Exercises — Determinant of 3×3 matrix — cofactor expansion

HUM KYA KARTE HAIN: dono indices add karein, phir poochhen "even hai ya odd?" Even $\to +$, odd $\to -$.

• $(1,3)$: $1+3=4$, even $\to +$
• $(2,3)$: $2+3=5$, odd $\to -$
• $(3,2)$: $3+2=5$, odd $\to -$
• $(2,2)$: $2+2=4$, even $\to +$

YAISE DIKHTA HAI: inhe checkerboard picture se check karein — chaar cells exactly $+,-,-,+$ light up hote hain. Signs kabhi bhi matrix ke numbers par depend nahi karte, sirf position par.

HUM KYA DHOONDH RAHE HAIN: woh line (row ya column) jisme sabse zyada zeros hain. Har zero entry apne poore cofactor term ko khatam kar deta hai (zero times kuch bhi zero hota hai), isliye zero ka matlab hai "ek kam $2\times 2$ determinant compute karna."

Scan karein: column 2 hai $(0,0,6)$ — do zeros. Koi aur row ya column usse beat nahi karta.

Answer: column 2 ke along expand karein. Sirf ek term survive karta hai: $\det (A)=6\cdot C_{32}.$ ISKA KYUN MATLAB HAI: lazy-but-correct choice teen minors ko ek tak cut kar deta hai.

ROW 2 KYUN? Isme position $(2,2)$ par $0$ hai, isliye ek term vanish ho jaata hai.

Row-2 entries hain $a_{21}=-1,a_{22}=0,a_{23}=4$. Unke checkerboard signs (pattern ki middle row padhein) hain $-,+,-$.

$\det (A)=-1\cdot C_{21}+0\cdot C_{22}+4\cdot C_{23}.$

Cofactor $C_{21}$ — sign $(-1{)}^{2+1}=-1$; row 2, column 1 cross out karein: $M_{21}=\left|\begin{array}{cc}3& 2\\ 5& 1\end{array}\right|=3\cdot 1-2\cdot 5=3-10=-7,C_{21}=(-1)(-7)=7.$

Cofactor $C_{23}$ — sign $(-1{)}^{2+3}=-1$; row 2, column 3 cross out karein: $M_{23}=\left|\begin{array}{cc}1& 3\\ 2& 5\end{array}\right|=1\cdot 5-3\cdot 2=5-6=-1,C_{23}=(-1)(-1)=1.$

Sum: $\det (A)=(-1)(7)+4(1)=-7+4=-3.$

Row 1 se check karein (KYUN: invariance feel karne ke liye — alag raasta, same answer). Entries $1,3,2$, signs $+,-,+$:

• $M_{11}=\left|\begin{array}{cc}0& 4\\ 5& 1\end{array}\right|=0-20=-20$
• $M_{12}=\left|\begin{array}{cc}-1& 4\\ 2& 1\end{array}\right|=-1-8=-9$
• $M_{13}=\left|\begin{array}{cc}-1& 0\\ 2& 5\end{array}\right|=-5-0=-5$ $\det =1(-20)-3(-9)+2(-5)=-20+27-10=-3.✓$

COLUMN 1 KYUN? Isme $(2,1)$ par $0$ hai, isliye teen mein se ek term disappear ho jaata hai.

Column-1 entries $a_{11}=2,a_{21}=0,a_{31}=1$, signs (checkerboard ka left column) $+,-,+$.

$C_{11}$ — sign $+$; row 1, column 1 delete karein: $M_{11}=\left|\begin{array}{cc}3& 4\\ 0& 2\end{array}\right|=6-0=6,C_{11}=+6.$

$C_{31}$ — sign $(-1{)}^{3+1}=+$; row 3, column 1 delete karein: $M_{31}=\left|\begin{array}{cc}1& -1\\ 3& 4\end{array}\right|=4-(-3)=7,C_{31}=+7.$

Sum: $\det (C)=2(6)+1(7)=12+7=19.$

Column 1 ke along. Entries $1,0,0$ signs $+,-,+$ ke saath. Sirf pehla survive karta hai: $\det (U)=1\cdot C_{11}=1\cdot \left|\begin{array}{cc}4& 5\\ 0& 6\end{array}\right|=1\cdot (24-0)=24.$

Row 1 ke along. Entries $1,2,3$ signs $+,-,+$ ke saath:

• $M_{11}=\left|\begin{array}{cc}4& 5\\ 0& 6\end{array}\right|=24$
• $M_{12}=\left|\begin{array}{cc}0& 5\\ 0& 6\end{array}\right|=0-0=0$
• $M_{13}=\left|\begin{array}{cc}0& 4\\ 0& 0\end{array}\right|=0-0=0$ $\det (U)=1(24)-2(0)+3(0)=24.$

Same answer, $24$ — parent ki "different path, same destination" baat confirm hoti hai.

General pattern: kisi bhi triangular matrix ke liye (diagonal ke ek taraf sab zeros) determinant diagonal entries ka product hota hai. Yahan $1\cdot 4\cdot 6=24$. KYUN: zero column ke neeche expand karne par hamesha exactly ek nonzero term bachta hai — diagonal entry aur chhote triangular block ka product — aur woh recursion diagonal ko poora neeche multiply karta hai. Dekhen properties of determinants.

Pehle structural observation. Rows 1 aur 2 dekhein: $(2,4,6)=2\cdot (1,2,3)$. Row 2, row 1 ka scalar multiple hai — dono rows space mein ek hi line ke along point karti hain.

Confirm karne ke liye row 3 ke along expand karein. Entries $5,1,0$, signs $+,-,+$:

• $M_{31}=\left|\begin{array}{cc}2& 3\\ 4& 6\end{array}\right|=12-12=0$
• $M_{32}=\left|\begin{array}{cc}1& 3\\ 2& 6\end{array}\right|=6-6=0$ $\det (D)=5(0)-1(0)+0\cdot C_{33}=0.$

Geometric meaning (YAISE DIKHTA HAI): teen row-vectors normally ek solid box (parallelepiped) span karte hain kuch volume ke saath. Yahan unme se do ek line par hain, isliye "box" flat ho jaata hai — zero thickness, isliye zero volume, zero determinant. Neeche figure dekhein.

$\det =0$ KYUN SET KAREIN? Matrix exactly tab singular hoti hai jab uska determinant zero hota hai — tab uska koi inverse nahi hota; dekhen determinant and inverse.

Column 1 ke along expand karein (isme $(3,1)$ par zero hai). Entries $k,1,0$.

$C_{11}$ — sign $(-1{)}^{1+1}=+1$; row 1, column 1 delete karein: $M_{11}=\left|\begin{array}{cc}k& 1\\ 2& k\end{array}\right|=k^2-2,C_{11}=+(k^2-2).$

$C_{21}$ — sign $(-1{)}^{2+1}=-1$; row 2, column 1 delete karein: $M_{21}=\left|\begin{array}{cc}2& 0\\ 2& k\end{array}\right|=2k-0=2k,C_{21}=(-1)(2k)=-2k.$

Teesri entry $a_{31}=0$ apne term ko khatam kar deti hai. Entry-times-cofactor assemble karein: $\det =k\cdot C_{11}+1\cdot C_{21}+0=k(k^2-2)+1(-2k)=k^3-2k-2k=k^3-4k.$

Zero set karein aur factor karein: $k^3-4k=k(k^2-4)=k(k-2)(k+2)=0.$ $\boxed{k=0,k=2,k=-2.}$

Sanity check $k=2$: $\det =8-8=0.✓$ Aur $k=1$ (nonzero hona chahiye): $1-4=-3\ne 0$, isliye $M(1)$ invertible hai.

ROW 1 KYUN EXPAND KAREIN? Row 1 mein basis vectors $\mathbf{i},\mathbf{j},\mathbf{k}$ hain; wahan expand karne se har cofactor se result ka ek component nikalta hai. Signs $+,-,+$.

$\mathbf{u}\times \mathbf{v}=\mathbf{i}\left|\begin{array}{cc}2& 3\\ 5& 6\end{array}\right|-\mathbf{j}\left|\begin{array}{cc}1& 3\\ 4& 6\end{array}\right|+\mathbf{k}\left|\begin{array}{cc}1& 2\\ 4& 5\end{array}\right|.$

Har $2\times 2$ compute karein:

• $\left|\begin{array}{cc}2& 3\\ 5& 6\end{array}\right|=12-15=-3$
• $\left|\begin{array}{cc}1& 3\\ 4& 6\end{array}\right|=6-12=-6$
• $\left|\begin{array}{cc}1& 2\\ 4& 5\end{array}\right|=5-8=-3$

Beech ke minus sign ka dhyan rakhte hue assemble karein: $\mathbf{u}\times \mathbf{v}=(-3)\mathbf{i}-(-6)\mathbf{j}+(-3)\mathbf{k}=(-3,6,-3).$

Check (perpendicularity): cross product dono inputs ke perpendicular hota hai, isliye $\mathbf{u}$ ke saath iska dot product $0$ hona chahiye: $(-3)(1)+(6)(2)+(-3)(3)=-3+12-9=0.✓$

CRAMER'S RULE KYA KEHTA HAI: $A\mathbf{x}=\mathbf{b}$ ke liye, unknown $y=\frac{\det (A_y)}{\det (A)}$, jahan $A_y$ woh $A$ hai jisme uska doosra column right-hand side $\mathbf{b}=(6,13,13)$ se replace kar diya gaya ho. KYUN KAAM KARTA HAI: ek column ko $\mathbf{b}$ se replace karke aur determinants lene se determinant ki linearity ke through woh ek variable isolate hota hai.

Coefficient matrix aur uska determinant. ROW 1 KE ALONG KYUN EXPAND KAREIN? $A$ ki koi row ya column mein zero nahi hai, isliye koi choice kaam nahi bachaati; hum consistency ke liye row 1 par default karte hain (koi bhi row same number deta hai, jaise L3.1 ne dikhaya). Signs $+,-,+$: $A=\left[\begin{array}{ccc}1& 2& 1\\ 2& 1& 3\\ 1& 3& 2\end{array}\right]$

• $M_{11}=\left|\begin{array}{cc}1& 3\\ 3& 2\end{array}\right|=2-9=-7$
• $M_{12}=\left|\begin{array}{cc}2& 3\\ 1& 2\end{array}\right|=4-3=1$
• $M_{13}=\left|\begin{array}{cc}2& 1\\ 1& 3\end{array}\right|=6-1=5$ $\det (A)=1(-7)-2(1)+1(5)=-7-2+5=-4.$

Kyunki $\det (A)=-4\ne 0$, system ka unique solution hai — Cramer's rule valid hai.

$A_y$ banayein (column 2 ko $(6,13,13)$ se replace karein). PHIR ROW 1 KYUN? Same reason — kaan zeros nahi, isliye clean parallel comparison ke liye same row rakhte hain. Signs $+,-,+$: $A_y=\left[\begin{array}{ccc}1& 6& 1\\ 2& 13& 3\\ 1& 13& 2\end{array}\right]$

• $M_{11}=\left|\begin{array}{cc}13& 3\\ 13& 2\end{array}\right|=26-39=-13$
• $M_{12}=\left|\begin{array}{cc}2& 3\\ 1& 2\end{array}\right|=4-3=1$
• $M_{13}=\left|\begin{array}{cc}2& 13\\ 1& 13\end{array}\right|=26-13=13$ $\det (A_y)=1(-13)-6(1)+1(13)=-13-6+13=-6.$

Isliye: $y=\frac{\det (A_y)}{\det (A)}=\frac{-6}{-4}=\frac{3}{2}.$

Full-solution check (zaroori nahi, lekin reassuring — KYUN: ek independent route ko $y$ reproduce karna chahiye). Teeno equations simultaneously solve karne par $x=\frac{5}{2},y=\frac{3}{2},z=\frac{3}{2}.$ In values se har equation verify karein:

• Eq 1: $\frac{5}{2}+2(\frac{3}{2})+\frac{3}{2}=\frac{5}{2}+3+\frac{3}{2}=7$… yani $\frac{5}{2}+\frac{6}{2}+\frac{3}{2}=\frac{14}{2}=7\ne 6$, isliye yeh values galat hain — ek warning ki round numbers guess karna dangerous hai. Sahi se solve karte hain.

Sahi se solve karein (elimination). Eq 1 se, $x=6-2y-z$. Eq 2 aur Eq 3 mein substitute karein:

• Eq 2: $2(6-2y-z)+y+3z=13\Rightarrow 12-4y-2z+y+3z=13\Rightarrow -3y+z=1.$
• Eq 3: $(6-2y-z)+3y+2z=13\Rightarrow 6+y+z=13\Rightarrow y+z=7.$

Pehli reduced equation ko doosri se subtract karein: $(y+z)-(-3y+z)=7-1\Rightarrow 4y=6\Rightarrow y=\frac{3}{2}.$ Phir $z=7-y=7-\frac{3}{2}=\frac{11}{2}$, aur $x=6-2(\frac{3}{2})-\frac{11}{2}=6-3-\frac{11}{2}=3-\frac{11}{2}=-\frac{5}{2}.$

Sahi solution hai $x=-\frac{5}{2},y=\frac{3}{2},z=\frac{11}{2}$. Eq 2 check karein: $2(-\frac{5}{2})+\frac{3}{2}+3(\frac{11}{2})=-5+\frac{3}{2}+\frac{33}{2}=-5+\frac{36}{2}=-5+18=13.✓$ Eq 3 check karein: $-\frac{5}{2}+3(\frac{3}{2})+2(\frac{11}{2})=-\frac{5}{2}+\frac{9}{2}+11=\frac{4}{2}+11=2+11=13.✓$

Dono routes agree karte hain: $\boxed{y=\frac{3}{2}}$. Cramer's rule ne seedha sahi value di, aur full elimination ne usse confirm kiya.

Determinant is par depend nahi karta ki aap kis row/column ke along expand karte hain. ::: Sach — yeh matrix ki ek fixed property hai; expansion choice sirf arithmetic route badalta hai. Ek row jo doosri row ka multiple hoti hai determinant ko… bana deti hai ::: $0$ (box flat hai, zero volume). Cramer's rule se $k$-ve variable ke liye solve karne ke liye aap… replace karte hain ::: $A$ ka $k$-va column right-hand-side vector $\mathbf{b}$ se. Entry $(i,j)$ ke saath attached sign hai… ::: $(-1{)}^{i+j}$, sirf position par depend karta hai, entry ki value par kabhi nahi.