Question

Let $A =\left[ a _{i j}\right], a _{i j} \in Z \cap[0,4], 1 \leq i, j \leq 2$ .
The number of matrices $A$ such that the sum of all entries is a prime number $p \in(2,13)$ is _____.

Answer 1

Correct Answer: 204

To solve the problem, we need to determine how many matrices \( A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \), where each \( a_{ij} \in \{0, 1, 2, 3, 4\} \), have entries summing to a prime \( p \) within the range \( (2,13) \). The possible prime numbers are 3, 5, 7, and 11. We compute the number of ways for each prime:

First, combine the entries such that \( a_{11} + a_{12} + a_{21} + a_{22} = p \) where \( p \) is a prime. Each \( a_{ij} \) can take values 0 through 4.
 

Let \(A = \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22}\end{bmatrix}\) \(a_{11}, a_{12}, a_{21}, a_{22} \text{ } \in \{0,1,2,3,4\}\)

Now, \(a_{11} + a_{12} + a_{23}+ a_{22} = \text{Prime no. = 3,5,7,11}\)

⇒ \((x^0 + x^1 + x^2 + x^3 + x^4)^4\)

⇒\(\left(\frac{1-x^5}{1-x}\right)^4\)

⇒ \((1-x^5). (1-x)^{-4}\)

⇒ \((4C_{r_1}.(-x^5)^{r_1} ). (^{4+r_2-1}C_{r_2}.x^{r_2})\)

⇒ \(4C_{r_1}\,.^{3+r_2}C_{r_2}.(-\.1)^{r_1}\,.(x^{5r_1+r_2})\)

∴ \(5r_1 + r_2 = 3,5,7,11\)

When,  \(5r_1 +r_2 = 3\)

⇒ \(r_1=0, \, r_2=3\)

\(5r_1 +r_2 = 5\)

⇒ \(r_1=0, \, r_2=5\)    Or   \(r_1=1, \, r_2=0\)

\(5r_1 +r_2 = 7\)

⇒ \(r_1=1, \, r_2=2\)    Or   \(r_1=0, \, r_2=7\)

\(5r_1 +r_2 = 11\)

⇒ \(r_1=0, \, r_2=11\)    Or   \(r_1=1, \, r_2=6\) Or   \(r_1=2, \, r_2=1\)

\(\text{Sum of all coefficients = }\)
    \(^4C_0 \times ^6C_3\times ^4C_0 \times ^8C_5 \times ^4C_1 \times ^3C_0 +\)\(^4C_0 \times ^{10}C_7  - ^4C_1\times ^5C_2 + ^4C_0\times ^{14}C_{11} - ^4C_1 \times ^9C_6 + ^4C_2 \times ^4C_1\)

\(= \frac{4!}{4!} \times \frac{6!}{3! \times 3!} + \frac{4!}{4!} \times \frac{8!}{5!\times3!} -  \frac{4!}{3!} \times \frac{3!}{3!}+  \frac{4!}{4!} \times \frac{10!}{7!\times3!}\)\(-  \frac{4!}{3!} \times \frac{5!}{3!\times2!}+  \frac{4!}{4!} \times \frac{14!}{11!\times3!} -  \frac{4!}{3!} \times \frac{9!}{6!\times3!}+   \frac{4!}{2!\times2!}\times \frac{4!}{3!}\)

\(= 20 + 56 - 4 + \left(\frac{10\times 9 \times 8}{3 \times 21}\right) - 4 \times 10 + \frac{14 \times 13 \times 12 }{6} - \frac{4 \times 9 \times 8 \times 7}{3 \times 2} + 24\)

\(=204\)