JEE Main 2023 — Matrices Question with Solution

Given,

$A=\left[\begin{array}{ll}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right] \& a_{11},a_{12},a_{21},a_{22}\in \left\{0,1,2,3,4\right\}$

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 $x^3, x^5, x^7 \& x^{11}$in expansion of ${\left(x^0+x^1+x^2+x^3+x^4\right)}^4$

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

$=\left({}^{4}C_{r_1}{\left(-x^5\right)}^{r_1}\right)\left({}^{4+r_2-1}C_{r_2}{x}^{r^2}\right)$

$={}^{4}C_{r1}^{3+r_2}{C}_{r2}{\left(-1\right)}^{r_1}{\left(x\right)}^{{}^{5r_1+r_2}}$

Now taking, $5r_1+r_2=3,5,7,11$

when $5r_1+r_2=3\Rightarrow r_1=0,r_2=3$

when $5r_1+r_2=5\Rightarrow r_1=0,r_2=5$ or $r_1=1,r_2=0$

when $5r_1+r_2=7\Rightarrow r_1=1,r_2=2$ or $r_1=0,r_2=7$

when $5r_1+r_2=11\Rightarrow r_1=0,r_2=11$ or $r_1=1,r_2=6$ or $r_1=2,r_2=1$

So, sum of all coefficient $={}^{4}C_0\times {}^{6}C_3+{}^{4}C_0{}^{8}C_5-{}^{4}C_1{}^{3}C_0+{}^{4}C_0{}^{10}C_7-{}^{4}C_1{}^{5}C_2+{}^{4}C_0{}^{14}C_{11}-{}^{4}C_1{}^{9}C_6+{}^{4}C_2{}^{4}C_1$

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

$=204$