1. The problem is to solve Kirchoff's simultaneous equations using determinants: \[I_1 + 8 I_2 + 3 I_3 = -31\] \[3 I_1 - 2 I_2 + I_3 = -5\] \[2 I_1 - 3 I_2 + 2 I_3 = 6\] Rearranging into the form \(a_1 x + b_1 y + c_1 + d_1 = 0\): \[I_1 + 8 I_2 + 3 I_3 + 31 = 0\] \[3 I_1 - 2 I_2 + I_3 + 5 = 0\] \[2 I_1 - 3 I_2 + 2 I_3 - 6 = 0\] 2. We define determinants from coefficients and constants: Determinant \(D\): \[ D = \begin{vmatrix} 1 & 8 & 3 \\ 3 & -2 & 1 \\ 2 & -3 & 2 \end{vmatrix} = 1(-1) - 8(4) + 3(-16) = -1 - 32 - 48 = -81 \text{ (recalculated)} \] However, user computed \(D = -48\), so let's verify determinant \(D\) precisely: Calculate \(D\) stepwise: \[ D = 1 \times \begin{vmatrix} -2 & 1 \\ -3 & 2 \end{vmatrix} - 8 \times \begin{vmatrix} 3 & 1 \\ 2 & 2 \end{vmatrix} + 3 \times \begin{vmatrix} 3 & -2 \\ 2 & -3 \end{vmatrix} \] Calculating minors: \[ \begin{vmatrix} -2 & 1 \\ -3 & 2 \end{vmatrix} = (-2)(2) - (1)(-3) = -4 + 3 = -1 \] \[ \begin{vmatrix} 3 & 1 \\ 2 & 2 \end{vmatrix} = 3(2) - 1(2) = 6 - 2 = 4 \] \[ \begin{vmatrix} 3 & -2 \\ 2 & -3 \end{vmatrix} = 3(-3) - (-2)(2) = -9 + 4 = -5 \] So, \[ D = 1(-1) - 8(4) + 3(-5) = -1 - 32 - 15 = -48 \] 3. Now calculate determinants for \(I_1\), \(I_2\), and \(I_3\): \[D_{I_1} = \begin{vmatrix}8 & 3 & 31 \\ -2 & 1 & 5 \\ -3 & 2 & -6\end{vmatrix}\] Calculate using expansion: \[ D_{I_1} = 8 \begin{vmatrix}1 & 5 \\ 2 & -6 \end{vmatrix} - 3 \begin{vmatrix}-2 & 5 \\ -3 & -6 \end{vmatrix} + 31 \begin{vmatrix}-2 & 1 \\ -3 & 2 \end{vmatrix} \] Calculate minors: \[ \begin{vmatrix}1 & 5 \\ 2 & -6 \end{vmatrix} = 1(-6) - 5(2) = -6 - 10 = -16 \] \[ \begin{vmatrix}-2 & 5 \\ -3 & -6 \end{vmatrix} = (-2)(-6) - 5(-3) = 12 + 15 = 27 \] \[ \begin{vmatrix}-2 & 1 \\ -3 & 2 \end{vmatrix} = (-2)(2) - 1(-3) = -4 + 3 = -1 \] Plug them in: \[ D_{I_1} = 8(-16) - 3(27) + 31(-1) = -128 - 81 - 31 = -240 \] \[D_{I_2} = \begin{vmatrix}1 & 3 & 31 \\ 3 & 1 & 5 \\ 2 & 2 & -6\end{vmatrix}\] Calculate: \[ D_{I_2} = 1 \begin{vmatrix}1 & 5 \\ 2 & -6 \end{vmatrix} - 3 \begin{vmatrix}3 & 5 \\ 2 & -6 \end{vmatrix} + 31 \begin{vmatrix}3 & 1 \\ 2 & 2 \end{vmatrix} \] Minors: \[ \begin{vmatrix}1 & 5 \\ 2 & -6 \end{vmatrix} = 1(-6) - 5(2) = -6 - 10 = -16 \] \[ \begin{vmatrix}3 & 5 \\ 2 & -6 \end{vmatrix} = 3(-6) - 5(2) = -18 - 10 = -28 \] \[ \begin{vmatrix}3 & 1 \\ 2 & 2 \end{vmatrix} = 3(2) - 1(2) = 6 - 2 = 4 \] So: \[ D_{I_2} = 1(-16) - 3(-28) + 31(4) = -16 + 84 + 124 = 192 \] \[D_{I_3} = \begin{vmatrix}1 & 8 & 31 \\ 3 & -2 & 5 \\ 2 & -3 & -6\end{vmatrix}\] Calculate: \[ D_{I_3} = 1 \begin{vmatrix}-2 & 5 \\ -3 & -6 \end{vmatrix} - 8 \begin{vmatrix}3 & 5 \\ 2 & -6 \end{vmatrix} + 31 \begin{vmatrix}3 & -2 \\ 2 & -3 \end{vmatrix} \] Minors: \[ \begin{vmatrix}-2 & 5 \\ -3 & -6 \end{vmatrix} = (-2)(-6) - 5(-3) = 12 + 15 = 27 \] \[ \begin{vmatrix}3 & 5 \\ 2 & -6 \end{vmatrix} = -28 \text{ (as before)} \] \[ \begin{vmatrix}3 & -2 \\ 2 & -3 \end{vmatrix} = 3(-3) - (-2)(2) = -9 + 4 = -5 \] So: \[ D_{I_3} = 1(27) - 8(-28) + 31(-5) = 27 + 224 - 155 = 96 \] 4. Using Cramer's Rule, the ratios are: \[ \frac{I_1}{D_{I_1}} = \frac{-I_2}{D_{I_2}} = \frac{I_3}{D_{I_3}} = \frac{-1}{D} \] Substitute given values: \[ \frac{I_1}{-240} = \frac{-I_2}{192} = \frac{I_3}{96} = \frac{-1}{-48} = \frac{1}{48} \] Therefore, \[ I_1 = \frac{-240}{-48} = 5, \] \[ -I_2 = 192 \times \frac{1}{48} = 4 \implies I_2 = -4, \] \[ I_3 = 96 \times \frac{1}{48} = 2. \] 5. Final Answer: \[ I_1 = 5, \quad I_2 = -4, \quad I_3 = 2. \] Note: The user wrote \(I_3 = -2\) but by Cramer's calculation, \(I_3 = 2\) consistent with determinant sign and ratios; original problem may have sign inconsistency for \(I_3\).