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  • Let <img alt="\mathrm{\mathrm{A}=\left[\mathrm{a}_{i j}\right], \mathrm{a}_{i j} \in Z \cap[0,4], 1 \leq i, j \leq 2}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cmathrm%7BA

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

Option: 1

204


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{A=\left[a_{i j}\right], \text { aij } \in Z \cap[0,4], \text { so } \mathrm{a}_{\mathrm{ij}}=\{0,1,2,3,4\} }
\mathrm{ A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right]_{2 \times 2}}
\mathrm{ a_{11}+a_{12}+a_{21}+a_{22}=P \quad where\: P \: is \: a \: prime\: number }
\mathrm{a_{11}+a_{12}+a_{21}+a_{22}=3,5,7,11 }
\mathrm{\left(1+x+x^2+\mathrm{x}^3+\mathrm{x}^4\right)^4 }
\mathrm{\left(1-x^5\right)^4(1-\mathrm{x})^{-4} }

\mathrm{\left(1-4 x^5+6 x^{10}-4 x^{15}+x^{20}\right)\left(1+{ }^4 \mathrm{C}_1 \mathrm{x}+{ }^5 \mathrm{C}_2 \mathrm{x}^2+{ }^6 \mathrm{C}_3 \mathrm{x}^3+{ }^7 \mathrm{C}_4 \mathrm{x}^7 \ldots\right)}
\mathrm{For \: 3{ }^6 \mathrm{C}_3=\frac{6 \times 5 \times 4}{6}=20.}
\mathrm{For \: 5(-4)+{ }^8 \mathrm{C}_5=-4+56=52.}
\mathrm{For \: 7(-4){ }^5 \mathrm{C}_2+{ }^{10} \mathrm{C}_7=-40+120=80.}
\mathrm{For \: 11{ }^{14} \mathrm{C}_{11}+(-4){ }^9 \mathrm{C}_6+6\left({ }^4 \mathrm{C}_1\right)=364-336+24=52}

Total number of matrices =204

 

\mathrm{ a_{11}+a_{12}+a_{22}+a_{12}=3 }

\begin{array}{lllll} 0 & 0 & 0 & 3 & =\frac{4 !}{3 !}=4 \\\\ 0 & 0 & 2 & 1 & =\frac{4 !}{2 !}=12 \end{array}\\

0 \quad 1 \quad 1 \quad 1=\frac{\frac{4 !}{3 !}=4}{20}

 

a_{11}+a_{12}+a_{21}+a_{22}=5

\begin{array}{lcccc} 0 & 0 & 1 & 4 & =\frac{4 !}{2 !}=12 \\\\ 0 & 0 & 2 & 3 & =\frac{4 !}{2 !}=12 \\\\ 0 & 1 & 1 & 3 & =\frac{4 !}{2 !}=12 \\\\ 0 & 1 & 2 & 2 & =\frac{4 !}{2 !}=12 \end{array}

1 \quad 1 \quad 1 \quad 2=\frac{\frac{4 !}{3 !}=4}{52}

 

a_{11}+a_{12}+a_{21}+a_{22}=7

\begin{array}{lllll} 0 & 0 & 3 & 4 & =\frac{4 !}{2 !}=12 \\\\ 0 & 1 & 3 & 3 & =\frac{4 !}{2 !}=12 \\\\ 0 & 1 & 2 & 4 & =4 !=24 \\\\ 1 & 1 & 1 & 4 & =\frac{4 !}{3 !}=4 \end{array}

\begin{array}{lllll} 0 & 2 & 2 & 3 & =\frac{4 !}{3 !}=12 \\\\ 1 & 1 & 2 & 3 & =\frac{4 !}{2 !}=12 \end{array}

1 \quad 2 \quad 2 \quad 2=\frac{\frac{4 !}{3 !}=4}{80}

 

a_{11}+a_{12}+a_{21}+a_{22}=11

\begin{array}{lllll} 0 & 3 & 4 & 4 & =\frac{4 !}{2 !}=12 \\\\ 1 & 2 & 4 & 4 & =\frac{4 !}{2 !}=12 \end{array}

\begin{array}{lllll} 1 & 3 & 3 & 4 & =\frac{4 !}{2 !}=12 \\\\ 2 & 3 & 3 & 4 & =\frac{4 !}{3 !}=12 \end{array}

2 \quad 3 \quad 3 \quad 3=\frac{\frac{4 !}{3 !}=4}{52}


 

\mathrm{ \text { total matrix is }=20+52+80+52=204 \text {. } }

                                  

 

 

 

Posted by

Ritika Kankaria

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