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The voltage in an electric circuit are related by following equations : $V_{1}+V_{2}+V_{3}= 9;V_{1}-V_{2}+V_{3} = 1find V_{1},V_{2} and V_{3} by using cramer's rule.$
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Solution:

$D = \quad \begin{vmatrix} 1 & 1 & 1\\ 1 & -1 & 1\\ 1 & 1 & -1 \end{vmatrix}$

=1(1-1)-1(-1-1)+1(1+1)= 4

$D_{V1} = \quad \begin{vmatrix} 9 & 1 & 1\\ 3 & -1 & 1\\ 1 & 1 & -1 \end{vmatrix}$

= 9(1-1)-1(-3-1)+1(3+1) = 8

$V_{1} = \frac{D_{V1}}{D} = \frac{8}{4} = 2$

$D_{V2} = \quad \begin{vmatrix} 1 & 9 & 1\\ 1 & 3 & 1\\ 1 & 1 & -1 \end{vmatrix}$

= 1(-3-1)-9(-1-1)+1(1-3) = 12

$V_{2} = \frac{D_{V2}}{D} = \frac{12}{4} = 3$

$D_{V3} = \quad \begin{vmatrix} 1 & 1 & 9\\ 1 & -1 & 3\\ 1 & 1 & -1 \end{vmatrix}$

$V_{3} = \frac{D_{V3}}{D} = \frac{16}{4} = 4$

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