Strassen's Matrix Multiplication Recurrence:
$T(n) = 7T(n/2) + n^{2}$ = $n^{\log_{2}7}$
Given $T(n)=aT(n/4)+Θ(n2)T(n)=aT(n/4)+Θ(n2)$ = $n^{\log_{4}a}$
For Professor Caesar’s algorithm to be asymptotically faster than Strassen’s algorithm
-> $n^{\log_{4}a}$ $<$ $n^{\log_{2}7}$
-> ${\log_{4}a}$ $<$ ${\log_{2}7}$
solving above equation we get $a$ $<$ $49$ , hence largest value of $a = 48$