Mumbai University > First Year Engineering > Sem 1 > Applied Maths 1

Marks : 8 M

Year : Dec 2013

The given system of equations can be written in matrix form AX=B as follows

$ \left[ \begin{array}{ccc} 2& 3& 1\\ 5& 1& 1\\ 3& 2& 4 \end{array}\right] \left[ \begin{array}{ccc} x\\ y \\ z \end{array}\right] \;=\; \left[ \begin{array}{ccc} -1 \\ 9 \\ 11 \end{array}\right] \\ \; \\ \; \\ $

Let, A=LU such that

$ \left[ \begin{array}{ccc} 2& 3& 1\\ 5& 1& 1\\ 3& 2& 4 \end{array}\right] \;=\; \left[ \begin{array}{ccc} a& 0& 0\\ b& c& 0\\ d& e& f \end{array}\right] \left[ \begin{array}{ccc} 1& g& h\\ 0& 1& i\\ 0& 0& 1 \end{array}\right] \;=\; \left[ \begin{array}{ccc} a& ag& ah\\ b& bg+c& bh+ci\\ d& dg+e& dh+ei+f \end{array}\right] \\ \; \\ \; \\ \; \\ \; \\ \therefore a=2 , b=5 , d=3 \\ \; \\ \; \\ ag=3 \; \therefore g\;=\;\dfrac{3}{2} \\ \; \\ \; \\ bg+c=1 \; \; \therefore c=1-5(3/2) \;=\; \dfrac{-13}{2} \\ \; \\ \; \\ dg+e\;=\; 2 \; \; \therefore e\;=\; 2- 3 . \dfrac{3}{2} \;=\; \dfrac{-5}{2} \\ \; \\ \; \\ ah=1 \; \; \therefore h=1/2 \\ \; \\ \; \\ bh+ci=-3 \; \; \; \therefore (5)(1/2)+(-13/2)i=1 \\ \; \\ \therefore i=\dfrac{1-5/2}{-13/2} \;=\; \dfrac{-3}{2} \times \dfrac{-2}{12} \;=\; \dfrac{3}{13} \\ \; \\ \; \\ dh+ei+f=4 \\ \; \\ \therefore 3(\dfrac{1}{2})+\dfrac{-5}{2} \cdot \dfrac{3}{13} +f=4 \; \; \therefore f \;=\; 4-\dfrac{3}{2} +\dfrac{15}{26} \;=\; \dfrac{104-39+15}{26} \;=\; \dfrac{80}{26} \;=\; \dfrac{40}{13} \\ \; \\ \; \\ \therefore L \;=\; \left[ \begin{array}{ccc} 2& 0& 0\\ 5& -13/2& 0\\ 3& -5/2& 40/3 \end{array}\right] \; \; and \; U \;=\; \left[ \begin{array}{ccc} 1& 3/2& 1/2\\ 0& 1& 3/13\\ 0& 0& 1 \end{array}\right] \\ \; \\ \; \\ \; \\ Now, \; \; LY=B \; \; Where Y\;=\;UX \\ \; \\ \; \\ \therefore \left[ \begin{array}{ccc} 2& 0& 0\\ 5& -13/2& 0\\ 3& -5/2& 40/3 \end{array}\right] \left[ \begin{array}{ccc} y_1 \\ y_2 \\ y_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} -1 \\ 9 \\ 11 \end{array}\right] \\ \; \\ \; \\ \therefore 2y_1=-1 \; \; \therefore y_1=-1/2 \\ \; \\ \; \\ 5y_1-\dfrac{13}{2}y_2=19 \; \therefore -\dfrac{13}{2}y_2\;=\; 9-5(-1/2) \;=\; 9+\dfrac{5}{2} \;=\; \dfrac{23}{2} \\ \; \therefore y_2= \dfrac{23}{2} \cdot \dfrac{2}{-13} \;=\; \dfrac{-23}{13} \\ \; \\ \; \\ 3y_1 - \dfrac{5}{2} \cdot \dfrac{-23}{13} + \dfrac{40}{13} y_3 \;=\; 11 \\ \therefore \dfrac{40}{13} y_3 \;=\; 11+\dfrac{3}{2} - \dfrac{115}{26} \;=\; \dfrac{210}{26} \; \; \; \; \therefore y_3 \;=\; \dfrac{210}{26} \times \dfrac{13}{40} \;=\; \dfrac{21}{8} \\ \; \\ \; \\ Now \; UX=Y \\ \; \\ \; \\ \; \\ \therefore \left[ \begin{array}{ccc} 1& 3/2& 1/2\\ 0& 1& 3/13\\ 0& 0& 1 \end{array}\right] \left[ \begin{array}{ccc} x \\ y \\ z \end{array}\right] \;=\; \left[ \begin{array}{ccc} -1/2 \\ -23/12 \\ 21/8 \end{array}\right] \\ \; \\ \; \\ \; \\ \therefore z=\dfrac{21}{8} \;=\; 2.625 \\ \; \\ \; \\ y+\dfrac{3}{13}z=-\dfrac{23}{13} \; \; \; \therefore y\;=\;\dfrac{-23}{13} - \dfrac{3}{13} . \dfrac{21}{8} \;=\; \dfrac{-247}{104} \;=\; -2.375 \\ \; \\ \; \\ x-\dfrac{3}{2}y+\dfrac{1}{2}z=\dfrac{-1}{2} \; \; x= \dfrac{-3}{2} . \dfrac{-247}{104} \;-\; \dfrac{1}{2} . \dfrac{21}{8} \;-\; \dfrac{1}{2} \\ \therefore x\;=\; 1.75 \\ \; \\ \; \\ \; \\ \therefore x=1.75, y=-2.375, z=2.625 $