Let A = ⌊ a _ ^ i ^ j ⌋ · a _ ij ∈ Z [0, 4], 1 ≤ i, j ≤ 2 . The...

Given,

A = \left[\begin{cases}{a}_{11} & {a}_{12} \\ {a}_{21} & {a}_{22}\end{cases}\right] \text{ and } {a}_{11}, {a}_{12}, {a}_{21}, {a}_{22} \in \left\{0, 1, 2, 3, 4\right\}

Now given ${a}_{11} + {a}_{12} + {a}_{23} + {a}_{22} =$ prime number,

So sum can be, $p = 3, 5, 7, 11$

Now using multinomial theorem we get,

Sum will be sum of coefficient of ${x}^{3}, {x}^{5}, {x}^{7} \text{ and } {x}^{11}$in expansion of ${\left({x}^{0} + {x}^{1} + {x}^{2} + {x}^{3} + {x}^{4}\right)}^{4}$

$⇒ {\left(\dfrac{1 - {x}^{5}}{1 - x}\right)}^{4} = {\left(1 - {x}^{5}\right)}^{4}{\left(1 - x\right)}^{- 4}$

$= \left(C{r}_{1} 4 {\left(- {x}^{5}\right)}^{{r}_{1}}\right)\left(C{r}_{2} 4 + {r}_{2} - 1 {x}^{{r}^{2}}\right)$

$= Cr13 + {r}_{2} 4 {C}_{r2}{\left(- 1\right)}^{{r}_{1}}{\left(x\right)}^{{}^{5{r}_{1} + {r}_{2}}}$

Now taking, $5{r}_{1} + {r}_{2} = 3, 5, 7, 11$

when $5{r}_{1} + {r}_{2} = 3 ⇒ {r}_{1} = 0, {r}_{2} = 3$

when $5{r}_{1} + {r}_{2} = 5 ⇒ {r}_{1} = 0, {r}_{2} = 5$ or ${r}_{1} = 1, {r}_{2} = 0$

when $5{r}_{1} + {r}_{2} = 7 ⇒ {r}_{1} = 1, {r}_{2} = 2$ or ${r}_{1} = 0, {r}_{2} = 7$

when $5{r}_{1} + {r}_{2} = 11 ⇒ {r}_{1} = 0, {r}_{2} = 11$ or ${r}_{1} = 1, {r}_{2} = 6$ or ${r}_{1} = 2, {r}_{2} = 1$

So, sum of all coefficient $= C0 4 \times C3 6 + C0 4 C5 8 - C1 4 C0 3 + C0 4 C7 10 - C1 4 C2 5 + C0 4 C11 14 - C1 4 C6 9 + C2 4 C1 4$

$= 20 + 56 - 4 + 120 - 40 + 364 - 336 + 24$

$= 204$