Exercises — Solving systems using matrix inversion

KYA karte hain: numbers seedha equations se padhte hain. Coefficient matrix unknowns ko multiply karne waale numbers collect karta hai, row by row; constant matrix right-hand sides collect karta hai. $A=\left[\begin{array}{cc}3& 2\\ 4& -1\end{array}\right],X=\left[\begin{array}{c}x\\ y\end{array}\right],B=\left[\begin{array}{c}7\\ 1\end{array}\right].$ Check karo ki system reproduce hoti hai: $A$ ki row 1 times $X$ deta hai $3x+2y$, jo $B$ ki pehli entry $7$ ke barabar set hoti hai — exactly equation one. ✓

$\det A=(5)(1)-(2)(3)=5-6=-1$. Kyunki $\det A=-1\ne 0$, machine reversible hai, isliye $A^{-1}$ exist karta hai.

$\det A=(1)(-2)-(-1)(2)=-2+2=0$. Gate closed hai ($\det A=0$), isliye $A^{-1}$ exist nahi karta aur $X=A^{-1}B$ use nahi ho sakta. Hume consistency branch par fall back karna hoga.

$A=\left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right],B=\left[\begin{array}{c}5\\ 10\end{array}\right].$ Gate: $\det A=(2)(4)-(1)(3)=8-3=5\ne 0$ — invert karna safe hai. Inverse. Yeh moves kyun? $2\times 2$ formula $\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$ kehta hai: do diagonal entries swap karo ($a\leftrightarrow d$), do off-diagonal entries negate karo ($b,c$), phir $\det$ se divide karo. Har move "transpose the cofactor matrix and divide by the determinant" ka $2\times 2$ special case hai. $A^{-1}=\frac{1}{5}\left[\begin{array}{cc}4& -1\\ -3& 2\end{array}\right].$ Multiply ($X=A^{-1}B$, $B$ ko right par rakhte hain kyunki $X=A^{-1}B$ mein $B$ right par hai): $X=\frac{1}{5}\left[\begin{array}{cc}4& -1\\ -3& 2\end{array}\right]\left[\begin{array}{c}5\\ 10\end{array}\right]=\frac{1}{5}\left[\begin{array}{c}20-10\\ -15+20\end{array}\right]=\frac{1}{5}\left[\begin{array}{c}10\\ 5\end{array}\right]=\left[\begin{array}{c}2\\ 1\end{array}\right].$ Toh $x=2,y=1$. Check: $2(2)+1=5$ ✓, $3(2)+4(1)=10$ ✓.

$A=\left[\begin{array}{cc}1& 2\\ 3& -1\end{array}\right],B=\left[\begin{array}{c}4\\ 5\end{array}\right].$ Gate: $\det A=(1)(-1)-(2)(3)=-1-6=-7\ne 0$. Inverse. Yeh moves kyun? $2\times 2$ inverse formula $\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$ bas "transpose the cofactor matrix, divide by $\det$" ka miniature version hai: ek $2\times 2$ ke liye, $a$ ka cofactor $+d$ hai, $b$ ka $-c$ hai, $c$ ka $-b$ hai, $d$ ka $+a$ hai; us grid ko transpose karne se $d$ aur $a$ diagonal par aa jaate hain (swapped) aur $-b,-c$ off it (negated). Toh diagonal $1\leftrightarrow -1$ swap karo, off-diagonal $2,3$ negate karo, aur $-7$ se divide karo: $A^{-1}=\frac{1}{-7}\left[\begin{array}{cc}-1& -2\\ -3& 1\end{array}\right].$ $X=\frac{1}{-7}\left[\begin{array}{cc}-1& -2\\ -3& 1\end{array}\right]\left[\begin{array}{c}4\\ 5\end{array}\right]=\frac{1}{-7}\left[\begin{array}{c}-4-10\\ -12+5\end{array}\right]=\frac{1}{-7}\left[\begin{array}{c}-14\\ -7\end{array}\right]=\left[\begin{array}{c}2\\ 1\end{array}\right].$ Toh $x=2,y=1$. Check: $2+2(1)=4$ ✓, $3(2)-1=5$ ✓.

$A=\left[\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 1& 0& 1\end{array}\right],B=\left[\begin{array}{c}3\\ 5\\ 4\end{array}\right].$ Gate — $\det A$ ko row 1 ke along expand karo. Row 1 kyun? Koi bhi row (ya column) same determinant deta hai, isliye hum sabse convenient numbers wali row choose karte hain; yahan row 1 $(1,1,0)$ hai aur woh $0$ ek poora term kill kar deta hai, kaam bachata hai. Row 1 par sign pattern $+,-,+$ lagaate hue: $\det A=1(1\cdot 1-1\cdot 0)-1(0\cdot 1-1\cdot 1)+0=1(1)-1(-1)+0=2\ne 0.$ Cofactor matrix, phir adjugate. Har cofactor $(-1{)}^{i+j}$ times minor hai (apni row aur column delete karo, $2\times 2$ determinant lo). Nau sab karke transpose karte hue: $\mathrm{adj}A=\left[\begin{array}{ccc}1& -1& 1\\ 1& 1& -1\\ -1& 1& 1\end{array}\right].$ Phir $A^{-1}=\frac{1}{2}\mathrm{adj}A$ aur $X=\frac{1}{2}\left[\begin{array}{ccc}1& -1& 1\\ 1& 1& -1\\ -1& 1& 1\end{array}\right]\left[\begin{array}{c}3\\ 5\\ 4\end{array}\right]=\frac{1}{2}\left[\begin{array}{c}3-5+4\\ 3+5-4\\ -3+5+4\end{array}\right]=\frac{1}{2}\left[\begin{array}{c}2\\ 4\\ 6\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right].$ Toh $x=1,y=2,z=3$. Check: $1+2=3$ ✓, $2+3=5$ ✓, $1+3=4$ ✓.

$A=\left[\begin{array}{cc}1& -2\\ 2& 1\end{array}\right],B=\left[\begin{array}{c}3\\ 4\end{array}\right]$. $\det A=(1)(1)-(-2)(2)=1+4=5\ne 0$ — gate open. Non-zero determinant ka matlab hai machine reversible hai, isliye exactly ek solution hai — unique case. Do lines parallel nahi hain, isliye ek single point par cross karti hain (left panel mein orange dot). Completeness ke liye, solve karo: $\mathrm{adj}A=\left[\begin{array}{cc}1& 2\\ -2& 1\end{array}\right]$, toh $X=\frac{1}{5}\left[\begin{array}{cc}1& 2\\ -2& 1\end{array}\right]\left[\begin{array}{c}3\\ 4\end{array}\right]=\frac{1}{5}\left[\begin{array}{c}3+8\\ -6+4\end{array}\right]=\frac{1}{5}\left[\begin{array}{c}11\\ -2\end{array}\right]=\left[\begin{array}{c}11/5\\ -2/5\end{array}\right].$ Unique solution $(x,y)=(11/5,-2/5)$. Check: $11/5-2(-2/5)=11/5+4/5=15/5=3$ ✓.

$A=\left[\begin{array}{cc}1& -2\\ 2& -4\end{array}\right],B=\left[\begin{array}{c}3\\ 6\end{array}\right]$. $\det A=(1)(-4)-(-2)(2)=-4+4=0$ — gate closed. Ab $(\mathrm{adj}A)B$ test karo. Ek $2\times 2$ ke liye, $\mathrm{adj}A=\left[\begin{array}{cc}-4& 2\\ -2& 1\end{array}\right]$ (diagonal swap karo, off-diagonal negate karo — $2\times 2$ adjugate). $(\mathrm{adj}A)B=\left[\begin{array}{cc}-4& 2\\ -2& 1\end{array}\right]\left[\begin{array}{c}3\\ 6\end{array}\right]=\left[\begin{array}{c}-12+12\\ -6+6\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]=O.$ Yahan $O$ woh $2\times 1$ zero vector hai. $\det A=0$ aur $(\mathrm{adj}A)B=O$ ⇒ infinitely many solutions. Sanity check: doosri equation exactly pehli ki double hai, toh woh ek hi line hain — middle panel.

Same $A$ jaise Ex 3.2, toh $\det A=0$. Ab $B=\left[\begin{array}{c}3\\ 7\end{array}\right]$: $(\mathrm{adj}A)B=\left[\begin{array}{cc}-4& 2\\ -2& 1\end{array}\right]\left[\begin{array}{c}3\\ 7\end{array}\right]=\left[\begin{array}{c}-12+14\\ -6+7\end{array}\right]=\left[\begin{array}{c}2\\ 1\end{array}\right]\ne O.$ $\det A=0$ aur $(\mathrm{adj}A)B\ne O$ ⇒ no solution (inconsistent). Geometry: do parallel lekin alag lines — "3" vs "7" ek line ko doosri se push kar deta hai, right panel.

$A=\left[\begin{array}{cc}3& k\\ 1& 2\end{array}\right]$. Unique solution exactly tab fail hota hai jab $\det A=0$: $\det A=(3)(2)-(k)(1)=6-k=0\Rightarrow k=6.$ $k=6$ par: $A=\left[\begin{array}{cc}3& 6\\ 1& 2\end{array}\right]$, $B=\left[\begin{array}{c}6\\ 4\end{array}\right]$, $\mathrm{adj}A=\left[\begin{array}{cc}2& -6\\ -1& 3\end{array}\right]$. $(\mathrm{adj}A)B=\left[\begin{array}{cc}2& -6\\ -1& 3\end{array}\right]\left[\begin{array}{c}6\\ 4\end{array}\right]=\left[\begin{array}{c}12-24\\ -6+12\end{array}\right]=\left[\begin{array}{c}-12\\ 6\end{array}\right]\ne O.$ Toh $k=6$ par system ka no solution hai. (Equation one ban jaati hai $3x+6y=6\Rightarrow x+2y=2$, jo $x+2y=4$ se contradict karti hai.)

KYA/KYUN: hume code $X$ aur rule $A$ pata hai; forward machine output deta hai $B=AX$. Yahan koi inverse ki zaroorat nahi — inversion sirf backwards jaane ke liye hai. $B=AX=\left[\begin{array}{cc}2& 3\\ 1& -1\end{array}\right]\left[\begin{array}{c}8/5\\ 3/5\end{array}\right]=\left[\begin{array}{c}16/5+9/5\\ 8/5-3/5\end{array}\right]=\left[\begin{array}{c}25/5\\ 5/5\end{array}\right]=\left[\begin{array}{c}5\\ 1\end{array}\right].$ Toh $B=\left[\begin{array}{c}5\\ 1\end{array}\right]$. Yeh round trip confirm karta hai $B\to X\to AX=B$: $B=\left[\begin{array}{c}5\\ 1\end{array}\right]$ ko reverse machine $A^{-1}$ mein daalo toh $X=\left[\begin{array}{c}8/5\\ 3/5\end{array}\right]$ milta hai, aur forward machine $A$ $B$ return karta hai — exactly system $2x+3y=5,x-y=1$.

Saath kyun karte hain: expensive step $A^{-1}$ find karna hai; ek baar mil jaaye toh har right-hand side sirf ek multiplication hai. Yahi efficiency reason hai ki inversion elimination baar baar karne se better hai. Ex 2.1 se, $\det A=5$ aur $A^{-1}=\frac{1}{5}\left[\begin{array}{cc}4& -1\\ -3& 2\end{array}\right]$. $X_1=A^{-1}{B}_1=\frac{1}{5}\left[\begin{array}{cc}4& -1\\ -3& 2\end{array}\right]\left[\begin{array}{c}5\\ 10\end{array}\right]=\frac{1}{5}\left[\begin{array}{c}10\\ 5\end{array}\right]=\left[\begin{array}{c}2\\ 1\end{array}\right].$ $X_2=A^{-1}{B}_2=\frac{1}{5}\left[\begin{array}{cc}4& -1\\ -3& 2\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]=\frac{1}{5}\left[\begin{array}{c}4\\ -3\end{array}\right]=\left[\begin{array}{c}4/5\\ -3/5\end{array}\right].$ $X_2$ check karo: $2(4/5)+(-3/5)=8/5-3/5=1$ ✓, $3(4/5)+4(-3/5)=12/5-12/5=0$ ✓.

Teeno variables dikhate hue rewrite karo: $2x+1y+0z=3,1x+0y+1z=2,0x+1y-1z=1$. $A=\left[\begin{array}{ccc}2& 1& 0\\ 1& 0& 1\\ 0& 1& -1\end{array}\right],B=\left[\begin{array}{c}3\\ 2\\ 1\end{array}\right].$ Gate — row 1 ke along expand karo (choose kiya kyunki column 3 mein iska $0$ ek term drop karta hai). Signs $+,-,+$ ke saath: $\det A=2(0\cdot (-1)-1\cdot 1)-1(1\cdot (-1)-1\cdot 0)+0=2(-1)-1(-1)=-2+1=-1\ne 0.$ Adjugate — transpose kyun? KYA karte hain: saare nau cofactors $C_{ij}=(-1{)}^{i+j}{M}_{ij}$ compute karo cofactor grid $[C_{ij}]$ banane ke liye, phir use transpose karo (rows columns ban jaate hain) taaki $\mathrm{adj}A=[C_{ji}]$ mile. Transpose kyun: sirf transposed grid hi $A\mathrm{adj}(A)=(\det A)I$ satisfy karta hai, woh identity jo $\frac{1}{\det A}\mathrm{adj}(A)$ ko actually $A$ undo karaati hai; ise skip karna $3\times 3$ mein sabse common slip hai. Yahan carry out karte hue: $\mathrm{adj}A=\left[\begin{array}{ccc}-1& 1& 1\\ 1& -2& -2\\ 1& -2& -1\end{array}\right].$ Phir $A^{-1}=\frac{1}{-1}\mathrm{adj}A=-\mathrm{adj}A$, aur $X=A^{-1}B=-\left[\begin{array}{ccc}-1& 1& 1\\ 1& -2& -2\\ 1& -2& -1\end{array}\right]\left[\begin{array}{c}3\\ 2\\ 1\end{array}\right]=-\left[\begin{array}{c}-3+2+1\\ 3-4-2\\ 3-4-1\end{array}\right]=-\left[\begin{array}{c}0\\ -3\\ -2\end{array}\right]=\left[\begin{array}{c}0\\ 3\\ 2\end{array}\right].$ Toh $x=0,y=3,z=2$. Check: $2(0)+3=3$ ✓, $0+2=2$ ✓, $3-2=1$ ✓.

$A=\left[\begin{array}{ccc}1& 2& 3\\ 2& 5& 1\\ 1& 1& 1\end{array}\right],B=\left[\begin{array}{c}14\\ 15\\ 6\end{array}\right].$ Gate — row 1 ke along expand karo (saari entries small hain, koi obvious zero nahi, toh koi bhi row chalega; row 1 arithmetic light rakhti hai). Signs $+,-,+$ ke saath: $\det A=1(5\cdot 1-1\cdot 1)-2(2\cdot 1-1\cdot 1)+3(2\cdot 1-5\cdot 1)$ $=1(4)-2(1)+3(-3)=4-2-9=-7\ne 0.$ Non-zero ⇒ unique price list exist karti hai, invert karna safe hai. Cofactor matrix, phir adjugate ke liye transpose. Yaad karo har cofactor sign $(-1{)}^{i+j}$ carry karta hai: $\mathrm{adj}A=\left[\begin{array}{ccc}4& 1& -13\\ -1& -2& 5\\ -3& 1& 1\end{array}\right].$ (Ek corner verify karo: $C_{11}=(-1{)}^{1+1}\det \left[\begin{array}{cc}5& 1\\ 1& 1\end{array}\right]=+4$; yeh adjugate ki position $(1,1)$ par baithta hai.) Phir $A^{-1}=\frac{1}{-7}\mathrm{adj}A$ aur $X=A^{-1}B=\frac{1}{-7}\left[\begin{array}{ccc}4& 1& -13\\ -1& -2& 5\\ -3& 1& 1\end{array}\right]\left[\begin{array}{c}14\\ 15\\ 6\end{array}\right].$ Divide karne se pehle product compute karo: $\left[\begin{array}{c}4(14)+1(15)-13(6)\\ -1(14)-2(15)+5(6)\\ -3(14)+1(15)+1(6)\end{array}\right]=\left[\begin{array}{c}56+15-78\\ -14-30+30\\ -42+15+6\end{array}\right]=\left[\begin{array}{c}-7\\ -14\\ -21\end{array}\right].$ $X=\frac{1}{-7}\left[\begin{array}{c}-7\\ -14\\ -21\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right].$ Prices $x=1,y=2,z=3$ — saare non-negative, physically sensible. Check: $1+2(2)+3(3)=1+4+9=14$ ✓, $2(1)+5(2)+3=15$ ✓, $1+2+3=6$ ✓.

$A=\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 2& \lambda \end{array}\right]$, $B=\left[\begin{array}{c}6\\ 10\\ \mu \end{array}\right]$. Gate — $\det A$ row 1 ke along expand karo (signs $+,-,+$): $\det A=1(2\lambda -6)-1(\lambda -3)+1(2-2)=2\lambda -6-\lambda +3+0=\lambda -3.$ (a) Unique. $\det A\ne 0\Rightarrow \lambda \ne 3$. Kisi bhi $\mu$ ke liye, ek unique solution exist karta hai.

Degenerate branch: $\lambda =3$. $\lambda =3$ set karo; phir $A$ ki row 3 $(1,2,3)$ ban jaati hai, jo exactly row 2 jaisi hai. Toh last do equations ka same left side $x+2y+3z$ hai: $x+2y+3z=10\text{aur}x+2y+3z=\mu .$ Yeh dono tab hi hold kar sakte hain jab $\mu =10$ ho. Yeh branch saaf split ho jaata hai:

(b) No solution. Agar $\lambda =3$ aur $\mu \ne 10$, toh do equations $x+2y+3z$ ko ek saath do alag numbers ke barabar demand karti hain — flat contradiction. Koi $X$ exist nahi karta. (Consistency test se confirm karo: $\lambda =3$ par, $\det A=0$, aur check karo ki $(\mathrm{adj}A)B\ne O$ precisely tab jab $\mu \ne 10$.)

(c) Infinitely many. Agar $\lambda =3$ aur $\mu =10$, toh teesri equation doosri ki exact copy hai — yeh koi nayi information nahi deti. Teen unknowns mein genuinely do independent equations rah jaate hain, ek free parameter chodke, toh infinitely many solutions hote hain. (Consistency test: $\det A=0$ aur $(\mathrm{adj}A)B=O$.)

Summary: $\lambda \ne 3$ ke liye unique (koi bhi $\mu$); $\lambda =3,\mu \ne 10$ ke liye no solution; $\lambda =3,\mu =10$ ke liye infinitely many.