Q : 16         Solve the system of equations

                    \frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4

                    \frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1

                    \frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2

 

Answers (1)
D Divya Prakash Singh

We have a system of equations;

   \frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4

  \frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1

  \frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2

So, we will convert the given system of equations in a simple form to solve the problem by the matrix method;

Let us take, \frac{1}{x} = a\frac{1}{y} = b\ and\ \frac{1}{z} = c

Then we have the equations;

  2a +3b+10c = 4

  4a-6b+5c =1

  6a+9b-20c = 2

We can write it in the matrix form as AX =B , where

A= \begin{bmatrix} 2 &3 &10 \\ 4& -6 & 5\\ 6 & 9 & -20 \end{bmatrix} , X = \begin{bmatrix} a\\b \\c \end{bmatrix}\ and\ B = \begin{bmatrix} 4\\1 \\ 2 \end{bmatrix}.

Now, Finding the determinant value of A;

|A| = 2(120-45)-3(-80-30)+10(36+36)

       =150+330+720

       =1200 \neq 0

Hence we can say that A is non-singular \therefore its invers exists;

Finding cofactors of A;

A_{11} = 75 , A_{12} = 110A_{13} = 72

A_{21} = 150A_{22} = -100A_{23} = 0

A_{31} =75A_{31} =30A_{33} =-24

\therefore as we know A^{-1} = \frac{1}{|A|}adjA

= \frac{1}{1200} \begin{bmatrix} 75 &150 &75 \\ 110& -100& 30\\ 72&0 &-24 \end{bmatrix}

Now we will find the solutions by relation X = A^{-1}B.

\Rightarrow \begin{bmatrix} a\\b \\ c \end{bmatrix} = \frac{1}{1200} \begin{bmatrix} 75 &150 &75 \\ 110& -100& 30\\ 72&0 &-24 \end{bmatrix}\begin{bmatrix} 4\\1 \\ 2 \end{bmatrix}

               = \frac{1}{1200}\begin{bmatrix} 300+150+150\\440-100+60 \\ 288+0-48 \end{bmatrix}

              = \frac{1}{1200}\begin{bmatrix} 600\\400\\ 240 \end{bmatrix} = \begin{bmatrix} \frac{1}{2}\\ \\ \frac{1}{3} \\ \\ \frac{1}{5} \end{bmatrix}

Therefore we have the solutions a = \frac{1}{2},\ b= \frac{1}{3},\ and\ c = \frac{1}{5}.

Or in terms of x, y, and z;

x =2,\ y =3,\ and\ z = 5

 

 

 

 

 

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