JEE Main 2023 — Mathematics Algebra

Given,

A=[\begin{matrix}{a}_{11} & {a}_{12} \\ {a}_{21} & {a}_{22}\end{matrix}]&{a}_{11},{a}_{12},{a}_{21},{a}_{22}\in {0,1,2,3,4}

Now given ${a}_{11}+{a}_{12}+{a}_{23}+{a}_{22}=$a11​+a12​+a23​+a22​= prime number,

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

Now using multinomial theorem we get,

Sum will be sum of coefficient of {x}^{3},{x}^{5},{x}^{7}&{x}^{11}in expansion of ${({x}^{0}+{x}^{1}+{x}^{2}+{x}^{3}+{x}^{4})}^{4}$(x0+x1+x2+x3+x4)4

$$\Rightarrow {(\frac{1-{x}^{5}}{1-x})}^{4}={(1-{x}^{5})}^{4}{(1-x)}^{-4}$$⇒(1−x1−x5​)4=(1−x5)4(1−x)−4

$$=(C{r}_{1}4{(-{x}^{5})}^{{r}_{1}})(C{r}_{2}4+{r}_{2}-1{x}^{{r}^{2}})$$=(Cr1​4(−x5)r1​)(Cr2​4+r2​−1xr2)

$$=Cr13+{r}_{2}4{C}_{r2}{(-1)}^{{r}_{1}}{(x)}^{{}^{5{r}_{1}+{r}_{2}}}$$=Cr13+r2​4Cr2​(−1)r1​(x)5r1​+r2​

Now taking, $5{r}_{1}+{r}_{2}=3,5,7,11$5r1​+r2​=3,5,7,11

when $5{r}_{1}+{r}_{2}=3\Rightarrow {r}_{1}=0,{r}_{2}=3$5r1​+r2​=3⇒r1​=0,r2​=3

when $5{r}_{1}+{r}_{2}=5\Rightarrow {r}_{1}=0,{r}_{2}=5$5r1​+r2​=5⇒r1​=0,r2​=5 or ${r}_{1}=1,{r}_{2}=0$r1​=1,r2​=0

when $5{r}_{1}+{r}_{2}=7\Rightarrow {r}_{1}=1,{r}_{2}=2$5r1​+r2​=7⇒r1​=1,r2​=2 or ${r}_{1}=0,{r}_{2}=7$r1​=0,r2​=7

when $5{r}_{1}+{r}_{2}=11\Rightarrow {r}_{1}=0,{r}_{2}=11$5r1​+r2​=11⇒r1​=0,r2​=11 or ${r}_{1}=1,{r}_{2}=6$r1​=1,r2​=6 or ${r}_{1}=2,{r}_{2}=1$r1​=2,r2​=1

So, sum of all coefficient $=C04\times C36+C04C58-C14C03+C04C710-C14C25+C04C1114-C14C69+C24C14$=C04×C36+C04C58−C14C03+C04C710−C14C25+C04C1114−C14C69+C24C14

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

$$=204$$=204