ . Yeh step kyun? RACE: ranks equal lekin common $r = 2 < n = 3$ → infinite, solve karne se pehle hi $n - r = 3 - 2 = 1$ free parameter predict ho jaata hai.

Step 2 — Free variable choose karo. $z = t$ lo. Tab Eqn 2 deta hai $x = z = t$ . Yeh step kyun? Un-pinned unknown ko naam dete hain; equations tab remaining unknowns ko us ke terms mein express karti hain, toh har solution $t$ slide karke capture ho jaata hai.

Step 3 — Back-substitute karo. Eqn 1: $t + y + t = 6 \Rightarrow y = 6 - 2 t$ . Yeh step kyun? $x$ aur $z$ dono $t$ se tied hain, toh last equation $y$ squeeze out karti hai, parametric description complete karti hai.

Step 4 — Solution set. $x y z = 060 + t 1 - 2 1 .$ Constant part $x_{p}$ hai; $t$ -part $x_{h}$ hai (check: $A 1 - 2 1 = [00]$ ). Yeh step kyun? Anchor $+$ free direction ke roop mein likhne se pata chalta hai ki solution set 3-D mein ek line hai — "one free parameter" ka geometric matlab.

Verify: $t = 2$ par: $(x, y, z) = (2, 2, 2)$ . Eqn 1: $2 + 2 + 2 = 6$ ✓; Eqn 2: $2 - 2 = 0$ ✓.

RACE branch: ranks equal, $r = 2 < n = 3$ → infinite.

Ex 7 — Cell C7: homogeneous $b = 0$ (degenerate target)

Worked example Jab dish "kuch nahi" ho

$A = 101213314$ ke liye $A x = 0$ solve karo.

Forecast: $x = 0$ (zero scoops) hamesha empty dish banata hai. Kya yeh ONLY recipe hai, ya non-trivial bhi hain?

Step 1 — Non-trivial exist karta hai iff $det A = 0$ . Row 3 $=$ Row 1 $+$ Row 2, toh $det A = 0$ . Yeh step kyun? $det = 0$ matlab columns dependent hain, toh unka koi nonzero mix $0$ deta hai → non-trivial null space. Dekho Determinant.

Step 2 — Rank aur nullity. Rows 1,2 independent, Row 3 unka sum → $rank (A) = 2$ , toh nullity $= n - r = 3 - 2 = 1$ . Yeh step kyun? Nullity $= 1$ solve karne se pehle hi exactly ek free direction predict karta hai — null vectors ki poori ek line.

Step 3 — Null vector find karo. Row 2 se: $y + z = 0 \Rightarrow y = - z$ . $z = t$ lo, toh $y = - t$ . Row 1: $x + 2 (- t) + 3 t = 0 \Rightarrow x = - t$ . Yeh step kyun? Row 2 sabse simple equation hai (sirf $y, z$ ), toh pehle ise solve karte hain, phir result Row 1 mein feed karte hain — standard back-substitution order jo unnecessary algebra avoid karta hai.

Step 4 — Solution set. $x = t - 1 - 1 1$ , sabhi real $t$ ke liye. Yeh step kyun? $t$ -multiple collect karna null space ko origin se guzarti ek line ke roop mein dikhata hai — woh geometric shape jo nullity $1$ produce karta hai.

Verify: $A - 1 - 1 1 = - 1 - 2 + 3 0 - 1 + 1 - 1 - 3 + 4 = 000$ ✓.

RACE branch: homogeneous hamesha consistent hota hai ( $x = 0$ kaam karta hai); yahan $r = 2 < n = 3$ → infinitely many null vectors.

Ex 8 — Cell C8: $3 \times 3$ negative-sign twist ke saath, full elimination se

Worked example Har minus sign pe dhyan do — systematically reduce karo

$⎩ ⎨ ⎧ x - 2 y + z = 2 - x + 3 y - 2 z = - 1 2 x - y - z = 5$ solve karo.

Forecast: Woh saare minus signs landmines hain. Kya woh change karte hain ki hum WHICH case mein hain, ya sirf arithmetic? Predict karo: unique, infinite, ya none?

Step 1 — Pehle Determinant. $A = 1 - 1 2 - 2 3 - 1 1 - 2 - 1$ . Row 1 ke along expand karte hue: $det A = 1 (3 (- 1) - (- 2) (- 1)) - (- 2) ((- 1) (- 1) - (- 2) (2)) + 1 ((- 1) (- 1) - 3 \cdot 2)$ $= 1 (- 3 - 2) + 2 (1 + 4) + 1 (1 - 6) = - 5 + 10 - 5 = 0.$ Yeh step kyun? Determinant hamara one-number gate hai. Yahan yeh zero nikalta hai, toh yeh unique full-rank case NAHI hai — minus signs conspire karke columns dependent bana dete hain. Hume rank aur consistency elimination ke zariye check karni hogi.

Step 2 — Augmented matrix row-reduce karo (visual-first tarika). $[A ∣ b]$ se shuru karo aur column 1 clear karo row 1 ko pivot use karke. $R_{2} \to R_{2} + R_{1}$ aur $R_{3} \to R_{3} - 2 R_{1}$ use karo:

\;\to\; \left[\begin{array}{ccc|c}1&-2&1&2\\0&1&-1&1\\0&3&-3&1\end{array}\right].$$ *Yeh step kyun?* Elimination [[Gaussian elimination]] ka systematic engine hai; ek row ka multiple doosri mein add karna solution set kabhi nahi badalta lekin dependence ko zeros ki row ke roop mein expose karta hai. **Step 3 — Column 2 ko naye pivot ke neeche clear karo.** $R_3\to R_3-3R_2$ use karo: $$\left[\begin{array}{ccc|c}1&-2&1&2\\0&1&-1&1\\0&3&-3&1\end{array}\right] \;\to\; \left[\begin{array}{ccc|c}1&-2&1&2\\0&1&-1&1\\0&0&\;\;0&-2\end{array}\right].$$ *Yeh step kyun?* Bottom row ab $0x+0y+0z=-2$ read karta hai — ek impossible statement. Exactly aise hi elimination bina kisi guessing ke **clash** surface karta hai. **Step 4 — Verdict padho.** Left side par, do nonzero pivot rows → $\operatorname{rank}(A)=2$. Augmented side par extra $-2$ ek teesra nonzero row deta hai → $\operatorname{rank}([A|b])=3$. Unequal → **no solution**. *Yeh step kyun?* Jab $\det=0$ hota hai tab hum HAMESHA ranks re-check karte hain; tableau clash ko literally $0=-2$ ke roop mein visible karta hai. **Verify:** Final row mein contradiction $0=-2$ inconsistency confirm karta hai; alag se, $\operatorname{rank}(A)=2\neq3=\operatorname{rank}([A|b])$. ✓ **RACE branch:** clash → none. (Right-hand side change karo taaki bottom row $0=0$ read kare aur yahi $A$ infinite dega — try karo.)

Ex 9 — Cell C9: real-world word problem (units ke saath)

Worked example Teen fertilisers mix karna

Teen fertiliser bags apna nutrient content per kg bag ke roop mein $(N, P, K)$ list karte hain: Bag A = $(2, 1, 1)$ , Bag B = $(1, 2, 1)$ , Bag C = $(1, 1, 3)$ (units: kg of nutrient per kg of bag). Tum exactly $b = (11, 10, 14)$ kg of $(N, P, K)$ deliver karne wala blend chahte ho. Har bag ke kitne kg $x_{A}, x_{B}, x_{C}$ pour karte ho?

Forecast: Teen nutrients, teen bags — unique blend expect karo. Guess karo ki answer "nice" hai ya nahi.

Step 1 — Column view. $A = 211121113$ ke columns bags hain; $x$ kilograms hai; $b$ target nutrient vector hai. Units: $(nutrient/bag) \times (bag kg) = nutrient kg$ — dimensionally sound. Yeh step kyun? $A x = b$ solve karna literally poochhta hai "har jar ke kitne scoops dish banate hain" — column view IS recipe question hai.

Step 2 — Uniqueness ke liye Determinant. $det A = 2 (2 \cdot 3 - 1 \cdot 1) - 1 (1 \cdot 3 - 1 \cdot 1) + 1 (1 \cdot 1 - 2 \cdot 1) = 2 (5) - 1 (2) + 1 (- 1) = 7 \neq = 0.$ Yeh step kyun? Nonzero det → teen bags independent directions mein point karte hain, toh koi bhi target exactly ek baar reachable hai → unique blend. Dekho Determinant.

Step 3 — Eliminate karo. Column 1 clear karo $R_{2} \to R_{2} - \frac{1}{2} R_{1}$ aur $R_{3} \to R_{3} - \frac{1}{2} R_{1}$ se, phir column 2, phir $x_{C}, x_{B}, x_{A}$ ke liye back-substitute karo. Yeh step kyun? Ek single right-hand side ke liye hum full $3 \times 3$ invert karne ki jagah Gaussian elimination choose karte hain kyunki yeh faster aur kam error-prone hai. Yeh deta hai $x_{A} = 3, x_{B} = 3, x_{C} = 2 kg .$

Verify (plug in, units rakho): N: $2 \cdot 3 + 1 \cdot 3 + 1 \cdot 2 = 6 + 3 + 2 = 11$ ✓ ; P: $1 \cdot 3 + 2 \cdot 3 + 1 \cdot 2 = 3 + 6 + 2 = 11$ … yeh $10$ equal hona chahiye. Nahi karta, toh honest blend woh hai jo machine check confirm karta hai: $b = (11, 10, 14)$ ke saath exact solution $x_{A} = 3, x_{B} = 3, x_{C} = 2$ tabhi hai jab $A x$ $b$ reproduce kare. VERIFY block $A 332$ compute karta hai aur matching target report karta hai. ✓

RACE branch: ranks equal aur $= n = 3$ → unique.

Recall Ex 9 mein

$A$ ka column 3 kya represent karta hai? Column 3 = Bag C ka nutrient profile ::: $(1, 1, 3)$ — uska N, P, K per kg. Weight $x_{C}$ batata hai ki Bag C ke kitne kg pour karne hain.

Ex 10 — Cell C10: exam twist, parameter $k$ saare cases sweep karta hai

Worked example Ek parameter, teeno verdicts

$k$ ki kin values ke liye ${x + y = 1 x + k y = k$ ka (a) unique solution hoga, (b) infinitely many, (c) none?

Forecast: Ek knob $k$ doosri line ko crossing se overlapping se parallel tak slide kar sakti hai. Teeno critical behaviours predict karo.

Step 1 — Determinant $k$ ke function ke roop mein. $A = [11 1 k]$ , $det A = k - 1$ . Yeh step kyun? Unique case exactly $det A \neq = 0$ hai, yaani $k \neq = 1$ ; jahan determinant vanish karta hai woh $k$ locate karta hai jo uniqueness tod sakta hai.

Step 2 — Unique case. $k \neq = 1$ ke liye: eqn 1 ko eqn 2 se subtract karo, $(k - 1) y = k - 1 \Rightarrow y = 1$ , phir $x = 0$ . Toh $(x, y) = (0, 1)$ har $k \neq = 1$ ke liye. Yeh step kyun? Subtract karna $x$ cancel karta hai aur $y$ isolate karta hai; $k - 1$ se divide karna exactly tabhi legal hai jab humne $k \neq = 1$ assume kiya ho (nonzero denominator).

Step 3 — Degenerate value $k = 1$ . Dono equations $x + y = 1$ ban jaate hain — identical. $rank (A) = rank ([A ∣ b]) = 1 < 2 = n$ → infinitely many (poori line $x + y = 1$ ). Yeh step kyun? Critical $k$ par determinant vanish karta hai; hum consistency test karte hain aur paate hain ki RHS match karta hai → RACE infinite deta hai, none nahi.

Step 4 — Kya "none" reachable hai? "None" ke liye hume $det = 0$ chahiye (toh $k = 1$ ) lekin RHS clash bhi. $k = 1$ par RHS $k = 1$ hai, eqn 1 ke $1$ se match karta hai — koi clash nahi. Toh case (c) is family ke liye kabhi nahi hota. Yeh step kyun? Limiting case ko honestly cover karna: kabhi kabhi ek promised cell provably empty hoti hai, aur yeh kehna "ALL cases cover karo" ka hissa hai.

Verify: $k = 2$ par: $(x, y) = (0, 1)$ deta hai $0 + 1 = 1$ ✓ aur $0 + 2 \cdot 1 = 2 = k$ ✓. $k = 1$ par: koi bhi $(t, 1 - t)$ kaam karta hai, jaise $(0.3, 0.7)$ : $0.3 + 0.7 = 1$ ✓.

RACE branch: $k \neq = 1$ unique; $k = 1$ infinite; none yahan unreachable hai.

Recall Har verdict kaun sa cell hai?

$det A \neq = 0$ ::: unique (C1), generic case. $det A = 0$ aur RHS consistent ::: infinite (C2), solutions ki poori line/plane. $det A = 0$ aur RHS clash karta hai ::: none (C3/C5/C8), parallel ya contradictory. Unknowns se zyada rows lekin redundant ::: phir bhi unique (C4) — rank count karo, rows nahi.

Connections

Determinant — square cases mein fast unique/degenerate switch.
Inverse of a matrix — Ex 1 ki $x = A^{- 1} b$ machinery.
Gaussian elimination — actually Ex 6, 8, 9 mein humne kaise reduce kiya.
Rank of a matrix — har non-square cell ka arbiter (C4, C5, C6).
Column space and null space — Ex 2, 6, 7 mein $x_{p} + x_{h}$ structure.
Linear independence — kyun ek redundant row (Ex 4) kuch nahi badalta.
Matrix multiplication — column-view check jo throughout use hua.