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 $=20+52+80+52=204$
