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The sum of three numbers is 6. If we multiply third number by 3 and add second number to it, we get 11. By adding first and third number, we get double of the second number. Represent it algebraically and find the numbers using matrix method.
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The sum of three numbers is 6. If we multiply third number by 3 and add second number to it, we get 11. By adding first and third number, we get double of the second number. Represent it algebraically and find the numbers using matrix method.

Answer

\(\mathrm{x}=1,\mathrm{y}=2,\mathrm{z}=3\)
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Solution

Let first ,second and third number be denoted by \(\mathrm{x},\mathrm{y}\)and \(\mathrm{z}\), respectively.Then, according to given conditions,we have,
\(\mathrm{x}+\mathrm{y}+\mathrm{z}=6\)
\(\mathrm{y}+3\mathrm{z}=11\)
\(\mathrm{x}+\mathrm{z}=2\mathrm{y}\Rightarrow \mathrm{x}-2\mathrm{y}+\mathrm{z}=0\)
This system can be written in matrix form as \(\mathrm{A}\mathrm{X}=\mathrm{B}\) where
\(\mathrm{A}=\left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 3\\ 1& -2& 1\end{array}\right]\mathrm{X}=\left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]\mathrm{B}=\left[\begin{array}{c}6\\ 11\\ 0\end{array}\right]\)
\(\left\vert \mathrm{A}\right\vert =1(1+6)-(0-3)+(0-1)=9\neq 0\)
Since matrix is non singular, it's inverse exists.
\({\mathrm{A}}_{11}=(-1){}^{1+1}\left\vert \begin{array}{ll}1& 3\\ -2& 1\end{array}\right\vert =7,{\mathrm{A}}_{12}=(-1){}^{1+2}\left\vert \begin{array}{ll}0& 3\\ 1& 1\end{array}\right\vert =3,{\mathrm{A}}_{13}=(-1){}^{1+3}\left\vert \begin{array}{ll}0& 1\\ 1& -2\end{array}\right\vert =-1\)
\({\mathrm{A}}_{21}=(-1){}^{2+1}\left\vert \begin{array}{ll}1& 1\\ -2& 1\end{array}\right\vert =-3,{\mathrm{A}}_{22}=(-1){}^{2+2}\left\vert \begin{array}{ll}1& 1\\ 1& 1\end{array}\right\vert =0,{\mathrm{A}}_{23}=(-1){}^{2+3}\left\vert \begin{array}{ll}1& 1\\ 1& -2\end{array}\right\vert =3\)
\({\mathrm{A}}_{31}=(-1){}^{3+1}\left\vert \begin{array}{ll}1& 1\\ 1& 3\end{array}\right\vert =2,{\mathrm{A}}_{32}=(-1){}^{3+2}\left\vert \begin{array}{ll}1& 1\\ 0& 3\end{array}\right\vert =-3,{\mathrm{A}}_{33}=(-1){}^{3+3}\left\vert \begin{array}{ll}1& 1\\ 0& 1\end{array}\right\vert =1\)
\(\mathrm{a}\mathrm{d}\mathrm{j}\mathrm{A}=\left[\begin{array}{ccc}7& -3& 2\\ 3& 0& -3\\ -1& 3& 1\end{array}\right]\)
\({\mathrm{A}}^{-1}=\frac{1}{\left\vert \mathrm{A}\right\vert }\mathrm{a}\mathrm{d}\mathrm{j}\mathrm{A}=\frac{1}{9}\left[\begin{array}{ccc}7& -3& 2\\ 3& 0& -3\\ -1& 3& 1\end{array}\right]\)
\(\mathrm{X}={\mathrm{A}}^{-1}\mathrm{B}\)
\(\left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=\frac{1}{9}\left[\begin{array}{ccc}7& -3& 2\\ 3& 0& -3\\ -1& 3& 1\end{array}\right]\left[\begin{array}{c}6\\ 11\\ 0\end{array}\right]\)
\(\left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=\frac{1}{9}\left[\begin{array}{c}42-33+0\\ 18+0+0\\ -6+33+0\end{array}\right] \begin{array}{l}=\frac{1}{9}\left[\begin{array}{c}9\\ 18\\ 27\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]\end{array}\)
Hence, \(\mathrm{x}=1,\mathrm{y}=2,\mathrm{z}=3\)
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