JEE MAIN - Mathematics (2023 - 31st January Evening Shift - No. 18)
Let $\mathrm{A}=\left[\mathrm{a}_{i j}\right], \mathrm{a}_{i j} \in \mathbb{Z} \cap[0,4], 1 \leq i, j \leq 2$.
The number of matrices A such that the sum of all entries is a prime number $\mathrm{p} \in(2,13)$ is __________.
The number of matrices A such that the sum of all entries is a prime number $\mathrm{p} \in(2,13)$ is __________.
Answer
204
Explanation
$A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{13} & a_{14}\end{array}\right]$
Such that $\Sigma a_{i i}=3,5,7$ or 11
Then for sum 3, the possible entries are $(0,0,0,3)$, $(0,0,1,2),(0,1,1,1)$.
Then total number of possible matrices $$ =4+12+4 $$ $=20$
For sum 5 the possible entries are $(0,0,1,4)$, $(0,0,2,3),(0,1,2,2),(0,1,1,3)$ and $(1,1,1,2)$.
$\therefore $ Total possible matrices $=12+12+12+12+4=52$
For sum 7 the possible entries are $(0,0,3,4)$, $(0,2,2,3),(0,1,2,4),(0,1,3,3),(1,2,2,2)$, $(1,1,2,3)$ and $(1,1,1,4)$.
$\therefore $ Total possible matrices $=80$
For sum 11 the possible entries are $(0,3,4,4)$, $(1,2,4,4),(2,3,3,3),(2,2,3,4)$.
$\therefore $ Total number of matrices $=52$
$\therefore $ Total required matrices $=20+52+80+52$ $$ =204 $$
Such that $\Sigma a_{i i}=3,5,7$ or 11
Then for sum 3, the possible entries are $(0,0,0,3)$, $(0,0,1,2),(0,1,1,1)$.
Then total number of possible matrices $$ =4+12+4 $$ $=20$
For sum 5 the possible entries are $(0,0,1,4)$, $(0,0,2,3),(0,1,2,2),(0,1,1,3)$ and $(1,1,1,2)$.
$\therefore $ Total possible matrices $=12+12+12+12+4=52$
For sum 7 the possible entries are $(0,0,3,4)$, $(0,2,2,3),(0,1,2,4),(0,1,3,3),(1,2,2,2)$, $(1,1,2,3)$ and $(1,1,1,4)$.
$\therefore $ Total possible matrices $=80$
For sum 11 the possible entries are $(0,3,4,4)$, $(1,2,4,4),(2,3,3,3),(2,2,3,4)$.
$\therefore $ Total number of matrices $=52$
$\therefore $ Total required matrices $=20+52+80+52$ $$ =204 $$
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