Matrices Question 13

Question 13 - 31 January - Shift 2

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

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Answer: (196)

Solution:

Formula: Combinations under restrictions

As given $a+b+c+d=3$ or 5 or 7 or 11

if sum $=3$

$(1+x+x^{2}+\ldots .+x^{4})^{4} \to x^{3}$

$(1-x^{5})^{4}(1-x)^{-4} \to x^{3}$

$\therefore{ }^{4+3-1} C_3={ }^{6} C_3=20$

If sum $=5$

$(1-4 x^{5})(1-x)^{-4} \to x^{5}$

$\Rightarrow{ }^{4+5-1} C_5-4 x^{4.4+0-1} C_0={ }^{8} C_5-4=52$

If sum $=7$

$(1-4 x^{5})(1-x)^{-4} \to x^{7}$

$\Rightarrow{ }^{4+5-1} C_4-{ }^{4.4+0-1} C_0={ }^{8} C_5-4=52$

If sum $=11$

$\begin{aligned} & (1-4 x^{5}+6 x^{10})(1-x)^{-4} \to x^{11} \\ \Rightarrow & { }^{4+11-1} C _{11}-4 \cdot{ }^{4+6-4} C_6+6 \cdot{ }^{4+1-1} C_1 \\ = & { }^{14} C _{11}-4 .{ }^{9} C_6+6 \cdot 4=364-336+24=52\end{aligned}$

$\therefore$ Total matrices $=20+52+80+52=204$