Exercises — Inverse of 2×2 matrix

Hum kya karenge: $\det A=ad-bc$ apply karo $a=3,b=2,c=5,d=4$ ke saath. Kyun: invertibility poori tarah determinant se decide hoti hai — yes/no question ka jawab dene ke liye inverse banana zaroori nahi. $\det A=(3)(4)-(2)(5)=12-10=2.$ Kyunki $\det A=2\ne 0$, $A$ invertible hai (non-singular).

Hum kya karenge: har determinant compute karo; singular matlab woh $0$ ke barabar ho.

• $\det P=(1)(4)-(2)(2)=4-4=0$ → singular (row 2, row 1 ka $2\times$ hai).
• $\det Q=(6)(2)-(3)(4)=12-12=0$ → singular (row 1, row 2 ka $1.5\times$ hai).
• $\det R=(0)(0)-(5)(5)=0-25=-25\ne 0$ → invertible (negative determinant theek hai — iska matlab sirf itna hai ki transform orientation flip karta hai, dekho Linear transformations and area). $R$ log ko kyun surprised karta hai: diagonal par do zeros hona degenerate lagta hai, lekin degeneracy parallel columns ke baare mein hai, na ki zeros kisi bhi jagah hone ke baare mein.

Step 1 — determinant (hamesha pehle). $\det A=(4)(2)-(3)(3)=8-9=-1$. Non-zero ⟹ aage badho. Kyun pehle? Poori recipe $\det A$ se divide karti hai; agar woh $0$ hota to inverse exist hi nahi karta aur baad ke har step waste ho jaate (zero se division). Ise pehle compute karna dead end pakadne ka sabse sasta tarika hai. Step 2 — adjugate. Corners $4\leftrightarrow 2$ swap karo, middle $3\to -3$ flip karo (yahi upar define kiya hua $\mathrm{adj}A$ hai): $\mathrm{adj}A=\left(\begin{array}{cc}2& -3\\ -3& 4\end{array}\right).$ Step 3 — det se divide karo. $A^{-1}=\frac{1}{-1}\left(\begin{array}{cc}2& -3\\ -3& 4\end{array}\right)=\left(\begin{array}{cc}-2& 3\\ 3& -4\end{array}\right).$ Step 4 — verify (sasti insurance): $\left(\begin{array}{cc}4& 3\\ 3& 2\end{array}\right)\left(\begin{array}{cc}-2& 3\\ 3& -4\end{array}\right)=\left(\begin{array}{cc}-8+9& 12-12\\ -6+6& 9-8\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=I$ ✓.

Step 1 — pehle determinant. $\det B=(1)(5)-(2)(3)=5-6=-1$ (non-zero ⟹ inverse exist karta hai). Step 2 — adjugate phir divide. $B^{-1}=\frac{1}{-1}\left(\begin{array}{cc}5& -2\\ -3& 1\end{array}\right)=\left(\begin{array}{cc}-5& 2\\ 3& -1\end{array}\right).$ Step 3 — solve karo. $B\vec{x}=\vec{b}$ ko left-multiply karo $B^{-1}$ se taaki $\vec{x}=B^{-1}\vec{b}$ mile (dekho Solving linear systems): $\vec{x}=\left(\begin{array}{cc}-5& 2\\ 3& -1\end{array}\right)\left(\genfrac{}{}{0ex}{}{4}{7}\right)=\left(\genfrac{}{}{0ex}{}{(-5)(4)+2(7)}{3(4)+(-1)(7)}\right)=\left(\genfrac{}{}{0ex}{}{-20+14}{12-7}\right)=\left(\genfrac{}{}{0ex}{}{-6}{5}\right).$ Check: $1(-6)+2(5)=-6+10=4$ ✓, $3(-6)+5(5)=-18+25=7$ ✓. To $x=-6,y=5$.

$\det A=(2)(3)-(0)(1)=6$. $A^{-1}=\frac{1}{6}\left(\begin{array}{cc}3& 0\\ -1& 2\end{array}\right)=\left(\begin{array}{cc}1/2& 0\\ -1/6& 1/3\end{array}\right).$ $\det (A^{-1})=\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)-\left(0\right)\left(-\frac{1}{6}\right)=\frac{1}{6}.$ Relationship: $\det (A^{-1})=\frac{1}{\det A}$. Kyun yeh sach hona hi chahiye (area picture): $A$ har area ko $\det A=6$ factor se scale karta hai. Undo-machine $A^{-1}$ ko area ko reciprocal $\frac{1}{6}$ se scale karna chahiye taaki dono karne par original area wapas aaye — dekho Linear transformations and area.

Singular condition: $\det M=k\cdot k-2\cdot 8=k^2-16=0\Rightarrow k=\pm 4.$ To $M$ ka koi inverse nahi hoga exactly jab $==k=4==$ ya $==k=-4==$ (do rows proportional ho jaati hain). Otherwise ($k\ne \pm 4$): $M^{-1}=\frac{1}{k^2-16}\left(\begin{array}{cc}k& -2\\ -8& k\end{array}\right).$ Sanity check $k=6$ par: $\det =36-16=20$, $M^{-1}=\frac{1}{20}\left(\begin{array}{cc}6& -2\\ -8& 6\end{array}\right)$; wapas multiply karo $I$ confirm karne ke liye.

Proof. Yeh dikhane ke liye ki matrix $X$, $AB$ ka inverse hai, yeh kaafi hai ki $(AB)X=I$ dikhao (dekho Identity matrix). $X=B^{-1}{A}^{-1}$ lo: $(AB)(B^{-1}{A}^{-1})=A(B{B}^{-1})A^{-1}=AI{A}^{-1}=A{A}^{-1}=I.$ Humne associativity use ki regroup karne ke liye, phir $B{B}^{-1}=I$ aur $I$ multiplication ke under disappear ho jaata hai. To $(AB{)}^{-1}=B^{-1}{A}^{-1}$. Order kyun flip hota hai (glove picture): "pehle socks, phir shoes" ko undo karne ke liye tumhe "pehle shoes utaro, phir socks" karna hoga — pehle last operation reverse karo. Numeric check. $AB=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)\left(\begin{array}{cc}2& 0\\ 0& 2\end{array}\right)=\left(\begin{array}{cc}2& 2\\ 0& 2\end{array}\right)$, $\det =4$, to $(AB{)}^{-1}=\frac{1}{4}\left(\begin{array}{cc}2& -2\\ 0& 2\end{array}\right)=\left(\begin{array}{cc}1/2& -1/2\\ 0& 1/2\end{array}\right).$ Alag se $A^{-1}=\left(\begin{array}{cc}1& -1\\ 0& 1\end{array}\right)$, $B^{-1}=\left(\begin{array}{cc}1/2& 0\\ 0& 1/2\end{array}\right)$, aur $B^{-1}{A}^{-1}=\left(\begin{array}{cc}1/2& 0\\ 0& 1/2\end{array}\right)\left(\begin{array}{cc}1& -1\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}1/2& -1/2\\ 0& 1/2\end{array}\right).$ Match ✓.

Key idea: $(A^{-1}{)}^{-1}=A$ — undo ka undo original return karta hai. To bas di gayi matrix ko invert karo. Maano $C=A^{-1}$. $\det C=(2)(2)-(-1)(-3)=4-3=1.$ $A=C^{-1}=\frac{1}{1}\left(\begin{array}{cc}2& 1\\ 3& 2\end{array}\right)=\left(\begin{array}{cc}2& 1\\ 3& 2\end{array}\right).$ Check: $A{A}^{-1}=\left(\begin{array}{cc}2& 1\\ 3& 2\end{array}\right)\left(\begin{array}{cc}2& -1\\ -3& 2\end{array}\right)=\left(\begin{array}{cc}4-3& -2+2\\ 6-6& -3+4\end{array}\right)=I$ ✓.

Step 1 — $A^2$ compute karo $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ ke liye: $A^2=\left(\begin{array}{cc}{a}^2+bc& ab+bd\\ ca+dc& cb+d^2\end{array}\right)=\left(\begin{array}{cc}{a}^2+bc& b(a+d)\\ c(a+d)& d^2+bc\end{array}\right).$ Step 2 — $(a+d)A$ subtract karo: $A^2-(a+d)A=\left(\begin{array}{cc}{a}^2+bc-a(a+d)& 0\\ 0& d^2+bc-d(a+d)\end{array}\right)=\left(\begin{array}{cc}bc-ad& 0\\ 0& bc-ad\end{array}\right)=-(\det A)I.$ Step 3 — rearrange karo: $A^2-(a+d)A+(\det A)I=O$ ✓ (yeh identity har $2\times 2$ matrix ke liye hold karti hai). Step 4 — inverse extract karo. Maano $\det A\ne 0$. Pehle do terms mein se $A$ factor out karo: $A(A-(a+d)I)=-(\det A)I\Rightarrow A\cdot \frac{(a+d)I-A}{\det A}=I.$ Isliye $A^{-1}=\frac{(a+d)I-A}{\det A}=\frac{1}{\det A}\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right),$ jo bilkul adjugate formula hai — pure algebra se recover hua, koi hand-swapping nahi. Yeh beautiful kyun hai: wahi swap/negate pattern jo parent note ne chaar equations solve karke nikala tha, woh ek polynomial identity se nikal aata hai.

Step 1 — determinant. $\det A_{\epsilon }=(1)(1+\epsilon )-(1)(1)=1+\epsilon -1=\epsilon .$ Step 2 — inverse (corners swap, middle flip, $\epsilon$ se divide karo): $A_{\epsilon }^{-1}=\frac{1}{\epsilon }\left(\begin{array}{cc}1+\epsilon & -1\\ -1& 1\end{array}\right)=\left(\begin{array}{cc}\frac{1+\epsilon }{\epsilon }& -\frac{1}{\epsilon }\\ -\frac{1}{\epsilon }& \frac{1}{\epsilon }\end{array}\right).$ Step 3 — limiting behaviour. Jaise $\epsilon \to 0^{+}$, divisor $\epsilon$ zero ki taraf shrink hota hai, to har entry $\frac{1}{\epsilon },\frac{1+\epsilon }{\epsilon }\to +\infty$ (aur $-\frac{1}{\epsilon }\to -\infty$): inverse bina bound ke blow up ho jaata hai. Step 4 — singular limit. Exactly $\epsilon =0$ par hamare paas $A_0=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)$ hai jiska $\det A_0=0$ hai aur do identical (isliye parallel) columns hain, to $A_0$ singular hai: koi inverse exist hi nahi karta. Step 3 mein blow-up algebra ka woh warning hai, jaise tum us wall ke paas aate ho, ki undo-machine impossible hone wali hai. The story (edge picture): near-singular matrices ke inverses huge, unstable hote hain — input data mein thodi si wobble $1/\epsilon$ factor se amplify ho jaati hai. Wall $\epsilon =0$ par area ek line mein collapse ho jaata hai aur inverse bilkul exist karna band kar deta hai. Isliye tiny determinant se divide karna numerically dangerous hai — dekho Singular vs non-singular matrices.

Neeche wali figure exactly is collapse ko plot karti hai.