Exercises — Systems of linear equations — matrix form Ax = b

Hum kya karte hain: coefficient matrix ke determinant ke zariye dono lines ki slopes compare karo. $A=\left[\begin{array}{cc}1& 2\\ 3& -1\end{array}\right],\det A=(1)(-1)-(2)(3)=-1-6=-7.$ Determinant kyun? Ek square $2\times 2$ system ke liye, $\det A\ne 0$ ka matlab hai columns genuinely alag directions mein point karte hain, toh transformation $x\mapsto Ax$ har target ko exactly ek baar hit karta hai — ek unique solution (parent note: invertible-matrix shortcut). Answer: $\det A=-7\ne 0\Rightarrow$ unique solution. Upar diye figure ka left picture.

Hum kya karte hain: check karo ki kya ek equation doosri ki scalar multiple hai. Row 1 exactly $2\times$ Row 2 hai (LHS: $2x+4y=2(x+2y)$; RHS: $6=2\cdot 3$). Toh $\det A=(2)(2)-(4)(1)=0$, aur augmented right-hand side bhi agree karti hai. Yeh kyun matter karta hai: $\det A=0$ invertible shortcut ko khatam kar deta hai, toh hum rank par aate hain. Yahan $\mathrm{rank}(A)=1=\mathrm{rank}([A|b])$, lekin $n=2>1$. Answer: infinitely many solutions — ek hi line do baar bani (middle picture).

Hum kya karte hain: left side same hai, right side alag. Left sides se $\mathrm{rank}(A)=1$ milta hai; lekin augmented matrix $[A|b]=\left[\begin{array}{cccc}1& -1& |& 1\\ 1& -1& |& 4\end{array}\right]$ ke do independent rows hain ($1\ne 4$ ek contradiction force karta hai), toh $\mathrm{rank}([A|b])=2$. Yeh "no solution" kyun hai: augmented matrix ka rank $A$ ke rank se upar jump kar gaya. Right-hand side $b$ ek aisi direction maangta hai jo columns supply nahi kar sakte — $b\notin \mathrm{Col}(A)$. Answer: no solution — parallel lines (right picture).

L1.1 se, $\det A=-7\ne 0$, toh inverse exist karta hai. Step 1 — $A^{-1}$ banao. $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ ke liye inverse hai $\frac{1}{\det }\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$ (diagonal swap karo, off-diagonal negate karo, determinant se divide karo). $A^{-1}=\frac{1}{-7}\left[\begin{array}{cc}-1& -2\\ -3& 1\end{array}\right].$ Step 2 — $b$ se multiply karo. Left-multiply kyun? Kyunki $A^{-1}Ax=Ix=x$ unknown ko isolate karta hai. $x=\frac{1}{-7}\left[\begin{array}{cc}-1& -2\\ -3& 1\end{array}\right]\left[\begin{array}{c}4\\ 5\end{array}\right]=\frac{1}{-7}\left[\begin{array}{c}-4-10\\ -12+5\end{array}\right]=\frac{1}{-7}\left[\begin{array}{c}-14\\ -7\end{array}\right]=\left[\begin{array}{c}2\\ 1\end{array}\right].$ Answer: $x=2,y=1$. Check (column view): $2\left[\begin{array}{c}1\\ 3\end{array}\right]+1\left[\begin{array}{c}2\\ -1\end{array}\right]=\left[\begin{array}{c}4\\ 5\end{array}\right]=b$. ✓

Step 1 — row 3 se $x$ eliminate karo. Row3 $\leftarrow$ Row3 $-2\times$Row1: $(2x+5y-z)-2(x+y+z)=27-12\Rightarrow 3y-3z=15.$ System ab: $x+y+z=6,2y+5z=-4,3y-3z=15$. Step 2 — row 3 se $y$ eliminate karo. Row3 $\leftarrow$ Row3 $-\frac{3}{2}\times$Row2: $3y-3z-\frac{3}{2}(2y+5z)=15-\frac{3}{2}(-4)\Rightarrow -3z-\frac{15}{2}z=15+6\Rightarrow -\frac{21}{2}z=21.$ Toh $z=-2$. Step 3 — back-substitute karo. Row 2 se: $2y+5(-2)=-4\Rightarrow 2y=6\Rightarrow y=3$. Row 1 se: $x+3+(-2)=6\Rightarrow x=5$. Answer: $x=5,y=3,z=-2$. Check: $2(5)+5(3)-(-2)=10+15+2=27$. ✓

Hum kya notice karte hain: Row 2 $=2\times$ Row 1, toh $\mathrm{rank}(A)=1$, $n=3$. Free parameters $=n-r=2$. Homogeneous hamesha solvable kyun hota hai: $x=0$ hamesha kaam karta hai, toh hum kabhi existence nahi poochte — sirf yeh ki null space kitna bada hai. Ek real equation $x=-2y+z$ solve karo. Maano $y=s,z=t$: $x=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=s\left[\begin{array}{c}-2\\ 1\\ 0\end{array}\right]+t\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right].$ Answer: null space hai $\mathrm{span}\left\{\left[\begin{array}{c}-2\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]\right\}$ (origin se guzarti ek plane).

$A=\left[\begin{array}{cc}1& 1\\ 1& k\end{array}\right]$, $\det A=k-1$. (a) Unique jab $\det A\ne 0$, yaani $k\ne 1$. (Aise har $k$ ke liye lines ek baar cross karti hain.) (b)/(c) The knife-edge $k=1$. Tab system hai $x+y=2$ aur $x+y=4$ — same left side, alag right sides → contradiction → no solution. Kya koi $k$ hai jo infinitely many deta hai? Iske liye chahiye $\det A=0$ aur consistent RHS, yaani $k=1$ aur doosri equation pehli ki multiple ho. Lekin $k=1$ par RHS $4\ne 2$ hai, toh consistency fail hoti hai. Isliye koi bhi $k$ yahan infinitely many nahi deta. Answer: $k\ne 1$ ke liye unique; $k=1$ ke liye no solution; kabhi infinitely many nahi.

Step 1 — $x$ eliminate karo. Row2$-$Row1: $y+2z=1$. Row3$-$Row1: $3y+(a-1)z=b-1$. Step 2 — $y$ eliminate karo. Row3$-3\times$(naya Row2): $(a-1)z-6z=(b-1)-3$, yaani $(a-7)z=b-4.$ Last pivot ke baare mein reasoning:

• Agar $a\ne 7$: $z$ par genuine pivot → unique solution.
• Agar $a=7$ aur $b\ne 4$: $0=b-4\ne 0$, ek contradiction → no solution (augmented rank jump karta hai).
• Agar $a=7$ aur $b=4$: $0=0$, ek equation vanish ho jaati hai, $\mathrm{rank}=2<3=n$ → infinitely many (ek free parameter). Answer: $(a,b)=(7,4)$.

Step 1 — dependence pakdo. Row 2 $=2\times$ Row 1 (RHS bhi: $6=2\cdot 3$). Toh $\mathrm{rank}(A)=1=\mathrm{rank}([A|b])$, aur $n-r=3-1=2$ free parameters → infinitely many. Step 2 — ek particular solution $x_p$. Free variables $y=0,z=0$ set karo $x+2y-z=3$ mein: $x=3$. Toh $x_p=(3,0,0)$. Step 3 — null space $x_h$. Yeh exactly L2.3 ka answer hai: $\mathrm{span}\{(-2,1,0),(1,0,1)\}$. Step 4 — assemble karo parent ke rule $x=x_p+x_h$ se (valid hai kyunki $A(x_p+x_h)=b+0=b$): $x=\left[\begin{array}{c}3\\ 0\\ 0\end{array}\right]+s\left[\begin{array}{c}-2\\ 1\\ 0\end{array}\right]+t\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right],s,t\in \mathbb{R}.$ Check $s=1,t=1$ se: $x=(3-2+1,1,1)=(2,1,1)$, aur $2+2-1=3$. ✓

Column space kyun? Parent ke existence condition se, $Ax=b$ solvable hai iff $b\in \mathrm{Col}(A)$ — $b$ do columns ka combination $x_1{a}_{(1)}+x_2{a}_{(2)}$ hona chahiye. Step 1 — combination set up karo. Chahiye $x_1\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]+x_2\left[\begin{array}{c}0\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 4\end{array}\right].$ Rows 1 aur 2 se $x_1=1,x_2=2$ milta hai. Step 2 — last row test karo. Yeh $x_1+x_2=4$ maangta hai, lekin $1+2=3\ne 4$. Answer: required row-3 condition fail hoti hai, toh $b\notin \mathrm{Col}(A)$ aur $Ax=b$ ka koi solution nahi. (Yahan rank$(A)=2$ lekin rank$([A|b])=3$.)

Design logic: "one-parameter family" ka matlab $n-r=1$, toh $n=3$ ke saath $\mathrm{rank}(A)=\mathrm{rank}([A|b])=2$ chahiye. Recipe: do genuinely independent equations lo, phir teesri row ko unki ek consistent combination banao (taaki woh koi naya rank na jodte). Construction. Independent rows lo $x+y+z=1$ aur $y+z=1$. Teesri row ko Row1 $+$ Row2 banao taaki woh redundant aur consistent ho: $\{\begin{array}{l}x+y+z=1\\ y+z=1\\ x+2y+2z=2\end{array}$ Rank verify karo. Row3 $=$ Row1 $+$ Row2 exactly hai (LHS aur RHS dono: $2=1+1$), toh $\mathrm{rank}(A)=\mathrm{rank}([A|b])=2$, jo $3-2=1$ free parameter deta hai. ✓ Line dikhane ke liye solve karo. Row2 se, $y+z=1$; Row1 se, $x=1-(y+z)=0$. Maano $z=t$, toh $y=1-t$: $x=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]+t\left[\begin{array}{c}0\\ -1\\ 1\end{array}\right].$ Ek genuine line of solutions — exactly ek parameter, jaisa chahiye tha.

Maano $Ax=b$ ke do alag solutions $u\ne v$ hain, toh $Au=b$ aur $Av=b$. Key step — subtract karo. Matrix multiplication ki linearity se (distributivity): $A(u-v)=Au-Av=b-b=0.$ Toh $w:=u-v$ ek nonzero null-space vector hai. Infinitely many banao. Kisi bhi scalar $\lambda$ ke liye, $u+\lambda w$ consider karo: $A(u+\lambda w)=Au+\lambda Aw=b+\lambda \cdot 0=b.$ Har $\lambda$ ek solution deta hai, aur $w\ne 0$ hone se yeh saare distinct hain. Conclusion: jis moment tumhare paas do solutions hain, automatically infinitely many aa jaate hain. Isliye "exactly two" impossible hai — ek linear system ke $0$, $1$, ya $\infty$ solutions hote hain. $■$

Ranks $r_A=\mathrm{rank}(A)$, $r_{aug}=\mathrm{rank}([A|b])$, $n$ unknowns diye hain — verdict batao. Verify: $r_A<r_{aug}$ → none; $r_A=r_{aug}=n$ → unique; $r_A=r_{aug}<n$ → infinite ($n-r_A$ free params).