Solution:

A matrix is said to be singular if its determinant is equal to zero. i.e., $\left| \text{A} \right|=0$

 

(i) A = $\left[ \begin{matrix} 3 & 6 \\ 2 & 4 \\ \end{matrix} \right]$

Determinant of matrix A is calculated as:

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

= $\left| \begin{matrix} 3 & 6 \\ 2 & 4 \\ \end{matrix} \right|$

= $3 \times 4 - 2 \times 6$

= $12 – 12$

= $0$

 

 

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

As determinant of A is equal to zero so, A is a singular matrix.

 

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

Determinant of matrix B is calculated as:

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

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

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

= $8 - 3$

= $5$

 

Therefore, $\left| \text{B} \right|$ $\ne $ $0$

As determinant of B is not equal to zero, so B is a non-singular matrix.

 

(iii) C = $\left[ \begin{matrix} 7 & -9 \\ 3 & 5 \\ \end{matrix} \right]$

Determinant of matrix C is calculated as:

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

= $\left| \begin{matrix} 7 & -9 \\ 3 & 5 \\ \end{matrix} \right|$

= $7 \times 5 - 3 \times (-9)$

= $35 + 27$

= $62$

 

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

As determinant of C is not equal to zero, so C is a non-singular matrix.

 

(iv) D = $\left[ \begin{matrix} 5 & -10 \\ -2 & 4 \\ \end{matrix} \right]$

Determinant of matrix D is calculated as:

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

= $\left[ \begin{matrix} 5 & -10 \\ -2 & 4 \\ \end{matrix} \right]$

= $5 \times 4 - (-10) \times (-2)$

= $20 - 20$

= $0$

 

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

As determinant of D is equal to zero so, D is a singular matrix.