\(A \& B \text { are symmetric }\)
\(A^{T}=A \text {, and } B^{T}=B\)
\(\text { 1) }(A+B)^{T}=B^{T}+A^{T}=B+A=A+B \Rightarrow A+B\text { is symmetric }\)
\(\text { 2) }(A B)^{T}=B^{T} A^{T}=B A \neq A B\)
\(\therefore A B \text { is not symmetric }\)
\(\text { 3) } A B=A^{T} B^{T}=(B A)^{T}\)
\(\text { 4) } A^{T} B=A B=A B^{T}\)

\(A+I =\left[\begin{array}{cc} 3 & -2 \\ 4 & 1 \end{array}\right] \)
\(A =\left[\begin{array}{cc} 3 & -2 \\ 4 & 1 \end{array}\right]-\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 2 & -2 \\ 4 & 0 \end{array}\right] \)
\(A-I =\left[\begin{array}{cc} 2 & -2 \\ 4 & 0 \end{array}\right]-\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & -2 \\ 4 & -1 \end{array}\right] \)
\((A+I)(A-I) =\left[\begin{array}{ll} 3 & -2 \\ 4 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & -2 \\ 4 & -1 \end{array}\right] \)
\(=\left[\begin{array}{ll} 3-8 & -6+2 \\ 4+4 & -8-1 \end{array}\right] \)
\(=\left[\begin{array}{cc} -5 & -4 \\ 8 & -9 \end{array}\right] \)

\(\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right] A=\left[\begin{array}{cc} 1 & 1 \\ 0 & -1 \end{array}\right]\)
\(\text { Let } A=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\)
\(\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{cc} 1 & 1 \\ 0 & -1 \end{array}\right]\)
\(\left[\begin{array}{cc} a+3 c & b+3 d \\ 0+c & 0+d \end{array}\right]=\left[\begin{array}{cc} 1 & 1 \\ 0 & -1 \end{array}\right]\)
\(c=0,d=-l\)
\(a+3 c=1 \quad b+3 d=1\)
\(a+0=1 \quad b-3=1\)
\(a=1 \quad b=4\)
\(\therefore A=\left[\begin{array}{cc} 1 & 4 \\ 0 & -1 \end{array}\right]\)

\(-1 \leq x<0 \Rightarrow\lfloor x\rfloor=-1 ;\)
\(0 \leq y<1 \Rightarrow\lfloor y\rfloor=0 ; 1 \leq z<2 \Rightarrow\lfloor z\rfloor=1 \)
\(\therefore|\cdot A|=\left|\begin{array}{ccc} -1+1 & 0 & 1 \\ -1 & 0+1 & 1 \\ -1 & 0 & 1+1 \end{array}\right|=\left|\begin{array}{ccc} 0 & 0 & 1 \\ -1 & 1 & 1 \\ -1 & 0 & 2 \end{array}\right|=1(0+1)=1 \)
\(|A|=1=\lfloor z\rfloor \)

\(B =(-)\left|\begin{array}{ccc} -2 & 4 & 8 \\ 6 & 2 & 0 \\ -2 & 4 & 2 \end{array}\right| \quad R_{1} \leftrightarrow R_{3} \)
\(=-(2 \times 2)\left|\begin{array}{ccc} -1 & 2 & 4 \\ 3 & 1 & 0 \\ -2 & 4 & 2 \end{array}\right| \)
\(\text { Take } 2 \text { from } R_{1} \& R_{2}\)
\(B=-4 A\)

\(\left|\begin{array}{ccc} a & 2 b & 2 c \\ 3 & b & c \\ 4 & a & b \end{array}\right| =0 \)
\(\Rightarrow \frac{1}{2}\left|\begin{array}{lll} a & 2 b & 2 c \\ 6 & 2 b & 2 c \\ 4 & a & b \end{array}\right| =0 \quad R_{2} \rightarrow 2 R_{2} \)
\(\frac{1}{2}\left|\begin{array}{ccc} a-6 & 0 & 0 \\ 6 & 2 b & 2 c \\ 4 & a & b \end{array}\right| =0 \quad R_{1} \rightarrow R_{1}-R_{2} \)
\(\Rightarrow \frac{1}{2}\left[(a-6)\left(2 b^{2}-2 a c\right)\right] =0 \)
\((a-6)\left(2 b^{2}-2 a c\right) =0 \)
\(a=6,2 b^{2} =2 a c \)
\(b^{2} =a c \quad \therefore b^{3}=a b c \)

\(|A|=\left|\begin{array}{ccc} 0 & a & -b \\ -a & 0 & c \\ b & -c & 0 \end{array}\right|\)
\(=0-a(-b c)-b(a c)=a b c-a b c=0\)

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.