Ten students were selected from a school on the basis of values for giving awards and were divided into three groups. The first group comprises hard workers,the second group has honest and law abiding students and the third group consists vigilant and obedient students. Double the number of students of the first group added to the number in the second group gives 13, while the combined strength of first and second group is 4 times that of the third group. Find the number of students in each group.

Hence from the question x + y + z = 10

And 2x + y = 13

And x + y = 4z

Hence forming a matrix ,we get

$\left[\begin{array}{ccc}1& 1& 1\\ 2& 1& 0\\ 1& 1& -4\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}10\\ 13\\ 0\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

AX = B

Hence X = A

^{-1}B

And A

^{ -1}= adj(A)/|A|

And |A| = 1(-4-0) -1(-8 -0) + 1(2-1) = -4 + 8 + 1 = 5

And A

_{11}= ( -4 -0 ) = -4 , A

_{12}= -1(-8 -0 ) = 8 , A

_{13}= 1(2-1) = 1

A

_{21}= -1(-4 -1) = 5 , A

_{22}= 1(-4 -1) = -5 , A

_{23}= 0

A

_{31}= 1(0-1) = -1 , A

_{32}= -1(0-2) = 2 and A

_{33}

_{ }= -1(1-2) = 1

So A

^{-1 }= $\frac{1}{5}\left[\begin{array}{ccc}-4& 5& -1\\ 8& -5& 2\\ 1& 0& 1\end{array}\right]$

So $X=\frac{1}{5}\left[\begin{array}{ccc}-4& 5& -1\\ 8& -5& 2\\ 1& 0& 1\end{array}\right]\left[\begin{array}{c}10\\ 13\\ 0\end{array}\right]=\left[\begin{array}{c}5\\ 3\\ 2\end{array}\right]\phantom{\rule{0ex}{0ex}}Hence\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}5\\ 3\\ 2\end{array}\right]\phantom{\rule{0ex}{0ex}}Orx=5,y=3,z=2$

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