Obtain the following matrices and describe them
(a)$A=[a_{ij}]$ is a $3\times3$ matrix such that $a_{ij}=a$ if $i\leq{j}$ and
zero otherwise.
(b)$B=[b{ij}]$ is a $3\times3$ matrix such that $b_{ij}=i+j$ (Hint observe $B^T$).
(c)$C=[c_{ij}]$ is a $2\times2$ matrix such that $c_{11}=c_{22}=a$ and
$c_{12}=b$ and $c_{21}=-b,$ verify the orthogonality of C.
(d)$D=[d_{ij}]$ is $3\times3$ matrix such that $d_{12}=a,$  $d_{13}=a^2$,
and $c_{23}=a$, and others are zero ,then compute $D^3$ and  comment on the
resulting matrix.
SOLUTION :
(a)$A=[a_{ij}]$ is a $3\times3$ matrix such that
$a_{ij}=a$.
since A is defined as a $3\times3$ matrix and the elements are
defined as $a_{ij}=a$.
in as much as $i\leq{j}$
then \begin{equation}
a_{11}=a,\quad a_{12}=a,\quad a_{13}=a, 
\end{equation}
\begin{equation}
a_{21}=0,\quad a_{22}=a,\quad a_{23}=a,
\end{equation}
\begin{equation}
a_{31}=0,\quad a_{32}=0,\quad a_{33}=a,
\end{equation}
\begin{equation}A=\left[\begin{array}{ccc}
a & a & a\\ 0 & a & a\\0 & 0 & a\end{array}\right]     \end{equation}
Hence the resulting matrix is an upper triangular or echelon   matrix.
(b) $B=[b_{ij}]$ is a $3\times3$ such that $b_{ij}=i+j$,
here our matrix is defined as $B=b_{ij}=i+j$, taking values for
i,j yields \begin{equation}
a_{11}=1+1=2,\quad a_{12}=1+2=3,\quad a_{13}=1+3=4
\end{equation}
\begin{equation}
a_{21}=2+1=3,\quad a_{22}=2+2=4,\quad a_{23}=2+3=5,
\end{equation}
\begin{equation}
a_{31}=3+1=4,\quad a_{32}=3+2=5,\quad a_{33}=3+3=6
\end{equation}
hence the matrix B is
\begin{equation}
B=\left[\begin{array}{ccc}
2 & 3 & 4\\ 3 & 4 & 5\\4 & 5 & 6\end{array}\right]   
\end{equation}
(c) $C=[c_{ij}]$ is $2\times2$ matrix, such that
$c_{11}=c_{22}=a$,$c_{12}=b$ and $c_{21}=-b$
this matrix is simple to construct since the elements have
been defined and its a square $2\times2$ matrix. hence
\begin{equation}
C=\left[\begin{array}{cc}
a & b\\ -b & a\end{array}\right]   
\end{equation}
to verify the orthogonality of this matrix, we use the
formula for orthogonality which is $CC^T=I$ where is an identity matrix,
so we begin with $C^T$ \begin{equation}
C^T=\left[\begin{array}{cc}
a & -b\\ b & a\end{array}\right]\end{equation} so  \begin{equation}
CC^T=\left[\begin{array}{cc}
a & b\\ -b & a\end{array}\right] \times \left[\begin{array}{cc}
a & -b\\ b & a\end{array}\right] 
\end{equation}
\begin{equation}
=\left[\begin{array}{cc}
a^2+b^2 & 0\\ 0 & a^2+b^2\end{array}\right]
\end{equation}
hence the matrix is orthogonal as it yields a square matrix.
\begin{equation}
CC^T=a^2+b^2\left[\begin{array}{cc}
1 & 0\\ 0 & 1\end{array}\right]\end{equation}.
(d)$D=[{d_ij}]$ is $3\times3$ matrix such that
$d_{12}=a$, $d_{13}=a^2$, $d_{23}=a$ while others are zero,
now we are to compute $D^3$ after constructing the matrix D which is.
\begin{equation}
D=\left[\begin{array}{ccc}
0 & a & a^3\\ 0 & 0 & a\\0 & 0 & 0\end{array}\right]
\end{equation}
to compute $D^3=D^2\times{D}$
now for \begin{equation}D^2=\left[\begin{array}{ccc}
0 & a & a^3\\ 0 & 0 & a\\0 & 0 & 0\end{array}\right]\times\left[\begin{array}{ccc}
0 & a & a^3\\ 0 & 0 & a\\0 & 0 & 0\end{array}\right]\end{equation}
\begin{equation}
=\left[\begin{array}{ccc}
0 & 0 & a^2\\ 0 & 0 & 0\\0 & 0 & 0\end{array}\right]
\end{equation} 
now \begin{equation}
D^3=D^2\times{D}=\left[\begin{array}{ccc}
0 & 0 & a^2\\ 0 & 0 & 0\\0 & 0 & 0\end{array}\right]\times\left[\begin{array}{ccc}
0 & a & a^3\\ 0 & 0 & a\\0 & 0 & 0\end{array}\right] 
\end{equation}
\begin{equation}
=\left[\begin{array}{ccc}
0 & 0 & 0\\ 0 & 0 & 0\\0 & 0 & 0\end{array}\right]
\end{equation}
hence the resulting matrix is a NULL matrix.