JEE Mathematics 2023 Question with Solution

Given: $A$ is a $2 \times 2$ matrix with each entry chosen from $\{0,1,2,3,4\}$.

Find: The number of matrices for which the sum of all four entries is a prime number in the required range.

Let the four entries be $a,b,c,d$. Then we need the number of solutions of

$$a+b+c+d = s$$

with $a,b,c,d \in \{0,1,2,3,4\}$, where the relevant prime sums used in the solution are $s=3,5,7,11$.

Use the generating function

$$(1+x+x^2+x^3+x^4)^4$$

and extract coefficients of $x^s$.

For $s=3$,

$$(1+x+x^2+x^3+x^4)^4 = \left(\frac{1-x^5}{1-x}\right)^4$$

The coefficient of $x^3$ is

$$\binom{4+3-1}{3} = \binom{6}{3} = 20$$

So the number of matrices with sum $3$ is $20$.

For $s=5$, the solution gives

$$\binom{4+5-1}{5} - 4\binom{4+0-1}{0} = \binom{8}{5} - 4 = 56-4=52$$

So the number of matrices with sum $5$ is $52$.

For $s=7$, the solution totals this case as $80$.

For $s=11$, using the correction terms from the generating function expansion,

$$\binom{14}{11} - 4\binom{9}{6} + 6\binom{4}{1} = 364 - 336 + 24 = 52$$

So the number of matrices with sum $11$ is $52$.

Therefore, total matrices

$$20 + 52 + 80 + 52 = 204$$

Hence, the required number of matrices is $204$.

Given: Each entry of the $2 \times 2$ matrix belongs to $\{0,1,2,3,4\}$.

Find: The number of matrices whose total sum is one of the prime values counted in the solution.

Each matrix corresponds to an ordered quadruple $(a,b,c,d)$ with

$$a,b,c,d \in \{0,1,2,3,4\}$$

The generating function for one entry is

$$1+x+x^2+x^3+x^4$$

Hence for four entries it is

$$(1+x+x^2+x^3+x^4)^4$$

The coefficient of $x^s$ gives the number of matrices with total sum $s$.

1. For sum $3$, upper bounds do not interfere, so the count is the number of non-negative solutions of

$$a+b+c+d=3$$

which is

$$\binom{6}{3}=20$$

2. For sum $5$, first count all non-negative solutions:

$$\binom{8}{5}=56$$

Now subtract the four cases where one variable is at least $5$. That leaves

$$56-4=52$$

3. For sum $7$, the solution reports this case contributes $80$.

4. For sum $11$, inclusion-exclusion gives

$$\binom{14}{11} - 4\binom{9}{6} + 6\binom{4}{1}$$

that is,

$$364-336+24=52$$

Adding all reported valid cases,

$$20+52+80+52=204$$

Therefore, the answer is $204$.