Worked examples — Determinants — cofactor expansion along any row - column

4.5.21 · D3 · Maths › Linear Algebra (Full) › Determinants — cofactor expansion along any row - column

Shuru karne se pehle, do base facts jinpar sab kuch tika hua hai. Pehle $2\times 2$ determinant:

Doosra, cofactor symbol $C_{ij}$ jo hum neeche har formula mein use karenge:

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Scenario matrix

Yahan har class ka case hai jo cofactor expansion present kar sakta hai. Har row ek alag trap ya twist hai; last column bataata hai kaunsa worked example use handle karta hai.

#	Case class	Kya alag banaata hai	Killed by
A	Plain $3\times 3$, no zeros	Full 3-cofactor grind	Ex 1
B	Row/column with zeros	80/20: sparse line chuno	Ex 2
C	Sign-trap position $(i,j)$ with $i+j$ odd	Woh $(-1{)}^{i+j}=-1$ jo log bhool jaate hain	Ex 3
D	"Same answer any line" proof-in-practice	Do tarike se expand karo, compare karo	Ex 4
E	Make-zeros-first (row ops)	Row-add free hai; row-swap ka cost $-1$ hai	Ex 5
F	Degenerate / singular ($\det =0$)	Collapse detect karo, meaning samjho	Ex 6
G	$4\times 4$ recursion	Cofactor of a cofactor	Ex 7
H	Real-world word problem (area)	Determinant = signed area	Ex 8
I	Exam twist: symbolic / unknown entry	$x$ solve karo jo $\det =0$ banaye	Ex 9

Nau examples work karo aur aap har cell touch kar loge.

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Example 1 — Case A: plain $3\times 3$, koi free zeros nahi

1. Row-1 formula likho. Yeh step kyun? Row $i=1$ ke along cofactor expansion kehta hai top row mein chalo, har entry uske signed minor ke saath: $\det A=a_{11}{C}_{11}+a_{12}{C}_{12}+a_{13}{C}_{13}.$
2. Checkerboard se signs padhho. $(-1{)}^{1+1}=+$, $(-1{)}^{1+2}=-$, $(-1{)}^{1+3}=+$. Yeh step kyun? Sign sirf position $(i,j)$ par depend karta hai, entry ki value par kabhi nahi.
3. Har minor compute karo us entry ki row aur column cover karke. Yeh step kyun? Entry ki row aur column delete karna uske minor ki definition hai — problem ko ek aisi $2\times 2$ tak le aata hai jo hum pehle se solve karna jaante hain. $M_{11}=\det \left(\begin{array}{cc}4& 1\\ 2& -2\end{array}\right)=4(-2)-1(2)=-10,$ $M_{12}=\det \left(\begin{array}{cc}0& 1\\ 5& -2\end{array}\right)=0(-2)-1(5)=-5,$ $M_{13}=\det \left(\begin{array}{cc}0& 4\\ 5& 2\end{array}\right)=0(2)-4(5)=-20.$
4. Entries aur signs ke saath combine karo. Yeh step kyun? Ab $a_{1j}$, sign, minor ko saath plug karo: $\det A=2(+)(-10)+(-1)(-)(-5)+3(+)(-20).$ $=-20-5-60=-85.$

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Example 2 — Case B: zeros ke peeche bhaago

1. Column 2 pick karo aur formula likho. Yeh step kyun? 80/20: teen mein se do terms $0$ se multiply hote hain, toh pencil uthane se pehle hi gayab ho jaate hain. $\det A=a_{12}{C}_{12}+a_{22}{C}_{22}+a_{32}{C}_{32}=0+4C_{22}+0.$
2. $C_{22}$ ka sign: $(-1{)}^{2+2}=+$. Yeh step kyun? Position $(2,2)$ ek $+$ square par baith hai.
3. Ek minor: row 2, column 2 delete karo. Yeh step kyun? Surviving entry ki row aur column hatane ke baad woh $2\times 2$ block bachta hai jiska determinant hi minor hai. $M_{22}=\det \left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)=4-1=3.$
4. Combine karo: Yeh step kyun? Entry $\times$ sign $\times$ minor ek surviving term hai. $\det A=4\cdot (+)\cdot 3=12.$

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Example 3 — Case C: odd-position sign trap

1. Position signs list karo. Yeh step kyun? Numbers touch karne se pehle har sign settle karo taaki koi stray minus slip na kare. $(-1{)}^{1+3}=+,(-1{)}^{2+3}=-,(-1{)}^{3+3}=+.$
2. Teen minors (har entry ki row aur column 3 delete karo). Yeh step kyun? Har minor woh hai jo entry ki apni row aur column 3 cover karne ke baad bachta hai — same shading picture jaise intro figure mein hai. $M_{13}=\det \left(\begin{array}{cc}0& 4\\ 5& 2\end{array}\right)=-20,M_{23}=\det \left(\begin{array}{cc}2& -1\\ 5& 2\end{array}\right)=4-(-5)=9,$ $M_{33}=\det \left(\begin{array}{cc}2& -1\\ 0& 4\end{array}\right)=8-0=8.$
3. Entry $\times$ sign $\times$ minor combine karo. Yeh step kyun? Middle term wahi hai jahan log crash karte hain — entry $1$, sign $-$, minor $9$: $\det A=3(+)(-20)+1(-)(9)+(-2)(+)(8)=-60-9-16=-85.$

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Example 4 — Case D: fresh matrix par "any line" prove karo

1. Row 1 ke along expand karo. Yeh step kyun? Standard shuruat. $\det C=1(+)\det \left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)+2(-)\det \left(\begin{array}{cc}2& 2\\ 2& 1\end{array}\right)+2(+)\det \left(\begin{array}{cc}2& 1\\ 2& 2\end{array}\right).$
2. Minors. Yeh step kyun? Har $2\times 2$ woh block hai jo row-1 entry ki row aur column cover karne ke baad bachta hai. $\det \left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)=-3,\det \left(\begin{array}{cc}2& 2\\ 2& 1\end{array}\right)=-2,\det \left(\begin{array}{cc}2& 1\\ 2& 2\end{array}\right)=2.$
3. Combine karo. Yeh step kyun? Entries $\times$ signs $\times$ minors, summed: $1(-3)+2(-)(-2)+2(2)=-3+4+4=5.$

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Example 5 — Case E: pehle zeros banao (row ops), swap ka dhyan rakho

1. Row-add operations. Yeh step kyun? Ek row ka multiple doosri mein add karna $\det$ unchanged rehne deta hai — pure profit. $R_2\to R_2-2R_1$ aur $R_3\to R_3-R_1$ karo: $\to \left(\begin{array}{ccc}1& 2& 3\\ 0& 1& -3\\ 0& -2& 5\end{array}\right).$
2. Column 1 ke along expand karo — sirf top entry nonzero hai. Yeh step kyun? Do engineered zeros teen mein se do cofactors ko khatam kar dete hain, sirf ek bachta hai. $\det B=1(+)\det \left(\begin{array}{cc}1& -3\\ -2& 5\end{array}\right)=1(5-6)=-1.$
3. Ek cautionary variant. Yeh step kyun? Maano ki aapne kahin $R_1\leftrightarrow R_3$ swap kiya hota. Ek swap $\det$ ko $-1$ se multiply kar deta hai; ek row ko $k$ se scale karna $k$ se multiply karta hai. Row-adds hi ek maatra free operation hai.

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Example 6 — Case F: degenerate (singular) matrix

1. Row 1 ke along expand karo. Yeh step kyun? Koi zeros nahi, toh honestly grind karo collapse prove karne ke liye. $\det S=1(+)\det \left(\begin{array}{cc}4& 6\\ 8& 9\end{array}\right)+2(-)\det \left(\begin{array}{cc}2& 6\\ 7& 9\end{array}\right)+3(+)\det \left(\begin{array}{cc}2& 4\\ 7& 8\end{array}\right).$
2. Minors. Yeh step kyun? Har row-1 entry ki row aur column cover karo taaki surviving $2\times 2$ padh sako. $\det \left(\begin{array}{cc}4& 6\\ 8& 9\end{array}\right)=36-48=-12,\det \left(\begin{array}{cc}2& 6\\ 7& 9\end{array}\right)=18-42=-24,\det \left(\begin{array}{cc}2& 4\\ 7& 8\end{array}\right)=16-28=-12.$
3. Combine karo. Yeh step kyun? Entries $\times$ signs $\times$ minors, summed: $1(-12)+2(-)(-24)+3(-12)=-12+48-36=0.$

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Example 7 — Case G: ek $4\times 4$ (cofactor of a cofactor)

1. Row 2 ke along expand karo. Yeh step kyun? Ek nonzero entry $\Rightarrow$ ek cofactor. $\det D=3\cdot (-1{)}^{2+2}{M}_{22}=3M_{22},$ jahan $M_{22}$ row 2 aur column 2 delete karta hai: $M_{22}=\det \left(\begin{array}{ccc}1& 2& -1\\ 2& 1& 0\\ 0& 0& 4\end{array}\right).$
2. Ab is $3\times 3$ ko uski bottom row $(0,0,4)$ ke along expand karo — phir sparse. Yeh step kyun? Recursion: ek $4\times 4$ cofactor khud cofactor expansion se solve hota hai. $M_{22}=4\cdot (-1{)}^{3+3}\det \left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)=4(1-4)=4(-3)=-12.$
3. Combine karo. Yeh step kyun? Inner minor ko single row-2 term mein feed karo. $\det D=3\cdot (-12)=-36.$

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Example 8 — Case H: real-world word problem (signed area)

1. $P$ se edge vectors banao. Yeh step kyun? Ek determinant do vectors se bane parallelogram ka area measure karta hai; triangle iska aadha hota hai. $\overrightarrow{PQ}=(4,1),\overrightarrow{PR}=(1,3).$
2. $2\times 2$ determinant form aur expand karo (base case, koi cofactors ki zaroorat nahi). Yeh step kyun? $2\times 2$ par hum already base case $ad-bc$ par hain — kuch expand nahi karna. $\det \left(\begin{array}{cc}4& 1\\ 1& 3\end{array}\right)=4\cdot 3-1\cdot 1=11.$
3. Aadha karo. Yeh step kyun? Determinant parallelogram area deta hai; triangle exactly aadha hota hai. $\text{Area}=\frac{1}{2}(11)=5.5{\text{m}}^2.$ Positive kyun? Determinant ka sign $+$ hai, matlab $P\to Q\to R$ counter-clockwise chalta hai. Clockwise ordering se $-5.5$ milta; magnitude phir bhi area hai.

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Example 9 — Case I: unknown ke saath exam twist

1. Row 1 ke along expand karo (position $(1,3)$ par ek zero hai). Yeh step kyun? Ek term vanish ho jaata hai. $\det M=x(+)\det \left(\begin{array}{cc}x& 1\\ 1& x\end{array}\right)+1(-)\det \left(\begin{array}{cc}1& 1\\ 0& x\end{array}\right)+0.$
2. Minors. Yeh step kyun? Har row-1 entry ki row aur column cover karo taaki surviving $2\times 2$ padh sako. $\det \left(\begin{array}{cc}x& 1\\ 1& x\end{array}\right)=x^2-1,\det \left(\begin{array}{cc}1& 1\\ 0& x\end{array}\right)=x.$
3. Combine karo. Yeh step kyun? Entries $\times$ signs $\times$ minors se determinant ek polynomial in $x$ ban jaata hai: $\det M=x(x^2-1)-1\cdot x=x^3-x-x=x^3-2x=x(x^2-2).$
4. Zero set karo aur solve karo. Yeh step kyun? Singular $\iff \det =0$. $x(x^2-2)=0\Rightarrow x=0,x=\sqrt{2},x=-\sqrt{2}.$

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Recall Sab cells par self-test

Answers cover karo aur har number reproduce karo. Ex 1 answer ::: $-85$ Ex 3 answer ::: $-85$ Ex 4 answer ::: $5$ Ex 5 answer ::: $-1$ Ex 6 answer (singular) ::: $0$ Ex 7 answer ::: $-36$ Ex 8 area ::: $5.5{\text{m}}^2$ Ex 9 singular values ::: $x=0,\pm \sqrt{2}$

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Connections

• Parent: Cofactor expansion
• Leibniz formula for determinants — jahan signs ultimately aate hain.
• Determinant via row reduction — Example 5 ki trick poori tarah push ki.
• Properties of determinants (multilinearity, alternation) — Examples 6 & 8 inhe use karte hain.
• Invertibility and singular matrices — Examples 6 & 9.
• Adjugate matrix and inverse aur Cramer's rule — dono cofactors par built hain.