Exercises — Determinants — cofactor expansion along any row - column

$M_{23}$ ka matlab: row 2 aur column 3 cover karo. Row 2 hai $(0,3,4)$; column 3 hai $(1,4,8)$. Dono delete karo. $M_{23}=\det \left(\begin{array}{cc}2& 5\\ 7& 6\end{array}\right)=2\cdot 6-5\cdot 7=12-35=-23.$ To $M_{23}=-23$. (Dhyan rakho: $M$ mein koi sign abhi nahi hai — yeh minor ka kaam hai, cofactor ka nahi.)

Cofactor signed minor hota hai: $C_{23}=(-1{)}^{2+3}{M}_{23}$. Kyunki $2+3=5$ odd hai, $(-1{)}^5=-1$. $C_{23}=(-1)\cdot (-23)=+23.$ Sign ne minor ka sign flip kar diya — yahi checkerboard apna kaam kar raha hai.

$\left(\begin{array}{ccc}+& -& +\\ -& +& -\\ +& -& +\end{array}\right)$

• $(1,2)$: $(-1{)}^{1+2}=(-1{)}^3=-$
• $(2,2)$: $(-1{)}^{2+2}=(-1{)}^4=+$
• $(3,1)$: $(-1{)}^{3+1}=(-1{)}^4=+$ Sign sirf $i+j$ par depend karta hai — actual position par, iss baat par kabhi nahi ki tum ise row ya column expansion keh rahe ho.

Row 1 ke saath expand karo: $\det A=1\cdot C_{11}+2\cdot C_{12}+3\cdot C_{13}$. $C_{11}=(+)\det \left(\begin{array}{cc}5& 6\\ 8& 10\end{array}\right)=50-48=2.$ $C_{12}=(-)\det \left(\begin{array}{cc}4& 6\\ 7& 10\end{array}\right)=-(40-42)=-(-2)=2.$ $C_{13}=(+)\det \left(\begin{array}{cc}4& 5\\ 7& 8\end{array}\right)=32-35=-3.$ $\det A=1(2)+2(2)+3(-3)=2+4-9=-3.$

Row 1 kyun choose karein: ismein do zeros hain, isliye teen mein se do cofactor terms turant khatam ho jaate hain. Sirf $a_{13}=4$ bachta hai. $\det =4\cdot C_{13}=4\cdot (-1{)}^{1+3}\det \left(\begin{array}{cc}1& 2\\ 5& 6\end{array}\right)=4\cdot (+)(6-10)=4\cdot (-4)=-16.$

Column 1 hai $(0,1,5{)}^T$; top ka zero drop ho jaata hai. $\det =1\cdot C_{21}+5\cdot C_{31}.$ $C_{21}=(-1{)}^{2+1}\det \left(\begin{array}{cc}0& 4\\ 6& 7\end{array}\right)=-(0-24)=24.$ $C_{31}=(-1{)}^{3+1}\det \left(\begin{array}{cc}0& 4\\ 2& 3\end{array}\right)=+(0-8)=-8.$ $\det =1(24)+5(-8)=24-40=-16.✓$ Same value — "kisi bhi row/column ke saath expand karo" theorem action mein.

Row-add free kyun hai: ek row ka multiple doosri row mein add karne se $\det$ unchanged rehta hai. Top entry ke neeche pehle column ko khatam karo. $R_2\to R_2-2R_1$ aur $R_3\to R_3-3R_1$ karo: $\to \left(\begin{array}{ccc}2& 1& 3\\ 0& 1& 2\\ 0& 2& 4\end{array}\right).$ Column 1 ke neeche expand karo (sirf top entry $2$ bachti hai): $\det B=2\cdot (-1{)}^{1+1}\det \left(\begin{array}{cc}1& 2\\ 2& 4\end{array}\right)=2\cdot (4-4)=2\cdot 0=0.$ $\det B=0$ — matrix singular hai (iske rows dependent hain: $R_3=R_1+R_2$).

Column 1 ke saath expand karo (bottom slot mein zero hai): $\det =k\cdot (-1{)}^{1+1}\det \left(\begin{array}{cc}k& 1\\ 1& k\end{array}\right)+1\cdot (-1{)}^{2+1}\det \left(\begin{array}{cc}2& 0\\ 1& k\end{array}\right).$ $=k(k^2-1)-(2k-0)=k^3-k-2k=k^3-3k.$ Zero set karo: $k^3-3k=k(k^2-3)=0$, to $k=0,k=\sqrt{3},k=-\sqrt{3}.$ In $k$ values par matrix space ko collapse kar deta hai → non-invertible.

Rows kyun subtract karein: $R_2\to R_2-R_1$, $R_3\to R_3-R_1$ karne se column 1 mein top ke neeche zeros aa jaate hain aur clean factors milte hain: $\to \left(\begin{array}{ccc}1& x& x^2\\ 0& y-x& y^2-x^2\\ 0& z-x& z^2-x^2\end{array}\right).$ Column 1 ke neeche expand karo (sirf top entry $1$ bachti hai): $\det A=1\cdot \det \left(\begin{array}{cc}y-x& y^2-x^2\\ z-x& z^2-x^2\end{array}\right).$ Har entry factor karo: $y^2-x^2=(y-x)(y+x)$, $z^2-x^2=(z-x)(z+x)$. Row 1 se $(y-x)$ aur row 2 se $(z-x)$ baahar nikalo: $=(y-x)(z-x)\det \left(\begin{array}{cc}1& y+x\\ 1& z+x\end{array}\right)=(y-x)(z-x)[(z+x)-(y+x)].$ $=(y-x)(z-x)(z-y).$

$\det (2A)$: poore $3\times 3$ matrix ko $2$ se scale karne par teeno 3 rows mein se har ek $2$ se scale hoti hai, aur har row-scaling $\det$ mein $2$ multiply karti hai. To $\det (2A)=2^3\det A=8\cdot 7=56$. $\det (A^T)$: Leibniz sum transpose ke under invariant hai, isliye $\det (A^T)=\det A=7$. (General rule: $\det (kA)=k^n\det A$ ek $n\times n$ matrix ke liye.)

$2\times 2$ ke liye, har minor ek akela bacha hua entry hota hai. $C_{11}=+(4)=4,C_{12}=-(3)=-3,C_{21}=-(2)=-2,C_{22}=+(1)=1.$ Adjugate cofactor matrix ka transpose hota hai: $\mathrm{adj}(A)=\left(\begin{array}{cc}{C}_{11}& C_{21}\\ C_{12}& C_{22}\end{array}\right)=\left(\begin{array}{cc}4& -2\\ -3& 1\end{array}\right).$ $\det A=1\cdot 4-2\cdot 3=-2$ ke saath: $A^{-1}=\frac{1}{\det A}\mathrm{adj}(A)=\frac{1}{-2}\left(\begin{array}{cc}4& -2\\ -3& 1\end{array}\right)=\left(\begin{array}{cc}-2& 1\\ \frac{3}{2}& -\frac{1}{2}\end{array}\right).$ Yeh Adjugate matrix and inverse se connect karta hai.

Coefficient matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 3\end{array}\right)$, $\det A=6-1=5$. Column 1 (woh $x$-column) ko right-hand side $(5,10{)}^T$ se replace karo: $A_x=\left(\begin{array}{cc}5& 1\\ 10& 3\end{array}\right),\det A_x=15-10=5.$ $x=\frac{\det A_x}{\det A}=\frac{5}{5}=1.$ (Phir doosri equation se $y=(10-1)/3=3$; check karo $2(1)+3=5$.✓) Dekho Cramer's rule.

Maano rows $2$ aur $3$ equal hain. Un do identical rows ko swap karna matrix ko visibly unchanged chhodta hai, isliye uska determinant unchanged hai: $\det A=\det A$. Lekin ek single row-swap determinant mein $-1$ multiply karta hai (ek alternation property): $\det A=-\det A$. Dono sides mein $\det A$ add karo: $2\det A=0\Rightarrow \det A=0$. Cofactor cross-check: $R_3-R_2$ subtract karo (free) taaki row 3 poori zeros ho jaaye; us zero row ke saath expand karne par milta hai ${\sum }_{j}0\cdot C_{3j}=0$. Dono raaste agree karte hain.

Pehle row-2 cofactors: $C_{21}=(-1{)}^3\det \left(\begin{array}{cc}2& 3\\ 8& 10\end{array}\right)=-(20-24)=4,$ $C_{22}=(-1{)}^4\det \left(\begin{array}{cc}1& 3\\ 7& 10\end{array}\right)=(10-21)=-11,$ $C_{23}=(-1{)}^5\det \left(\begin{array}{cc}1& 2\\ 7& 8\end{array}\right)=-(8-14)=6.$ Ab inhe row 1 $=(1,2,3)$ ke saath dot karo: $1(4)+2(-11)+3(6)=4-22+18=0.✓$ Zero kyun: yeh expression us matrix ka determinant hai jiska rows 1 aur 2 dono $A$ ki row 1 hain — ek repeated row, isliye L5.1 ke anusaar $\det =0$. Yeh "alien cofactor = 0" fact exactly wahi hai jo $A\cdot \mathrm{adj}(A)=(\det A)I$ ko kaam karaat hai.

Column 1 hai $(a_{11},0,0{)}^T$ — wahan expand karo. Sirf top term bachta hai: $\det =a_{11}\cdot (-1{)}^{1+1}\det \left(\begin{array}{cc}{a}_{22}& a_{23}\\ 0& a_{33}\end{array}\right).$ Andar ka $2\times 2$ hai $a_{22}{a}_{33}-a_{23}\cdot 0=a_{22}{a}_{33}$. To $\det =a_{11}{a}_{22}{a}_{33}$ — diagonal product. Numerically: $3\cdot 5\cdot 4=60$. Har step ne un zeros ko "chase" kiya jo triangularity ne hame free mein de diye.