Exercises — Determinant of 2×2 matrix

WHAT: $a=4,b=1,c=2,d=3$ ke saath $ad-bc$ apply karo. WHY: direct definition, kuch aur chahiye hi nahi. $\det =(4)(3)-(1)(2)=12-2=10$ Meaning: areas $10$ se scale hongi; sign $+$ hai, toh orientation preserved hai.

$\det =(5)(2)-(5)(2)=10-10=0$ Zero determinant ka matlab hai matrix singular hai (non-invertible). Columns dekho: $(5,2)$ aur $(5,2)$ identical hain, toh woh ek hi line par lie karte hain — parallelogram flat hai, area $0$. Dekho Linear independence: columns dependent hain.

$\det =(0)(0)-(7)(-3)=0-(-21)=21$ WHY sign flip andar: $bc=7\times (-3)=-21$, aur negative subtract karna add karta hai. Yahi woh jagah hai jahan students slip karte hain — har entry ka sign track karo.

WHY determinant: do vectors se bane parallelogram ki area $|ad-bc|$ hai — determinant ka yahi toh poora point hai (dekho Area of parallelogram). Vectors ko columns ki tarah rakho: $A=\left[\begin{array}{cc}3& 1\\ 1& 4\end{array}\right],\det =(3)(4)-(1)(1)=12-1=11.$ Area $=|11|=11$. Positive sign $\Rightarrow$ $\vec{v}$, $\vec{u}$ se counter-clockwise hai.

Part 1: Inverse of a 2×2 matrix se, $\det (A)\det (A^{-1})=\det (I)=1$, toh $\det (A^{-1})=\frac{1}{\det (A)}=\frac{1}{4}.$ Part 2: Pehle actual $A$ check karo: $\det =(3)(2)-(2)(1)=6-2=4$ (match karta hai). Inverse formula $a\leftrightarrow d$ swap karta hai, off-diagonal negate karta hai, phir determinant se divide karta hai: $A^{-1}=\frac{1}{4}\left[\begin{array}{cc}2& -2\\ -1& 3\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}& -\frac{1}{2}\\ -\frac{1}{4}& \frac{3}{4}\end{array}\right].$

Singular ka matlab $\det =0$. WHY: zero area $\Rightarrow$ koi inverse nahi. $\det =k\cdot k-3\cdot 2=k^2-6=0\Rightarrow k=\pm \sqrt{6}.$ Do answers hain — ek common galti sirf $+\sqrt{6}$ rakhna hai.

$\det (AB)$: multiplicativity, $\det (AB)=\det A\det B=5\cdot (-2)=-10$. $\det (3A)$: $2\times 2$ matrix ko $3$ se scale karna dono columns ko $3$ se stretch karta hai, toh area $3\times 3=9$ se scale hoti hai: $\det (3A)=3^2\det (A)=9\cdot 5=45.$ $\det (A^{T}B)$: $\det (A^T)=\det (A)=5$ (transpose invariance), toh $\det (A^{T}B)=5\cdot (-2)=-10$.

Rotation: $\det (R)=(0)(0)-(-1)(1)=0+1=1$. Ek rotation square ko rigidly move karta hai — area unchanged (factor $1$) aur orientation preserved (positive sign). Dekho Transformations and scaling. Reflection: $\det =(0)(0)-(1)(1)=-1$. Area unchanged (magnitude $1$) lekin sign negative hai — ek reflection orientation flip karta hai, counter-clockwise square ko clockwise bana deta hai, jaise mirror image.

Shuru karo $\det \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=ad-bc$ se. Columns swap karo taaki $\left[\begin{array}{cc}b& a\\ d& c\end{array}\right]$ mile: $\det =b\cdot c-a\cdot d=bc-ad=-(ad-bc).$ Toh determinant sign flip kar leta hai. Geometric reason: do spanning arrows swap karna ye flip karta hai ki kaun "counter-clockwise" hai, signed orientation reverse ho jaata hai.

Set up. Coefficient matrix $A=\left[\begin{array}{cc}2& 1\\ 1& 3\end{array}\right]$, right side $(5,10)$. Main determinant: $\det A=(2)(3)-(1)(1)=6-1=5$. Non-zero $\Rightarrow$ unique solution exist karta hai. $x$ ke liye column 1 ko right side se replace karo: $D_x=\det \left[\begin{array}{cc}5& 1\\ 10& 3\end{array}\right]=(5)(3)-(1)(10)=15-10=5.$ $y$ ke liye column 2 replace karo: $D_y=\det \left[\begin{array}{cc}2& 5\\ 1& 10\end{array}\right]=(2)(10)-(5)(1)=20-5=15.$ Divide karo: $x=\frac{D_x}{\det A}=\frac{5}{5}=1,y=\frac{D_y}{\det A}=\frac{15}{5}=3.$ Check: $2(1)+3=5$ ✓, $1+3(3)=10$ ✓.

Singular $\Rightarrow$ determinant zero: $\det =(4-\lambda )(3-\lambda )-(1)(2)=0.$ Expand karo: $12-4\lambda -3\lambda +{\lambda }^2-2={\lambda }^2-7\lambda +10=0.$ Factor karo: $(\lambda -2)(\lambda -5)=0\Rightarrow \lambda =2\text{ or }\lambda =5.$ Sanity check ek property se: eigenvalues ka product determinant ke barabar hota hai: $2\times 5=10$, aur $\det \left[\begin{array}{cc}4& 1\\ 2& 3\end{array}\right]=12-2=10$ ✓.

Strategy: $\det =0$ ka matlab columns parallel hain — doosra column pehle column ka scalar multiple hona chahiye. Pehla column $(2,3)$ lo aur $2$ se multiply karke doosra column $(4,6)$ banao: $A=\left[\begin{array}{cc}2& 4\\ 3& 6\end{array}\right],\det =(2)(6)-(4)(3)=12-12=0.$ Saari entries non-zero ✓, aur yeh all-ones matrix ka multiple nahi hai ✓. (Infinitely many valid answers exist hain — koi bhi parallel-column matrix kaam karega. Dekho Linear independence.)

Requirement: $ad-bc=-3$. Sabse aasaan reflection-then-scale hai: $\left[\begin{array}{cc}3& 0\\ 0& -1\end{array}\right]$ lo: $\det =(3)(-1)-(0)(0)=-3.$ Area factor $|-3|=3$ (tripled) ✓, negative sign = orientation reversed ✓.

$A=B=I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ lo. $\det (A)+\det (B)=1+1=2.$ Lekin $A+B=\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]$, toh $\det (A+B)=(2)(2)-0=4\ne 2.$ Conclusion: determinant additive nahi hai. Yeh multiplicative zaroor hai ($\det (AB)=\det A\det B$) kyunki area-scaling factors tab multiplication se compose hote hain jab tum ek transformation ke baad doosra apply karte ho (Matrix multiplication) — lekin matrices add karna "pehle ek phir doosra karna" nahi hai, toh sums ke liye koi aisi rule exist nahi karti.

L1 plug-in ::: $ad-bc$ apply karo, signs dekho L2 area & inverse ::: columns area span karte hain; inverse $=\frac{1}{\det }\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$ L3 properties ::: $\det (AB)=\det A\det B$, $\det (kA)=k^2\det A$, swap sign flip karta hai L4 systems ::: Cramer $x=D_x/\det A$; eigenvalues $\det =0$ se L5 build & prove ::: $\det$ ko target par engineer karo; $\det$ additive nahi hai