(1) Given A = [ − 1 − 1 3 3 ] . A=\begin{bmatrix}
-1 & -1 \\
3 & 3
\end{bmatrix}. A = [ − 1 3 − 1 3 ] . Let B = [ b 11 b 12 b 21 b 22 ] . B=\begin{bmatrix}
b_{11} & b_{12} \\
b_{21} & b_{22}
\end{bmatrix}. B = [ b 11 b 21 b 12 b 22 ] .
Then
A B = [ − 1 − 1 3 3 ] [ b 11 b 12 b 21 b 22 ] = AB=\begin{bmatrix}
-1 & -1 \\
3 & 3
\end{bmatrix}\begin{bmatrix}
b_{11} & b_{12} \\
b_{21} & b_{22}
\end{bmatrix}= A B = [ − 1 3 − 1 3 ] [ b 11 b 21 b 12 b 22 ] =
= [ − b 11 − b 21 − b 12 − b 22 3 b 11 + 3 b 21 3 b 12 + 3 b 22 ] =\begin{bmatrix}
-b_{11}-b_{21} & -b_{12}-b_{22} \\
3b_{11}+3b_{21} & 3b_{12}+3b_{22}
\end{bmatrix} = [ − b 11 − b 21 3 b 11 + 3 b 21 − b 12 − b 22 3 b 12 + 3 b 22 ] If A B = 0 , AB=0, A B = 0 , then
− b 11 − b 21 = 0 − b 12 − b 22 = 0 3 b 11 + 3 b 21 = 0 3 b 12 + 3 b 22 = 0 \begin{alignedat}{2}
-b_{11}-b_{21}=0 \\
-b_{12}-b_{22}=0 \\
3b_{11}+3b_{21}=0 \\
3b_{12}+3b_{22}=0
\end{alignedat} − b 11 − b 21 = 0 − b 12 − b 22 = 0 3 b 11 + 3 b 21 = 0 3 b 12 + 3 b 22 = 0
b 21 = − b 11 b 12 = − b 22 \begin{alignedat}{2}
b_{21}=-b_{11} \\
b_{12}=-b_{22}
\end{alignedat} b 21 = − b 11 b 12 = − b 22
B = [ c − d − c d ] , c , d ∈ R B=\begin{bmatrix}
c & -d \\
-c & d
\end{bmatrix}, c,d\in \R B = [ c − c − d d ] , c , d ∈ R (2)
A = [ 2 1 3 1 5 − 3 3 − 3 7 ] = > A T = [ 2 1 3 1 5 − 3 3 − 3 7 ] = A A=\begin{bmatrix}
2 & 1 & 3 \\
1 & 5 & -3 \\
3 & -3 & 7
\end{bmatrix}=>A^T=\begin{bmatrix}
2 & 1 & 3 \\
1 & 5 & -3 \\
3 & -3 & 7
\end{bmatrix}=A A = ⎣ ⎡ 2 1 3 1 5 − 3 3 − 3 7 ⎦ ⎤ => A T = ⎣ ⎡ 2 1 3 1 5 − 3 3 − 3 7 ⎦ ⎤ = A
Then the matrix A A A is symmetric.
B = [ 1 1 3 1 2 2 3 2 3 ] = > B T = [ 1 1 3 1 2 2 3 2 3 ] = B B=\begin{bmatrix}
1 & 1 & 3 \\
1 & 2 & 2 \\
3 & 2 & 3
\end{bmatrix}=>B^T=\begin{bmatrix}
1 & 1 & 3 \\
1 & 2 & 2\\
3 & 2 & 3
\end{bmatrix}=B B = ⎣ ⎡ 1 1 3 1 2 2 3 2 3 ⎦ ⎤ => B T = ⎣ ⎡ 1 1 3 1 2 2 3 2 3 ⎦ ⎤ = B
Then the matrix B B B is symmetric.