given:-

 $\text{ A. }\left[\begin{array}{l}2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\\ 4\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}5\end{array}\right]=\left[\begin{array}{cc}0& -4\\ 10& 3\end{array}\right]$

Step-1:- Let $\mathrm{A}=\left[\begin{array}{l}\mathrm{a}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{b}\\ \mathrm{c}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{d}\end{array}\right]$
Step-2:- $\left[\begin{array}{l}\mathrm{a}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{b}\\ \mathrm{c}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{d}\end{array}\right]\left[\begin{array}{l}2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\\ 4\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}5\end{array}\right]$
Step-3:- do multiplication of matrices

             $\left[\begin{array}{l}2\mathrm{a}+4\mathrm{b}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\mathrm{a}+5\mathrm{b}\\ 2\mathrm{c}+4\mathrm{d}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\mathrm{c}+5\mathrm{d}\end{array}\right]$

Step-4:- $\left[\begin{array}{cc}2\mathrm{a}+4\mathrm{b}& 3\mathrm{a}+5\\ 2\mathrm{c}+4\mathrm{d}& 3\mathrm{c}+5\mathrm{d}\end{array}\right]=\left[\begin{array}{cc}0& -4\\ 10& 3\end{array}\right]$
Step-5:- Equate them:-

              2a + 4b = 0    ;     3a + 5b = – 4

              2c + 4d = 0          3c + 5d = 3

Step-6:-

             2a + 4b = 0 

             a + 2b = 0     ⇒ a = – 2b

             3a + 5b = –4

           – 6b + 5b = – 4  ⇒  b = 4 ⇒ a = –8 

Step-7:-

             2c + 4d = 10

             c + 2d = 5 ⇒ c = – 2d + 5

             3c + 5d = 3

             3( –2d + 5) + 5d = 3

            – 6d + 1s + 5d = 3

             d = 12 ⇒ c = – 19

Solution!-

$\mathrm{A}=\left[\begin{array}{l}-8\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} \hspace{0.25em}4\\ -19\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}12\end{array}\right]$