Put x = 0
\(|A|=\left|\begin{array}{ccc} 3 & -6 & 3 \\ -6 & 3 & 3 \\ 3 & 3 & -6 \end{array}\right|=\left|\begin{array}{ccc} 0 & -6 & 3 \\ 0 & 3 & 3 \\ 0 & 3 & -6 \end{array}\right|\)
\(C_{1} \rightarrow C_{1}+C_{2}+C_{3}\)
= 0 
\(\therefore\) x = 0 is a root.

\(\left|\begin{array}{lll} k a & k b & k c \\ k x & k y & k z \\ k p & k q & k r \end{array}\right|=k^{3}\left|\begin{array}{lll} a & b & c \\ x & y & z \\ p & q & r \end{array}\right|=k^{3} \Delta\)
\(\text { Take } k \text { from } R_{1}, R_{2}, R_{3}\)

\(\text { Given } \text {matrix } A=\left[\begin{array}{cc} \alpha & \beta \\ \gamma & -\alpha \end{array}\right] \text { is unit matrix }\)
\(|A|=1 \)
\(-\alpha^{2}-\beta \gamma=1 \)
\(1+\alpha^{2}+\beta \gamma=0 \)

\(\text { Area of triangle }=\frac{1}{2}\left|\begin{array}{lll} \frac{x_{1}}{a} & \frac{y_{1}}{a} & 1 \\ \frac{x_{2}}{b} & \frac{y_{2}}{b} & 1 \\ \frac{x^{3}}{c} & \frac{y_{3}}{c} & 1 \end{array}\right|\)
\(=\frac{1}{2 a b c}\left|\begin{array}{lll} x_{1} & y_{1} & a \\ x_{2} & y_{2} & b \\ x_{3} & y_{3} & c \end{array}\right| \text { Multiply }R_{1}, R_{2} \& R_{3} \text { by } a, b, c\text { respectively }\)
\(=\frac{1}{2 a b c} \frac{1}{(2)}\left|\begin{array}{ccc} 2 a & x_{1} & y_{1} \\ 2 b & x_{2} & y_{2} \\ 2 c & x_{3} & y_{3} \end{array}\right|=\frac{1}{4 a b c}\left(\frac{a b c}{2}\right)\)
\(=\frac{1}{8}\)

\(|A|=0 \quad \text { [since } A \text { is singular] }\)
\(\left|\begin{array}{ll} e^{x-2} & e^{7+x} \\ e^{2+x} & e^{2 x+3} \end{array}\right| =0 \)
\(e^{x-2+2 x+3}-e^{2+x+7+x} =0 \)
\(e^{3 x+1} =e^{9+2 x} \)
\(3 x+1 =9+2 x \)
\(x =8 \)

\(A^{T}=\left[\begin{array}{ll} a & y \\ x & a \end{array}\right] \quad \therefore A A^{T}=\left[\begin{array}{ll} a & x \\ y & a \end{array}\right]\left[\begin{array}{ll} a & y \\ x & a \end{array}\right] \)
\(=\left[\begin{array}{ll} a^{2}+x^{2} & a y+a x \\ a y+a x & y^{2}+a^{2} \end{array}\right] \)
\(\operatorname{det}\left(A A^{T}\right)=\left|\begin{array}{ll} a^{2}+x^{2} & a y+a x \\ a y+a x & y^{2}+a^{2} \end{array}\right| \)
\(=\left(a^{2}+x^{2}\right)\left(a^{2}+y^{2}\right)-(a y+a x)(a y+a x) \)
\(=a^{4}+a^{2} y^{2}+a^{2} x^{2}+x^{2} y^{2}-a^{2} y^{2}-a^{2} x y -a^{2} x y-a^{2} x^{2}\)
\(=a^{4}+1-2 a^{2} x y=a^{4}-2 a^{2}+1(\because x y=1) \)
\(=\left(a^{2}-1\right)^{2} \)

\(2 X=\left[\begin{array}{ll} 3 & 8 \\ 7 & 2 \end{array}\right]-\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]=\left[\begin{array}{cc} 2 & 6 \\ 4 & -2 \end{array}\right]\)
\(X=\left[\begin{array}{cc} 1 & 3 \\ 2 & -1 \end{array}\right]\)

\(A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] \)
\(a_{11}=\frac{1}{2}(3-2)=\frac{1}{2} ; a_{12}=\frac{1}{2}(3-4)=\frac{-1}{2} \)
\(a_{21}=\frac{1}{2}(3(2)-2)=\frac{4}{2}=2 ; a_{22}=\frac{1}{2}(6-4)=\frac{2}{2}=1 \)
\(\therefore A=\left[\begin{array}{ll} \frac{1}{2} & -\frac{1}{2} \\ 2 & 1 \end{array}\right] \)