Wee have:
$3\left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right]=\left[\begin{array}{cc}\mathrm{a}& 6\\ -1& 2\mathrm{d}\end{array}\right]+\left[\begin{array}{cc}4& \mathrm{a}+\mathrm{b}\\ \mathrm{c}+\mathrm{d}& 3\end{array}\right]\mathrm{.}$
$\Rightarrow \left[\begin{array}{cc}3\mathrm{a}& 3\mathrm{b}\\ 3\mathrm{c}& 3\mathrm{d}\end{array}\right]=\left[\begin{array}{cc}\mathrm{a}+4& \mathrm{a}+\mathrm{b}+6\\ \mathrm{c}+\mathrm{d}-1& 2\mathrm{d}+3\end{array}\right]$
Now, comparing both matrices we get:
3a = a + 4  ⇒   2a = 4 ⇒ a = 2        ....(i)
3b = a + b + 6
putting a = 2 we get:
3b – b = 2 + 6 ⇒ 2b = 8 ⇒ b = 4    ....(ii)
3c = c + d – 1 ....(iii)
3d = 2d + 3 ⇒ d = 3      ...(iv)
putting the value of d in equation (iii) we get:
3c = c + 3 – 1 ⇒ 2c = 2 ⇒ c = 1
Hence, a = 2, b = 4, c = 1 and d = 3