Given matrices,
$\left[ \begin{matrix}
p+3q & 3r+s \\
2p-q & r-2s \\
\end{matrix} \right]=\left[ \begin{matrix}
5 & 8 \\
3 & 5 \\
\end{matrix} \right]$
$p+3q=5$ … (i)
$2p–q=3$ … (ii)
$q=2p–3$
Now, substitute the value of q in equation (i), we get
$p+3(2p–3)=5$
$p+6p–9=5$
On transposing,
$7p=5+9$
$7p=14$
$p=14/7$
$p=2$
Again substitute the value of p in equation (i),
$2+3q=5$
$3q=5–2$
$3q=3$
$q=3/3$
$q=1$
Then,
$3r+s=8$ … (iii)
$r–2s=5$ … (iv)
$r=5+2s$
Now substitute the value of r in equation (iii),
$3(5+2s)+s=8$
$15+6s+s=8$
$7s=8–15$
$7s=– 7$
$s=-7/7$
$s=-1$
substitute the value of s in equation (iv),
$r–2s=5$
$r=5+2s$
$r=5+2(-1)$
$r=5–2$
$r=3$
Hence, the value of $p=2$, $q=1$, $r=3$ and $s=-1$.