# Q : 16         Solve the system of equations                    $\dpi{100} \frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4$                    $\dpi{100} \frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1$                    $\dpi{100} \frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$

D Divya Prakash Singh

We have a system of equations;

$\dpi{100} \frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4$

$\dpi{100} \frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1$

$\dpi{100} \frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$

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;

$\dpi{100} 2a +3b+10c = 4$

$\dpi{100} 4a-6b+5c =1$

$\dpi{100} 6a+9b-20c = 2$

We can write it in the matrix form as $\dpi{100} AX =B$ , where

$\dpi{100} A= \begin{bmatrix} 2 &3 &10 \\ 4& -6 & 5\\ 6 & 9 & -20 \end{bmatrix} , X = \begin{bmatrix} a\\b \\c \end{bmatrix}\ and\ B = \begin{bmatrix} 4\\1 \\ 2 \end{bmatrix}.$

Now, Finding the determinant value of A;

$\dpi{100} |A| = 2(120-45)-3(-80-30)+10(36+36)$

$\dpi{100} =150+330+720$

$\dpi{100} =1200 \neq 0$

Hence we can say that A is non-singular $\dpi{100} \therefore$ its invers exists;

Finding cofactors of A;

$\dpi{100} A_{11} = 75$ , $\dpi{100} A_{12} = 110$$\dpi{100} A_{13} = 72$

$\dpi{100} A_{21} = 150$$\dpi{100} A_{22} = -100$$\dpi{100} A_{23} = 0$

$\dpi{100} A_{31} =75$$\dpi{100} A_{31} =30$$\dpi{100} A_{33} =-24$

$\dpi{100} \therefore$ as we know $\dpi{100} A^{-1} = \frac{1}{|A|}adjA$

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

Now we will find the solutions by relation $\dpi{100} X = A^{-1}B$.

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

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

$\dpi{100} = \frac{1}{1200}\begin{bmatrix} 600\\400\\ 240 \end{bmatrix} = \begin{bmatrix} \frac{1}{2}\\ \\ \frac{1}{3} \\ \\ \frac{1}{5} \end{bmatrix}$

Therefore we have the solutions $\dpi{100} a = \frac{1}{2},\ b= \frac{1}{3},\ and\ c = \frac{1}{5}.$

Or in terms of x, y, and z;

$\dpi{100} x =2,\ y =3,\ and\ z = 5$

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