Worked examples — LU decomposition — algorithm, applications

4.5.26 · D3 · Maths › Linear Algebra (Full) › LU decomposition — algorithm, applications

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Scenario matrix

Har $A$ jise tum factor karne ke liye kaho, in cells mein se kisi ek mein aayegi. Hum sab ko cover karenge.

#	Cell (scenario)	Kya special hai	Example
C1	Generic $2\times 2$, saare pivots nonzero	"Textbook happy path"	Ex 1
C2	Generic $3\times 3$, phir $A\mathbf{x}=\mathbf{b}$ solve karo	Factor + do solves combine karo	Ex 2
C3	Zero pivot top-left mein — row swap chahiye $\Rightarrow PA=LU$	Naive $A=LU$ fail hota hai	Ex 3
C4	Tiny pivot — mathematically theek hai par numerically dangerous	Stability ke liye pivot kyun karte hain	Ex 4
C5	Singular matrix ($\det =0$) — elimination ke baad pivot zero ho jaata hai	LU phir bhi exist karta hai, $U$ ke diagonal par zero hota hai	Ex 5
C6	LU se $\det A$ aur $A^{-1}$ nikalo	Same $L,U$ reuse karo	Ex 6
C7	Kai right-hand sides ek saath (real-world word problem)	80/20 payoff	Ex 7
C8	Negative & fractional entries — multipliers mein sign bookkeeping	Signs of $m_{ik}$	Ex 8
C9	Non-square (rectangular) matrix	LU column-by-column kaam karta hai; $L$ square, $U$ rectangular	Ex 9

Neeche har example apni cell ke saath tagged hai.

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Notation jo hum actually use karenge (zero se build ki hui)

Koi bhi arithmetic se pehle, har symbol ko naam dete hain taaki kuch bhi unexplained na lage.

Figure s01 (shabdon mein). $2\times 2$ grid of cells jisme $a_{11},a_{12}$ (top row) aur $a_{21},a_{22}$ (bottom row) label hain. Top-left cell $a_{11}$ amber mein highlight hai aur "PIVOT (isse divide karo)" label hai. Bottom-left cell $a_{21}$ cyan mein highlight hai aur "isko 0 karo" label hai. Daayein taraf formula $m_{21}=a_{21}/a_{11}$ likha hai caption ke saath "kitni pivot-rows subtract karni hain." Matlab: amber = woh number jisse tum divide karte ho, cyan = woh number jise tum zero karna chahte ho, aur unka ratio multiplier hai.

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C1 — happy path

Forecast: andaaza lagao — kya $L$ ke diagonal ke neeche whole number aayega ya fraction? (Pivot $3$ hai, neeche ka number $1$ hai, toh predict karo $m_{21}=\frac{1}{3}$.)

1. Pivot $=a_{11}=3$; multiplier $m_{21}=\frac{a_{21}}{a_{11}}=\frac{1}{3}$. Yeh step kyun? Pivot ke neeche ke $1$ ko khatum karna hai; $m_{21}$ exactly batata hai kitni pivot-rows subtract karni hain.
2. Row${}_2\leftarrow$Row${}_2-\frac{1}{3}$Row${}_1$: $\left(1,5\right)-\frac{1}{3}(3,6)=(0,3)$. Kyun? $(3,6)$ ka $\frac{1}{3}$ subtract karne se $(1,2)$ milta hai, jo $(1,5)$ se ghatane par $(0,3)$ deta hai — pehli entry design ke mutabik $0$ ho jaati hai.
3. Assemble karo: $U=\left(\begin{array}{cc}3& 6\\ 0& 3\end{array}\right)$, $L=\left(\begin{array}{cc}1& 0\\ \frac{1}{3}& 1\end{array}\right)$. Kyun? $U$ woh hai jo bacha; $L$ mein $+m_{21}=\frac{1}{3}$ store hota hai.

Verify: $LU=\left(\begin{array}{cc}1& 0\\ \frac{1}{3}& 1\end{array}\right)\left(\begin{array}{cc}3& 6\\ 0& 3\end{array}\right)=\left(\begin{array}{cc}3& 6\\ 1& 5\end{array}\right)=A.$ ✓

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C2 — pehle factor karo phir solve karo

Forecast: entries $2,4,8$ double ho rahe hain — expect karo saaf whole multipliers $m_{21}=2,m_{31}=4$.

1. Column 1. $m_{21}=\frac{4}{2}=2$, $m_{31}=\frac{8}{2}=4$. Row${}_2-2$Row${}_1=(0,1,1)$; Row${}_3-4$Row${}_1=(0,3,5)$. Kyun? Pehle pivot $2$ ke neeche sab kuch clear karo usse ratios use karke.
2. Column 2. Naya pivot $=1$ (position $(2,2)$). $m_{32}=\frac{3}{1}=3$. Row${}_3-3\cdot (0,1,1)=(0,0,2)$. Kyun? Ab doosre pivot ke neeche clear karo; hum hamesha current pivot row use karte hain.
3. Assemble karo. $U=\left(\begin{array}{ccc}2& 1& 1\\ 0& 1& 1\\ 0& 0& 2\end{array}\right),L=\left(\begin{array}{ccc}1& 0& 0\\ 2& 1& 0\\ 4& 3& 1\end{array}\right).$
4. Forward solve $L\mathbf{y}=\mathbf{b}$: $y_1=4;2y_1+y_2=10\Rightarrow y_2=2;4y_1+3y_2+y_3=26\Rightarrow 16+6+y_3=26\Rightarrow y_3=4.$ Forward kyun? $L$ lower triangular hai, toh top row turant solve ho jaati hai aur har nayi row mein sirf known values use hoti hain.
5. Back solve $U\mathbf{x}=\mathbf{y}$: $2x_3=4\Rightarrow x_3=2;x_2+x_3=2\Rightarrow x_2=0;2x_1+x_2+x_3=4\Rightarrow 2x_1=2\Rightarrow x_1=1.$ Back kyun? $U$ upper triangular hai, toh sabse neeche wala variable pehle isolate hota hai.

Verify: $A\mathbf{x}=\left(\begin{array}{ccc}2& 1& 1\\ 4& 3& 3\\ 8& 7& 9\end{array}\right)\left(\begin{array}{c}1\\ 0\\ 2\end{array}\right)=\left(\begin{array}{c}4\\ 10\\ 26\end{array}\right)=\mathbf{b}.$ ✓

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C3 — zero pivot, permute karna padega ($PA=LU$)

Forecast: top-left entry $0$ hai. Kya tum $m_{21}=\frac{3}{0}$ compute kar sakte ho? Nahi — zero se divide karna undefined hai. Toh naive $A=LU$ shuru hi nahi ho sakta. Pehle rows swap karni hongi.

Figure s02 (shabdon mein). Do matrices side by side. Baayen taraf, matrix $A=\left(\begin{array}{cc}0& 2\\ 3& 4\end{array}\right)$ jisme top-left entry ($0$) amber mein highlight hai aur "cannot divide by 0" label hai — yeh fatal zero pivot hai. Ek cyan arrow "swap rows, $P=[[0,1],[1,0]]$" label ke saath daayein taraf point karta hai, jahan swapped matrix $PA=\left(\begin{array}{cc}3& 4\\ 0& 2\end{array}\right)$ dikhti hai jiske top-left entry ($3$) green mein highlight hai aur "pivot = 3 (good)" label hai. Message: rows swap karne se ek dead zero pivot live nonzero pivot ban jaata hai.

1. Failure diagnose karo. Pivot $a_{11}=0$; $m_{21}=\frac{3}{0}$ impossible hai. Kyun? Elimination ke liye pivot se divide karna zaruuri hai. Zero pivot ka koi valid multiplier nahi hota. (Figure mein yeh amber cell hai.)
2. Rows 1 aur 2 swap karo. Swap ko ek permutation matrix $P=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ mein record karo, jisse milta hai $PA=\left(\begin{array}{cc}3& 4\\ 0& 2\end{array}\right).$ Kyun? $P$ sirf rows re-order karta hai; $A$ ko baayen se $P$ se multiply karna physically swap perform karta hai aur ek permanent receipt rakhta hai. (Figure mein yeh cyan arrow hai.)
3. Ab $PA$ factor karo. Yeh already upper triangular hai. Kisi pivot ke neeche sirf $(2,1)$ entry hai, jo already $0$ hai, toh $m_{21}=\frac{0}{3}=0$ — bilkul bhi elimination ki zarurat nahi. $L=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right),U=\left(\begin{array}{cc}3& 4\\ 0& 2\end{array}\right).$ $L$ identity kyun ban jaata hai? $L$ multipliers store karta hai, aur yahan har multiplier $0$ hai kyunki subtract karne ko kuch tha hi nahi — har pivot ke neeche wali entries already zero thi. Diagonal par $1$'s aur neeche $0$'s wala grid identity hi hota hai, toh $L=I$.

Verify: $PA=\left(\begin{array}{cc}3& 4\\ 0& 2\end{array}\right)=LU.$ Aur ulta karne par: $A=P^{-1}LU=P^{\top }LU$ original recover kar leta hai. ✓ (Check: $P^{\top }LU=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{cc}3& 4\\ 0& 2\end{array}\right)=\left(\begin{array}{cc}0& 2\\ 3& 4\end{array}\right)$.)

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C4 — tiny pivot, stability twist

Forecast: pivot $0.001$ zero nahi hai, toh naive LU mathematically kaam karta hai. Lekin $m_{21}=1/0.001=1000$ bahut bada hai — andaaza lagao ki is bade number mein hi rounding error chhup sakta hai.

1. Naive (no swap). $m_{21}=\frac{1}{0.001}=1000$. Row${}_2-1000\cdot$Row${}_1$: $(1,1)-1000(0.001,1)=(0,1-1000)=(0,-999)$. $L=\left(\begin{array}{cc}1& 0\\ 1000& 1\end{array}\right),U=\left(\begin{array}{cc}0.001& 1\\ 0& -999\end{array}\right).$ Size note kyun karein? Rounding wale computer par, $1-1000\approx -1000$ sach wala $-999$ daba sakta hai; $1000$ multiplier kisi bhi error ko amplify kar deta hai. Mathematically exact, numerically fragile.
2. Partial pivoting ke saath. Column mein sabse bade absolute value wali entry ($|1|>|0.001|$) ko pivot banane ke liye swap karo: $P=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, $PA=\left(\begin{array}{cc}1& 1\\ 0.001& 1\end{array}\right),m_{21}=\frac{0.001}{1}=\mathrm{0.001.}$ Row${}_2-0.001$Row${}_1=(0,1-0.001)=(0,0.999)$. $L=\left(\begin{array}{cc}1& 0\\ 0.001& 1\end{array}\right),U=\left(\begin{array}{cc}1& 1\\ 0& 0.999\end{array}\right).$ Behtar kyun? Ab har multiplier ki magnitude $\le 1$ hai, toh koi error amplify nahi hoti.

Verify: naive $LU=\left(\begin{array}{cc}0.001& 1\\ 1& 1\end{array}\right)=A$ ✓; pivoted $LU=\left(\begin{array}{cc}1& 1\\ 0.001& 1\end{array}\right)=PA$ ✓.

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C5 — singular matrix, $\det =0$

Forecast: rows $1$ aur $3$ proportional lagti hain; ${\text{col}}_2=2{\text{col}}_1$. Predict karo ki zero pivot aayega aur $\det A=0$.

1. Column 1. $m_{21}=2,m_{31}=3$. Row${}_2-2$Row${}_1=(0,0,1)$; Row${}_3-3$Row${}_1=(0,0,1)$. Kyun? Pehle pivot $1$ ke neeche standard clearing.
2. Column 2. Candidate pivot $(2,2)$ ab $0$ hai, aur uske neeche bhi sab $0$ hain. Eliminate karne ko kuch nahi. Formally $m_{32}=\frac{a_{32}}{a_{22}}=\frac{0}{0}$ — undefined, lekin kyunki $a_{32}$ already $0$ hai koi elimination zaruuri nahi, toh hum $m_{32}=0$ set karke aage badhte hain. $m_{32}=0$ kyun? Multiplier batata hai neeche ki entry ko zero karne ke liye kitni pivot-rows subtract karni hain; woh entry already zero hai, toh kuch subtract nahi karte, yaani multiplier $0$ hai. Koi swap help nahi kar sakta kyunki poora subcolumn zero hai — yeh singular matrix ki pehchaan hai.
3. Assemble karo. $U=\left(\begin{array}{ccc}1& 2& 3\\ 0& 0& 1\\ 0& 0& 1\end{array}\right),L=\left(\begin{array}{ccc}1& 0& 0\\ 2& 1& 0\\ 3& 0& 1\end{array}\right).$ Note karo pivot pattern broken hai: $(2,2)$ par diagonal mein zero baitha hai.
4. Determinant. $\det A=\prod u_{ii}=1\cdot 0\cdot 1=0.$ Kyun? $U$ ke diagonal par ek bhi zero pivot product ko zero kar deta hai — matrix singular hai, koi inverse nahi.

Verify: $LU=\left(\begin{array}{ccc}1& 2& 3\\ 2& 4& 7\\ 3& 6& 10\end{array}\right)=A$ ✓ aur $\det A=0$ ✓ (rows linearly dependent hain: Row${}_3$-Row${}_2$ = Row${}_1$).

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C6 — ek hi factorisation se determinant AUR inverse

Forecast: pivots the $2,1,2$; predict karo $\det A=2\cdot 1\cdot 2=4$.

1. Determinant $=\prod u_{ii}=2\cdot 1\cdot 2=4.$ Kyun? $\det (LU)=\det L\det U=1\cdot \prod u_{ii}$; $L$ ka diagonal sab $1$ hai toh contribute karta hai $1$.
2. Inverse columns setup karo. Yaad karo (upar definition mein) ${\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3$ $I$ ke columns hain. $A^{-1}$ ka $j$-th column $A\mathbf{x}={\mathbf{e}}_j$ ka solution hai, solve kiya gaya same $L,U$ se jo Ex 2 mein mila (dekho Matrix Inverse). Reuse kyun? Factorisation expensive part tha; har column sirf do saste triangular solves hai, forward phir back.
3. Column 1 — $A\mathbf{x}={\mathbf{e}}_1=(1,0,0{)}^{\top }$ solve karo. Forward $L\mathbf{y}={\mathbf{e}}_1$: $y_1=1;2y_1+y_2=0\Rightarrow y_2=-2;4y_1+3y_2+y_3=0\Rightarrow 4-6+y_3=0\Rightarrow y_3=2.$ Back $U\mathbf{x}=\mathbf{y}$: $2x_3=2\Rightarrow x_3=1;x_2+x_3=-2\Rightarrow x_2=-3;2x_1+x_2+x_3=1\Rightarrow 2x_1-3+1=1\Rightarrow x_1=\frac{3}{2}.$ Toh $A^{-1}$ ka column 1 hai ${\left(\frac{3}{2},-3,1\right)}^{\top }$. Kyun? Yeh akela column poora recipe dikhata hai; columns 2 aur 3 ise ${\mathbf{e}}_2,{\mathbf{e}}_3$ ke saath repeat karte hain.
4. Column 2 — $A\mathbf{x}={\mathbf{e}}_2=(0,1,0{)}^{\top }$ same tarah solve karo. Forward $L\mathbf{y}={\mathbf{e}}_2$: $y_1=0;2y_1+y_2=1\Rightarrow y_2=1;4y_1+3y_2+y_3=0\Rightarrow 0+3+y_3=0\Rightarrow y_3=-3.$ Back $U\mathbf{x}=\mathbf{y}$: $2x_3=-3\Rightarrow x_3=-\frac{3}{2};x_2+x_3=1\Rightarrow x_2=1-(-\frac{3}{2})=\frac{5}{2};2x_1+x_2+x_3=0\Rightarrow 2x_1+\frac{5}{2}-\frac{3}{2}=0\Rightarrow 2x_1=-1\Rightarrow x_1=-\frac{1}{2}.$ Toh column 2 hai ${\left(-\frac{1}{2},\frac{5}{2},-\frac{3}{2}\right)}^{\top }$. Column 3 — $A\mathbf{x}={\mathbf{e}}_3=(0,0,1{)}^{\top }$ solve karo. Forward $L\mathbf{y}={\mathbf{e}}_3$: $y_1=0;y_2=0;4y_1+3y_2+y_3=1\Rightarrow y_3=1.$ Back $U\mathbf{x}=\mathbf{y}$: $2x_3=1\Rightarrow x_3=\frac{1}{2};x_2+x_3=0\Rightarrow x_2=-\frac{1}{2};2x_1+x_2+x_3=0\Rightarrow 2x_1-\frac{1}{2}+\frac{1}{2}=0\Rightarrow x_1=0.$ Toh column 3 hai ${\left(0,-\frac{1}{2},\frac{1}{2}\right)}^{\top }$. Identical process kyun? $L,U$ kabhi nahi badalte — sirf right-hand side badalta hai, yahi toh LU ka poora point hai.
5. Inverse assemble karo har solved column ko side by side rakh kar: $A^{-1}=\left(\begin{array}{ccc}\frac{3}{2}& -\frac{1}{2}& 0\\ -3& \frac{5}{2}& -\frac{1}{2}\\ 1& -\frac{3}{2}& \frac{1}{2}\end{array}\right).$ Trust kyun karein? $A{A}^{-1}=I$ VERIFY mein check kiya gaya hai.

Verify: $\det A=4$ ✓, aur $A{A}^{-1}=I_3$ ✓ (dono neeche machine-checked hain).

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C7 — real-world word problem, kai right-hand sides

Forecast: $A$ already upper triangular hai — guess karo $L=I$ aur dono solves pure back-substitution hain.

1. Ek baar factor karo. $A$ ke diagonal ke neeche sab zeros hain, toh koi elimination zaruuri nahi: $L=I_3$, $U=A$. Ek baar kyun? Dono din same $A$ hai — ek hi baar factor karo, har $\mathbf{b}$ ke liye reuse karo.
2. Monday, forward solve $L\mathbf{y}={\mathbf{b}}_1$. Kyunki $L=I$, seedha $\mathbf{y}={\mathbf{b}}_1=(6,7,8{)}^{\top }$ milta hai. Trivial kyun? Identity kisi bhi vector ko unchanged chhod deta hai.
3. Monday, back solve $U\mathbf{x}=\mathbf{y}=(6,7,8{)}^{\top }$: $4x_3=8\Rightarrow x_3=2$; $2x_2+3x_3=7\Rightarrow 2x_2+6=7\Rightarrow x_2=\frac{1}{2}$; $x_1+x_2+x_3=6\Rightarrow x_1+\frac{1}{2}+2=6\Rightarrow x_1=\frac{7}{2}$. Bottom-up kyun? $U$ last variable pehle isolate karta hai. Monday batches: $\mathbf{x}={\left(\frac{7}{2},\frac{1}{2},2\right)}^{\top }$.
4. Tuesday, forward solve $L\mathbf{y}={\mathbf{b}}_2$. Phir se $L=I$, toh $\mathbf{y}={\mathbf{b}}_2=(3,4,4{)}^{\top }$. Re-factor kyun nahi? Same $L,U$ exactly reuse karte hain — sirf right-hand side badla hai.
5. Tuesday, back solve $U\mathbf{x}=\mathbf{y}=(3,4,4{)}^{\top }$: $4x_3=4\Rightarrow x_3=1$; $2x_2+3x_3=4\Rightarrow 2x_2+3=4\Rightarrow x_2=\frac{1}{2}$; $x_1+x_2+x_3=3\Rightarrow x_1+\frac{1}{2}+1=3\Rightarrow x_1=\frac{3}{2}$. Tuesday batches: $\mathbf{x}={\left(\frac{3}{2},\frac{1}{2},1\right)}^{\top }$.

Verify (units: batches): Monday $A{\left(\frac{7}{2},\frac{1}{2},2\right)}^{\top }=(6,7,8{)}^{\top }$ ✓; Tuesday $A{\left(\frac{3}{2},\frac{1}{2},1\right)}^{\top }=(3,4,4{)}^{\top }$ ✓. Har naye din mein sirf $\approx n^2$ kaam laga (ek forward + ek back solve), fresh $\frac{2}{3}{n}^3$ elimination nahi.

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C8 — negative & fractional entries (sign bookkeeping)

Forecast: pivot negative hai. Multiplier $m_{21}=3/(-2)=-1.5$ bhi negative hoga — guess karo $L$ ke diagonal ke neeche negative number aayega, aur ek negative multiple subtract karna = add karna.

1. Multiplier. $m_{21}=\frac{a_{21}}{a_{11}}=\frac{3}{-2}=-\mathrm{1.5.}$ Sign kyun? Formula $a_{ik}/a_{kk}$ signs ki parwah nahi karta — hamesha jawab deta hai "kitni pivot rows subtract karni hain." Yahan woh $-1.5$ hai.
2. Eliminate karo. Row${}_2-(-1.5)$Row${}_1=$ Row${}_2+1.5$Row${}_1$: $(3,-5)+1.5(-2,4)=(3-3,-5+6)=(0,1).$ "+" kyun? Negative multiplier subtract karna addition mein flip ho jaata hai — algebra apne aap theek ho jaata hai.
3. Assemble karo. $U=\left(\begin{array}{cc}-2& 4\\ 0& 1\end{array}\right)$, $L=\left(\begin{array}{cc}1& 0\\ -1.5& 1\end{array}\right)$.

Verify: $LU=\left(\begin{array}{cc}1& 0\\ -1.5& 1\end{array}\right)\left(\begin{array}{cc}-2& 4\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}-2& 4\\ 3& -5\end{array}\right)=A$ ✓. Also $\det A=(-2)(1)=-2$, jo $(-2)(-5)-(4)(3)=10-12=-2$ se match karta hai ✓.

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C9 — non-square (rectangular) matrix

Forecast: $A$ mein columns se zyada rows hain. Guess karo: $L$ square rehta hai ($3\times 3$, $A$ ki har row ke liye ek row) jabki $U$ rectangular shape inherit karta hai ($3\times 2$). Idea wahi hai — har column ke pivot ke neeche clear karo.

1. Column 1. Pivot $=a_{11}=2$; $m_{21}=\frac{1}{2}$, $m_{31}=\frac{3}{2}$. Row${}_2-\frac{1}{2}$Row${}_1=(0,5-2)=(0,3)$; Row${}_3-\frac{3}{2}$Row${}_1=(0,8-6)=(0,2)$. Kyun? Same rule: $m_{i1}$ pivot-rows subtract karo first column ko pivot ke neeche zero karne ke liye.
2. Column 2. Naya pivot $=3$ position $(2,2)$ par; $m_{32}=\frac{2}{3}$. Row${}_3-\frac{2}{3}$Row${}_2=(0,2-2)=(0,0)$. Kyun? Hamare paas sirf do columns hain, toh yeh last pivot hai; jo bacha usse neeche clear karo.
3. Assemble karo. $L$ $3\times 3$ hai (teen multipliers store karta hai), $U$ $3\times 2$ hai (surviving grid): $L=\left(\begin{array}{ccc}1& 0& 0\\ \frac{1}{2}& 1& 0\\ \frac{3}{2}& \frac{2}{3}& 1\end{array}\right),U=\left(\begin{array}{cc}2& 4\\ 0& 3\\ 0& 0\end{array}\right).$ $U$ mein bottom-zero row kyun? Sirf do columns hone ki wajah se, teesri row ka koi pivot nahi bacha — woh all-zero row ban jaati hai, jo ek "tall" matrix ke liye normal hai.

Verify: $LU=\left(\begin{array}{cc}2& 4\\ 1& 5\\ 3& 8\end{array}\right)=A$ ✓ (neeche machine-checked).

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Recall Kaun sa cell tumhe rokta?

Zero top-left entry — kaun sa cell aur kya fix? ::: C3; rows swap karo, $P$ mein record karo, $PA=LU$ factor karo. Pivot equals $10^{-9}$ but nonzero — kya LU galat hai? ::: C4; mathematically theek hai lekin numerically unstable hai, toh largest-magnitude entry par pivot karo. Pivot ke neeche poora subcolumn zero — swap karo? ::: C5; swap karne ke liye koi valid nonzero entry nahi — matrix singular hai, $\det =0$. Do alag $\mathbf{b}$'s, same $A$ — $L,U$ dobara compute karein? ::: C7; nahi — ek baar factor karo, har right-hand side ke liye $O(n^2)$ mein reuse karo. $3\times 2$ matrix — $L,U$ ke shapes kya hain? ::: C9; $L$ square rehta hai ($3\times 3$), $U$ rectangular shape inherit karta hai ($3\times 2$).

Decision map

Connections

• LU decomposition — algorithm, applications — woh parent jise yeh page drill karta hai.
• Gaussian Elimination — har example elimination hai jisme steps save hote hain.
• Permutation Matrices — C3, C4 mein row-swap receipts.
• Determinants — C5, C6, C8 mein pivots se padho.
• Matrix Inverse — C6 mein column-by-column reuse.
• Triangular Systems & Substitution — har jagah saste forward/back solves.
• Cholesky Decomposition — symmetric positive-definite cousin.