using matrices solve the system of equations
1/x + 1/y + 1/z= 2

2/x +1/y-3/z =0

1/x -1/y + 1/z =4

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using matrices solve the system of equations
1/x + 1/y + 1/z= 2
2/x +1/y-3/z =0
1/x -1/y + 1/z =4

Ajanta Trivedi · Accepted Answer

let $\frac{1}{x} = u, \frac{1}{y} = v a n d \frac{1}{z} = w$

therefore the equations become:

$u + v + w = 2 ............. (1) 2 u + v - 3 w = 0 .......... (2) u - v + w = 4 .............. (3)$

writing the equations in terms of matrices,

$[\begin{matrix} 1 & 1 & 1 \\ 2 & 1 & - 3 \\ 1 & - 1 & 1 \end{matrix}] [\begin{matrix} u \\ v \\ w \end{matrix}] = [\begin{matrix} 2 \\ 0 \\ 4 \end{matrix}] A X = B A = [\begin{matrix} 1 & 1 & 1 \\ 2 & 1 & - 3 \\ 1 & - 1 & 1 \end{matrix}], X = [\begin{matrix} u \\ v \\ w \end{matrix}], B = [\begin{matrix} 2 \\ 0 \\ 4 \end{matrix}] X = A^{- 1} B$

now we will find the inverse of matrix A and then will multiply it with B.

$| A | = | \begin{matrix} 1 & 1 & 1 \\ 2 & 1 & - 3 \\ 1 & - 1 & 1 \end{matrix} | = 1 (1 - 3) + 1 (- 3 - 2) + 1 (- 2 - 1) = - 2 - 5 - 3 = - 10$

now find the cofactor matrix

$[\begin{matrix} (1 - 3) & - (2 + 3) & (- 2 - 1) \\ - (1 + 1) & (1 - 1) & - (- 1 - 1) \\ (- 3 - 1) & - (- 3 - 2) & (1 - 2) \end{matrix}] \equiv [\begin{matrix} - 2 & - 5 & - 3 \\ - 2 & 0 & 2 \\ - 4 & 5 & 1 \end{matrix}]$

transpose of cofactor matrix is the adjoint matrix

$a d j o int (A) = [\begin{matrix} - 2 & - 2 & - 4 \\ - 5 & 0 & 5 \\ - 3 & 2 & - 1 \end{matrix}] A^{- 1} = \frac{1}{| A |} a d j o int A = - \frac{1}{10} [\begin{matrix} - 2 & - 2 & - 4 \\ - 5 & 0 & 5 \\ - 3 & 2 & - 1 \end{matrix}] A^{- 1} = [\begin{matrix} \frac{1}{5} & \frac{1}{5} & \frac{2}{5} \\ \frac{1}{2} & 0 & - \frac{1}{2} \\ \frac{3}{10} & - \frac{1}{5} & \frac{1}{10} \end{matrix}]$

$X = A^{- 1} B X = [\begin{matrix} \frac{1}{5} & \frac{1}{5} & \frac{2}{5} \\ \frac{1}{2} & 0 & - \frac{1}{2} \\ \frac{3}{10} & - \frac{1}{5} & \frac{1}{10} \end{matrix}] [\begin{matrix} 2 \\ 0 \\ 4 \end{matrix}] = [\begin{matrix} \frac{2}{5} + 0 + \frac{8}{5} \\ 1 + 0 - 2 \\ \frac{3}{5} + 0 + \frac{2}{5} \end{matrix}] = [\begin{matrix} 2 \\ - 1 \\ 1 \end{matrix}]$

$[\begin{matrix} u \\ v \\ w \end{matrix}] = [\begin{matrix} 2 \\ - 1 \\ 1 \end{matrix}]$

hence u=2, v=-1 and w=1

therefore $x = \frac{1}{2}, y = - 1, z = 1$

using matrices solve the system of equations1/x + 1/y + 1/z= 2 2/x +1/y-3/z =0 1/x -1/y + 1/z =4

using matrices solve the system of equations
1/x + 1/y + 1/z= 2

2/x +1/y-3/z =0

1/x -1/y + 1/z =4