Solution:

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

Determinant of matrix A is calculated as:

det A = $\left| \text{A} \right|$

= $\left| \begin{matrix} -1 & 1 \\ 2 & 0 \\ \end{matrix} \right|$

= $\left( -1 \right)\times 0-2\times 1$

= $0-2$

= $-2$

 

Therefore,

$\left| \text{A} \right|$ = $-2$

 

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

Determinant of matrix B is calculated as:

det B = $\left| \text{B} \right|$

= $\left| \begin{matrix} 1 & 3 \\ 2 & -2 \\ \end{matrix} \right|$

= $1\times \left( -2 \right)-3\times 2$

= $-2-6$

= $-8$

 

Therefore,

$\left| \text{B} \right|$ = $-8$

 

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

Determinant of matrix C is calculated as:

det C = $\left| \text{C} \right|$

= $\left| \begin{matrix} 3 & 2 \\ 3 & 2 \\ \end{matrix} \right|$

= $3 \times 2 - 2 \times 3$

= $6 - 6$

 

Therefore, $\left| \text{C} \right|$ = $0$

 

(iv) D = $\left[ \begin{matrix} 3 & 2 \\ 1 & 4 \\ \end{matrix} \right]$

Determinant of matrix D is calculated as:

det D = $\left| \text{D} \right|$

= $\left| \begin{matrix} 3 & 2 \\ 1 & 4 \\ \end{matrix} \right|$

= $3 \times 4 - 2 \times 1$

= $12 - 2$

 

Therefore,

$\left| \text{D} \right|$ = $10$