Exercises — Properties of determinants

Columns 1 aur 3 dekho — ye identical hain $(3,1,6)$. $P1$ kehta hai column properties row properties ke equal hain, aur $P3$ kehta hai do equal rows (isliye columns) determinant ko $0$ force kar dete hain. $\boxed{=0}$ Koi arithmetic zaroorat nahi — yahi Recognition ka poora point hai.

• $\det (A^{\top })=\det A=7$ by $P1$.
• $2A$ teeno rows ko $2$ se scale karta hai, toh $P4$ corollary se $\det (2A)=2^3\det A=8\cdot 7=56$. $\boxed{\det (A^{\top })=7,\det (2A)=56}$

Column 3 poora zeros hai. $P7$ se (via $P1$), ek zero row/column deta hai $\det =0$. $\boxed{=0}$

$P8$: ek triangular matrix ke liye determinant sirf diagonal ka product hota hai. $\det A=2\cdot (-3)\cdot 5=\boxed{-30}$

Step 1 (kyun): $P6$ use karo column 1 clear karne ke liye — ye operations value nahi badaltein. $R_2\to R_2-2R_1$, $R_3\to R_3-R_1$: $\left|\begin{array}{ccc}1& 2& -1\\ 0& -1& 3\\ 0& -3& 5\end{array}\right|.$ Step 2 (kyun): Diagonal ke neeche column 2 clear karo. $R_3\to R_3-3R_2$: $\left|\begin{array}{ccc}1& 2& -1\\ 0& -1& 3\\ 0& 0& -4\end{array}\right|.$ Step 3: Ab triangular hai. $P8$ se, $\det =1\cdot (-1)\cdot (-4)=\boxed{4}$.

Step 1 (kyun): Row 1 $=3\cdot (1,2,3)$ hai. $P4$ humein us ek row se $3$ factor out karne deta hai. $=3\left|\begin{array}{ccc}1& 2& 3\\ 1& 0& 2\\ 2& 2& 0\end{array}\right|.$ Step 2 (kyun): $P6$ se zeros banao. $R_2\to R_2-R_1$, $R_3\to R_3-2R_1$: $=3\left|\begin{array}{ccc}1& 2& 3\\ 0& -2& -1\\ 0& -2& -6\end{array}\right|.$ Step 3: $R_3\to R_3-R_2$: $\left|\begin{array}{ccc}1& 2& 3\\ 0& -2& -1\\ 0& 0& -5\end{array}\right|$. $P8$ se: $=3\cdot (1\cdot (-2)\cdot (-5))=3\cdot 10=\boxed{30}.$

• $P9$ corollary se, $\det (A^{-1})=\frac{1}{\det A}=\frac{1}{4}$.
• $2A^{-1}$ saare $3$ rows ko $2$ se scale karta hai, toh $P4$ corollary deta hai $2^3\cdot \frac{1}{4}=8\cdot \frac{1}{4}=\boxed{2}.$

Har operation ka factor track karo:

• $R_1$ ko $5$ se multiply karo: factor $\times 5$ ($P4$). Running $=5\cdot 6=30$.
• $R_2,R_3$ swap karo: factor $\times (-1)$ ($P2$). Running $=-30$.
• $R_1\to R_1+4R_3$: factor $\times 1$ ($P6$, no change). Running $=-30$. $\boxed{\det B=-30}$

Idea: row 2 ko $P6$ use karke collapse karo. $R_2\to R_2-R_1$ (row 1 subtract karo; $P6$, value unchanged): $\left|\begin{array}{ccc}a& b& c\\ 2& 2& 2\\ 1& 1& 1\end{array}\right|.$ Ab row 2 $=2\cdot (1,1,1)=2\cdot$ row 3 hai. $2$ bahar nikalo ($P4$): $=2\left|\begin{array}{ccc}a& b& c\\ 1& 1& 1\\ 1& 1& 1\end{array}\right|.$ Rows 2 aur 3 equal hain $\Rightarrow \det =0$ by $P3$. Isliye poori cheez $2\cdot 0=\boxed{0}$ hai.

$A=B=I_2$ lo (identity). Toh $\det A=\det B=1$, isliye $\det A+\det B=2$. Lekin $A+B=2I_2=\left(\begin{array}{cc}2& 0\\ 0& 2\end{array}\right)$ hai, aur $P8$ se (diagonal) $\det (2I_2)=2\cdot 2=4$. Kyunki $4\ne 2$, "rule" fail ho jaata hai. $P5$ ek ek baar mein ek row mein linearity hai; poore matrices add karne se har row change hoti hai ek saath, isliye wo factor nahi hota. $\boxed{\det (A+B)=4\ne 2=\det A+\det B}$

Step 1 (kyun): top ke neeche first column ko column operations se clear karo ($P1$ column ops ko row ops ki tarah mirror karne deta hai). $C_3\to C_3-C_2$ phir $C_2\to C_2-C_1$: $\left|\begin{array}{ccc}1& 0& 0\\ a& b-a& c-b\\ a^2& b^2-a^2& c^2-b^2\end{array}\right|.$ Step 2: nayi columns mein se factor karo ($P4$): $b^2-a^2=(b-a)(b+a)$, $c^2-b^2=(c-b)(c+b)$: $=(b-a)(c-b)\left|\begin{array}{ccc}1& 0& 0\\ a& 1& 1\\ a^2& b+a& c+b\end{array}\right|.$ Step 3: row 1 ke along expand karo (sirf top-left $1$ survive karta hai), $2\times 2$ bacha rehta hai: $=(b-a)(c-b)[(c+b)-(b+a)]=(b-a)(c-b)(c-a).$ $\boxed{=(b-a)(c-a)(c-b)}$

Teen properties chain karo:

• $\det (M^{\top })=\det M$ by $P1$.
• $M^{\top }=-M$ ka matlab hai humne saare $3$ rows ko $-1$ se multiply kiya, isliye $\det (M^{\top })=(-1{)}^3\det M=-\det M$ ($P4$ corollary). Combine karo: $\det M=-\det M\Rightarrow 2\det M=0\Rightarrow \boxed{\det M=0}$. (Key: $(-1{)}^3=-1$ kyunki $3$ odd hai — ye even-sized skew matrices ke liye fail hota hai, jahan $\det$ nonzero ho sakta hai.)

Product ko $P9$, $P1$, $P4$ corollary se break karo:

• $\det (A^2)=(\det A{)}^2=9$ ($P9$ do baar apply karo).
• $\det (B^{-1})=1/\det B=-\frac{1}{2}$.
• $\det (2A^{\top })=2^3\det (A^{\top })=8\cdot \det A=24$ ($P4$ corollary + $P1$). Multiply karo (order matter nahi karta, $\det$ of a product, products of $\det$s hota hai): $9\cdot \left(-\frac{1}{2}\right)\cdot 24=9\cdot (-12)=\boxed{-108}.$

Determinant map ka area-scaling factor hota hai (dekho Linear Transformations & Scaling of Area). $\det T=2\cdot 3-1\cdot 1=5.$ Nayi area $=|\det T|\times \text{old area}=5\times 6=\boxed{30}$. Kyunki $\det T=5>0$ hai, orientation preserved hai (koi flip nahi). Figure mein red parallelogram dekho: area $1$ wala unit square area $5$ wale region mein ban jaata hai — har shape same factor se scale hoti hai.

Agar $A^{-1}$ exist karta hai, toh $A{A}^{-1}=I$. $P9$ apply karo: $\det A\cdot \det (A^{-1})=\det I=1.$ Ek product jo $1$ ke equal hai, usmein zero factor nahi ho sakta, isliye $\det A\ne 0$ (aur $\det (A^{-1})=1/\det A$, $P9$ corollary). Iske ulta, agar $\det A=0$ hai toh map area ko $0$ par squash kar deta hai (rows linearly dependent hain, $P3$-style), isliye ise undo nahi kiya ja sakta — koi inverse exist nahi karta. Ye exactly Inverse of a Matrix via Adjoint se bridge hai. $■$