Let A be a matrix of order 2× 2, whose entries are from the set 0,1,2,3,4,5. If the sum of all the entries of A is a prime number p,2<p<8, then the…

Given, $A=[\begin{matrix}a & b \\ c & d\end{matrix}⌋$A=[ac​bd​⌋ and $a,b,c,d\in {0,1,2,3,4,5}$a,b,c,d∈0,1,2,3,4,5

Let, $a+b+c+d=p,p\in {3,5,7}$a+b+c+d=p,p∈3,5,7

Case-i

$a+b+c+d=3$a+b+c+d=3 where $a,b,c,d\in {0,1,2,3}$a,b,c,d∈0,1,2,3

Entries will be $(0,0,0,3)\text{or}(0,0,1,2)\text{or}(0,1,1,1)$(0,0,0,3)or(0,0,1,2)or(0,1,1,1)

Number of ways $=\frac{4!}{3!}+\frac{4!}{2!}+\frac{4!}{3!}=20\cdots (1)$=3!4!​+2!4!​+3!4!​=20⋯(1)

Case-ii

$a+b+c+d=5$a+b+c+d=5 where $a,b,c,d\in {0,1,2,3,4,5}$a,b,c,d∈0,1,2,3,4,5

Entries will be $(0,0,0,5)\text{or}(0,0,1,4)\text{or}(0,0,2,3)\text{or}(0,3,1,1)\text{or}(0,1,2,2)\text{or}(1,1,1,2)$(0,0,0,5)or(0,0,1,4)or(0,0,2,3)or(0,3,1,1)or(0,1,2,2)or(1,1,1,2)

Number of ways $=\frac{4!}{3!}+\frac{4!}{2!}+\frac{4!}{2!}+\frac{4!}{2!}+\frac{4!}{2!}+\frac{4!}{3!}=56\cdots (2)$=3!4!​+2!4!​+2!4!​+2!4!​+2!4!​+3!4!​=56⋯(2)

Case-iii

$$a+b+c+d=7$$a+b+c+d=7

Number of ways $=$= total number of ways when $a,b,c,d\in {0,1,2,3,4,5,6,7}$a,b,c,d∈0,1,2,3,4,5,6,7-total number of ways when $a,b,c,d\notin {6,7}$a,b,c,d∈/6,7

Number of ways ${=}^{7+4-1}{C}_{4-1}-(\frac{4!}{3!}+\frac{4!}{2!})$=7+4−1C4−1​−(3!4!​+2!4!​)

$$=C310-16=104\cdots (3)$$=C310−16=104⋯(3)

Hence adding all equation so total number of ways $=180$=180