Worked examples — Determinant of 3×3 matrix — cofactor expansion

2.6.8 · D3 · Maths › Matrices & Determinants — Introduction › Determinant of 3×3 matrix — cofactor expansion

Yeh page ek drill hall hai. Parent note ne tumhe method sikhaya; yahan hum har tarah ki matrix ko dhundte hain jo ek problem mein aa sakti hai aur har ek ko end tak solve karte hain. Shuru karne se pehle, vocabulary ka ek reminder, taaki koi bhi symbol use hone se pehle samjha ja sake.

Signs ka checkerboard jise tum baar baar dekhoge: $\left[\begin{array}{ccc}+& -& +\\ -& +& -\\ +& -& +\end{array}\right]$

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The scenario matrix

Har 3×3 determinant problem neeche diye cases mein se ek (ya ek mix) hoti hai. Har worked example un cells ka naam leta hai jo woh cover karta hai, isliye end tak har cell hit ho jaata hai.

#	Case class	Kya special hai	Kaun sa example hit karta hai
A	All-nonzero, mixed signs	Koi shortcut nahi — ek row carefully brute-force karo	Ex 1
B	Row/column full of zeros	Poora term vanish hota hai → kam kaam	Ex 2
C	Two expansions agree	Invariance prove karo (row vs column)	Ex 3
D	Triangular matrix	Answer = diagonal ka product	Ex 4
E	Degenerate (det = 0)	Rows linearly dependent → flat, zero volume	Ex 5
F	Sign / orientation flip	Negative determinant = mirrored orientation	Ex 6
G	Geometry / word problem	Ek real parallelepiped ka volume	Ex 7
H	Exam twist (unknown letter)	$k$ ke liye solve karo taaki $\det =0$	Ex 8

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Example 1 — Case A: all-nonzero, mixed signs

Forecast: Row 2 hai $(-1,0,4)$. Middle entry $0$ hai, isliye teen terms mein se ek apne aap khatam ho jaayega. Final answer ka sign padhne se pehle guess karo — positive hai ya negative?

1. Row-2 expansion checkerboard signs ke saath likho. Board se row 2 ke signs hain $-,+,-$. Toh $\det (A)=-(-1)M_{21}+(0)M_{22}-(4)M_{23}.$ Yeh step kyun? Row 2 cofactors $C_{21},C_{22},C_{23}$ deta hai jinke signs hain $(-1{)}^{2+1},(-1{)}^{2+2},(-1{)}^{2+3}=-,+,-$. Pehle signs pin karna sabse common error rokta hai (ek minus drop karna).

2. Middle term vanish ho jaata hai. Element $a_{22}=0$ hai, isliye $0\cdot M_{22}=0$. Hum $M_{22}$ compute bhi nahi karte. Kyun? Ek zero element apna poora term khatam kar deta hai — yahi wajah hai ki hum zeros pasand karte hain.

3. $M_{21}$ compute karo (row 2, column 1 delete karo): $M_{21}=\left|\begin{array}{cc}3& 2\\ 5& 1\end{array}\right|=3\cdot 1-2\cdot 5=3-10=-7.$ Yeh step kyun? Bachne wali 2×2 exactly woh chaar entries hain jo row 2 aur column 1 shade out karne par bachi hain.

4. $M_{23}$ compute karo (row 2, column 3 delete karo): $M_{23}=\left|\begin{array}{cc}1& 3\\ 2& 5\end{array}\right|=1\cdot 5-3\cdot 2=5-6=-1.$

5. Assemble karo. $\det (A)=-(-1)(-7)+0-(4)(-1)=-(7)+4=-3.$ Yeh step kyun? Dhyan se: $-(-1)\times (-7)=(+1)\times (-7)=-7$, phir $-(4)(-1)=+4$. Har term mein do sign flips — yahan beginners galti karte hain.

Verify: Ek independent path ke roop mein row 1 ke along expand karo. $1\left|\begin{array}{cc}0& 4\\ 5& 1\end{array}\right|-3\left|\begin{array}{cc}-1& 4\\ 2& 1\end{array}\right|+2\left|\begin{array}{cc}-1& 0\\ 2& 5\end{array}\right|=1(-20)-3(-9)+2(-5)=-20+27-10=-3.✓$ Same answer, alag route — exactly wahi jo parent note ne promise kiya tha.

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Example 2 — Case B: zeros se bhara ek column

Forecast: Poora ek column zero hai. Zero volume kaisa dikhta hai? Koi bhi arithmetic karne se pehle answer apne dimaag mein predict karo.

1. Zero column chuno. Column 2 ke along expand karo: entries $(0,0,0)$. Yeh step kyun? Wahan har element $0$ hai, isliye har term $0\times (\text{kuch bhi})=0$ ho jaati hai.
2. Inhe add karo. $\det (B)=0\cdot C_{12}+0\cdot C_{22}+0\cdot C_{32}=0.$ Yeh step kyun? Kisi bhi minor ko compute karne ki zaroorat nahi — yeh sabse fast possible determinant hai.

Geometric read: $B$ ke teen column-vectors sab ka $y$-component $0$ hai; woh $xz$-plane mein trapped hain. Ek plane mein squeeze ho gaye teen vectors ek flat parallelepiped span karte hain — zero volume.

Verify: Row 1 ke along expand karo: $3\left|\begin{array}{cc}0& 5\\ 0& 7\end{array}\right|-0+2\left|\begin{array}{cc}1& 0\\ 4& 0\end{array}\right|=3(0)+2(0)=0.✓$

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Example 3 — Case C: do expansions agree karni chahiye

Forecast: Column 1 hai $(1,0,0)$ — do zeros. Row 1 hai $(1,2,3)$ — koi zero nahi. Ek path short hai, ek lamba; theorem kehta hai dono same jagah pahunchte hain.

1. Column 1 ke along expand karo (easy path). Column-1 signs hain $+,-,+$. $\det (D)=1\cdot \left|\begin{array}{cc}4& 5\\ 0& 6\end{array}\right|-0+0=1(4\cdot 6-5\cdot 0)=24.$ Yeh step kyun? Do zeros teen terms mein se do khatam kar dete hain, ek clean 2×2 bachti hai.

2. Row 1 ke along expand karo (lamba path). Row-1 signs $+,-,+$. $\det (D)=1\left|\begin{array}{cc}4& 5\\ 0& 6\end{array}\right|-2\left|\begin{array}{cc}0& 5\\ 0& 6\end{array}\right|+3\left|\begin{array}{cc}0& 4\\ 0& 0\end{array}\right|.$ Yeh step kyun? Row 1 mein koi zero nahi, isliye hum honestly teeno minors compute karte hain.

3. Har 2×2 evaluate karo. $\left|\begin{array}{cc}4& 5\\ 0& 6\end{array}\right|=24$, $\left|\begin{array}{cc}0& 5\\ 0& 6\end{array}\right|=0\cdot 6-5\cdot 0=0$, $\left|\begin{array}{cc}0& 4\\ 0& 0\end{array}\right|=0$. Yeh step kyun? Unme se do mein zero column hai, isliye woh vanish ho jaate hain — matrix ka structure humari raksha karta hai.

4. Sum: $1(24)-2(0)+3(0)=24.$

Verify: Dono paths $24$ dete hain — invariance confirm hua. Yeh Case D neeche bhi demonstrate karta hai: $24=1\cdot 4\cdot 6$, diagonal ka product. Invariance hamesha kyun hold karta hai, uske liye properties of determinants dekho.

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Example 4 — Case D: triangular matrix

Forecast: Sirf diagonal $2,1,6$ use karke answer guess karo.

1. Row 1 ke along expand karo — yeh hai $(2,0,0)$, do free zeros. $\det (T)=2\left|\begin{array}{cc}1& 0\\ 4& 6\end{array}\right|-0+0.$ Yeh step kyun? Row 1 mein diagonal ke upar zeros do terms khatam kar dete hain.
2. Bachne wali 2×2 khud triangular hai. $\left|\begin{array}{cc}1& 0\\ 4& 6\end{array}\right|=1\cdot 6-0\cdot 4=6.$ Yeh step kyun? Anti-diagonal mein $0$ hai, isliye sirf main diagonal bachti hai — same trick ek level neeche.
3. Multiply karo: $\det (T)=2\cdot 6=12=2\cdot 1\cdot 6.$

Verify: $2\cdot 1\cdot 6=12.✓$

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Example 5 — Case E: degenerate matrix (det = 0)

Forecast: Rows suspiciously evenly spaced lag rahi hain. Koi hidden dependency smell aa rahi hai?

1. Row 1 ke along expand karo. $\det (C)=1\left|\begin{array}{cc}5& 6\\ 8& 9\end{array}\right|-2\left|\begin{array}{cc}4& 6\\ 7& 9\end{array}\right|+3\left|\begin{array}{cc}4& 5\\ 7& 8\end{array}\right|.$
2. Teen 2×2s. $\left|\begin{array}{cc}5& 6\\ 8& 9\end{array}\right|=45-48=-3$, $\left|\begin{array}{cc}4& 6\\ 7& 9\end{array}\right|=36-42=-6$, $\left|\begin{array}{cc}4& 5\\ 7& 8\end{array}\right|=32-35=-3$.
3. Sum: $1(-3)-2(-6)+3(-3)=-3+12-9=0.$ Yeh step kyun? Alternating signs pieces ko exactly cancel kar dete hain — ek collapsed shape ka signature.

Geometric read (figure dekho): row 3 = $2\times$row 2 $-$ row 1. Teen row-vectors ek akele tilted plane mein lie karte hain, isliye woh jo parallelepiped banate hain woh ek flat sheet hai — zero thickness, zero volume.

Verify: $\det =0.✓$

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Example 6 — Case F: sign / orientation flip

Forecast: Do axes swap karna ek glove ko inside-out palatne jaisa hai. Positive ya negative volume?

1. Row 3 ke along expand karo $=(0,0,1)$ — do zeros. $\det (P)=0-0+1\cdot (-1{)}^{3+3}\left|\begin{array}{cc}0& 1\\ 1& 0\end{array}\right|.$ Yeh step kyun? Row 3 sabse sasta hai.
2. Minor. $\left|\begin{array}{cc}0& 1\\ 1& 0\end{array}\right|=0\cdot 0-1\cdot 1=-1.$
3. Assemble karo: $\det (P)=1\cdot (+1)\cdot (-1)=-1.$

Verify: Identity ki do rows swap karna hamesha sign flip karta hai: $\det (I)=1\Rightarrow \det (P)=-1.✓$

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Example 7 — Case G: real-world volume (word problem)

Forecast: Volume determinant ka magnitude hai jiske rows $\mathbf{u},\mathbf{v},\mathbf{w}$ hain. Estimate karo: $2\cdot 3\cdot 4=24$ se bada? Chhota?

1. Determinant set up karo. Vectors ko rows ki tarah stack karo: $V_{\text{signed}}=\left|\begin{array}{ccc}2& 1& 0\\ 0& 3& 1\\ 1& 0& 4\end{array}\right|.$ Yeh step kyun? Edge-vectors ka determinant hi signed volume hai — yeh what is a determinant se poora geometric meaning hai. (Yeh scalar triple product bhi hai; dekho cross product and determinants.)
2. Column 1 ke along expand karo $(2,0,1)$ — ek zero help karta hai. $=2\left|\begin{array}{cc}3& 1\\ 0& 4\end{array}\right|-0+1\cdot (-1{)}^{3+1}\left|\begin{array}{cc}1& 0\\ 3& 1\end{array}\right|.$ Yeh step kyun? $a_{21}=0$ middle term remove karta hai; checkerboard column 1 ke neeche $+,-,+$ deta hai.
3. Do minors. $\left|\begin{array}{cc}3& 1\\ 0& 4\end{array}\right|=12-0=12$, $\left|\begin{array}{cc}1& 0\\ 3& 1\end{array}\right|=1-0=1.$
4. Assemble karo: $2(12)+1(1)=24+1=25.$
5. Physical volume ke liye magnitude lo: $|25|=25.$

Verify (units aur sanity): har vector nm mein hai, isliye volume ${\text{nm}}^3$ mein hai: $25{\text{nm}}^3$. Yeh naive $2\cdot 3\cdot 4=24$ se zyada hai kyunki vectors almost perpendicular hain — reasonable. Row 1 ke along expand karke cross-check karo: $2(12)-1(0\cdot 4-1\cdot 1)+0=24-1(-1)=25.✓$

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Example 8 — Case H: unknown wala exam twist

Forecast: Hume $k$ mein ek cubic milegi. Kitne real roots — ek, do, ya teen?

1. Column 1 ke along expand karo $(1,0,k)$ — middle zero ek minor bachata hai. $\det (K)=1\left|\begin{array}{cc}1& k\\ 0& 1\end{array}\right|-0+k\cdot (-1{)}^{3+1}\left|\begin{array}{cc}k& 2\\ 1& k\end{array}\right|.$ Yeh step kyun? Column 1 mein zero hai; uske neeche sign pattern $+,-,+$ hai.
2. Do minors. $\left|\begin{array}{cc}1& k\\ 0& 1\end{array}\right|=1-0=1$, $\left|\begin{array}{cc}k& 2\\ 1& k\end{array}\right|=k^2-2.$
3. $k$ mein polynomial ki tarah assemble karo. $\det (K)=1(1)+k(k^2-2)=k^3-2k+1.$ Yeh step kyun? Ab "singular" ban jaata hai "solve $k^3-2k+1=0$."
4. Factor karo. $k=1$ test karo: $1-2+1=0$ ✓, toh $(k-1)$ ek factor hai: $k^3-2k+1=(k-1)(k^2+k-1).$ Quadratic deta hai $k=\frac{-1\pm \sqrt{5}}{2}$. Yeh step kyun? Mila hua root divide out karne par ek quadratic bachti hai jise hum standard formula se solve karte hain.
5. Teeno roots: $k=1,k=\frac{-1+\sqrt{5}}{2}\approx 0.618,k=\frac{-1-\sqrt{5}}{2}\approx -\mathrm{1.618.}$

Verify: $k=1$ wapas plug karo: $\det =1^3-2(1)+1=0$ ✓. Aur $(k-1)(k^2+k-1)$ expand karo $=k^3+k^2-k-k^2-k+1=k^3-2k+1$ ✓. (Yeh $k$-values exactly wahan hain jahan the inverse fail karta hai aur jahan $K$ mein ek linear system ka koi unique Cramer's-rule solution nahi hota.)

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Recall Self-test: case ka naam lo, phir solve karo

Har ek mein kaun sa scenario cell hai, aur answer kya hai? Ek row of zeros — determinant? ::: $0$ (Case B), space collapse ho jaata hai. Diagonal $3,-2,5$ wali upper-triangular matrix — determinant? ::: $3\cdot (-2)\cdot 5=-30$ (Case D). $\det =7$ wali matrix ki do rows swap karo — naya determinant? ::: $-7$ (Case F), orientation flip hoti hai. $\det =0$ inverse ke baare mein kya bata ta hai? ::: Woh exist nahi karta; matrix singular hai (Case E).