Solution:

Matrix A is a null matrix (because all its entries are zero)

Matrix B is a row matrix (because it has only one row).

Matrix C is a column matrix (because it has only one column).

Matrix D is a unit matrix (because its diagonal entries are $1$ and non-diagonal entries are $0$).

Matrix E is a null matrix (because all its entries are $0$).

Matrix F is a column matrix (because it has only one column).

Solution:

(a)     (iii), (iv) and (viii) are square matrices because the number of rows are equal to number of columns.

(b)     (i), (ii), (v), (vi) (vii), (ix) are rectangular matrices because their rows and columns are not equal.

(c)     (vi) is a row matrix because it has only one row.

(d)     (ii) and (vii) are column matrices because they have only one column.

(e)     (iv) is an identity matrix as well because its diagonal elements are $1$ and all non-diagonal elements are $0$.

(f)      (ix) is a null matrix because its each entry is zero.

Solution:     

Matrix A is a scalar matrix (because its diagonal entries are same).

Matrix B is a diagonal matrix (because its diagonal entries are non-zero and non-diagonal entities are zero).

Matrix C is an identity matrix (because its diagonal entries are $1$).

Matrix D is a diagonal matrix (because its one diagonal entry is non-zero and non-diagonal entities are zero).

Matrix E is a scalar matrix (because its diagonal entries are same).

E = $\left[ \begin{matrix}    5-3 & 0  \\    0 & 1+1  \\ \end{matrix} \right]$            

E = $\left[ \begin{matrix}    2 & 0  \\    0 & 2  \\ \end{matrix} \right]$

Note:

All the matrices A, B, C, D and E are diagonal matrices because they have at least one non-zero diagonal entry and all their non-diagonal entries are zero. Scalar and unit matrices are further types of diagonal matrices.

Solution:

Negative of a matrix is obtained by inverting (changing) the signs of all its entries.

(i) $-~$A =$~~\left[ \begin{matrix} -1 \\ 0 \\ 1 \\ \end{matrix} \right]$

(ii) $-$B =$~\left[ \begin{matrix} -3 & 1 \\ -2 & -1 \\ \end{matrix} \right]$

(iii) $-~$C =$~\left[ \begin{matrix} -2 & -6 \\ -3 & -2 \\ \end{matrix} \right]$

(iv) $-$D = $\left[ \begin{matrix} 3 & -2 \\ 4 & -5 \\ \end{matrix} \right]$

(v) $-$E = $\left[ \begin{matrix} -1 & 5 \\ -2 & -3 \\ \end{matrix} \right]$

Solution:

Transpose of a matrix is obtained by converting all the columns of that matrix to the rows and all the rows to the columns. So, according to the definition

(i) $\rm{A}^{t}$ =$~~\left[ \begin{matrix} 0 & 1 & -2 \\ \end{matrix} \right]$

(ii) $\rm{B}^{t}$ =$~\left[ \begin{matrix} 5 \\ 1 \\ -6 \\ \end{matrix} \right]$

(iii) $\rm{C}^{t}$ =$~~$ $\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & -1 & 0 \\ \end{matrix} \right]$

(iv) $\rm{D}^{t}$ = $\left[ \begin{matrix} 2 & 0 \\ 3 & 5 \\ \end{matrix} \right]$

(v) $\rm{E}^{t}$ = $\left[ \begin{matrix} 2 & -4 \\ 3 & 5 \\ \end{matrix} \right]$

(vi) $\rm{F}^{t}$ = $\left[ \begin{matrix} 1 & 3 \\ 2 & 4 \\ \end{matrix} \right]$

Solution:

(i) To prove: $(\rm{A}^{t})^{t} = \rm{A}$

Given

$\rm{A} = \left[ \begin{matrix} 1 & 2 \\ 0 & 1 \\ \end{matrix} \right]$

$\rm{A}^{t} = \left[ \begin{matrix} 1 & 0 \\ 2 & 1 \\ \end{matrix} \right]$

Taking transpose of $\rm{A}^{t}$, we will get

$(\rm{A}^{t})^{t} = \left[ \begin{matrix} 1 & 2 \\ 0 & 1 \\ \end{matrix} \right]= \rm{A}$

Hence proved: $(\rm{A}^{t})^{t} = \rm{A}$

(ii) To prove: $(\rm{B}^{t})^{t} = \rm{B}$

Given

$\rm{B} = \left[ \begin{matrix} 1 & 1 \\ 2 & 0 \\ \end{matrix} \right]$

$\rm{B}^{t} = \left[ \begin{matrix} 1 & 2 \\ 1 & 0 \\ \end{matrix} \right]$

Taking transpose of $\rm{B}^{t}$, we will get

$(\rm{B}^{t})^{t} = \left[ \begin{matrix} 1 & 1 \\ 2 & 0 \\ \end{matrix} \right] = \rm{B}$

Hence proved: $(\rm{B}^{t})^{t} = \rm{B}$