JEE Main 2026 · 4 Apr · Shift 1, Mathematics question 5

Trace 4. $A=I$ (trace 2) and $A=3I$ (trace 6) are excluded, so $A$ has eigenvalues 1 and 3: $a+d=4$ and $ad-bc=3$ (then Cayley-Hamilton gives $A^2-4A+3I=0$).

Count with entries in $\{0,\dots,4\}$:

• $(a,d)=(1,3)$ or $(3,1)$: $bc=0$, giving $5+5-1=9$ each.
• $(2,2)$: $bc=1$, so $b=c=1$: 1.
• $(0,4)$, $(4,0)$: $bc=-3$, impossible.

Total $9+9+1=19$.

(D)