Construction of Matrices[MAT204 C.A Solution]

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.