1. **State the problem:** We are given a system of three simultaneous linear equations derived from Kirchhoff's laws: $$I_1 + 8I_2 + 3I_3 = -31$$ $$3I_1 - 2I_2 + I_3 = -5$$ $$2I_1 - 3I_2 + 2I_3 = 6$$ Our goal is to find the values of $I_1$, $I_2$, and $I_3$ using the determinant (Cramer's) method. 2. **Write the coefficient matrix and the constants vector:** Coefficient matrix $A$: $$\begin{bmatrix}1 & 8 & 3 \\ 3 & -2 & 1 \\ 2 & -3 & 2\end{bmatrix}$$ Constants vector $B$: $$\begin{bmatrix}-31 \\ -5 \\ 6\end{bmatrix}$$ 3. **Find the determinant of the coefficient matrix $D = \det(A)$:** Using the rule of Sarrus or cofactor expansion, $$D = 1((-2)(2) - 1(-3)) - 8(3(2) - 1(2)) + 3(3(-3) - (-2)(2))$$ Simplify step-by-step: $$= 1((-4) + 3) - 8(6 - 2) + 3(-9 + 4)$$ $$= 1(-1) - 8(4) + 3(-5)$$ $$= -1 - 32 - 15 = -48$$ 4. **Find determinant $D_1$ by replacing first column of $A$ with $B$:** $$D_1 = \det\begin{bmatrix}-31 & 8 & 3 \\ -5 & -2 & 1 \\ 6 & -3 & 2 \end{bmatrix}$$ Calculate: $$D_1 = (-31)((-2)(2) - 1(-3)) - 8((-5)(2) - 1(6)) + 3((-5)(-3) - (-2)(6))$$ $$= (-31)(-4 + 3) - 8(-10 -6) + 3(15 + 12)$$ $$= (-31)(-1) - 8(-16) + 3(27)$$ $$= 31 + 128 + 81 = 240$$ 5. **Find determinant $D_2$ by replacing second column of $A$ with $B$:** $$D_2 = \det\begin{bmatrix}1 & -31 & 3 \\ 3 & -5 & 1 \\ 2 & 6 & 2 \end{bmatrix}$$ Calculate: $$D_2 = 1((-5)(2) - 1(6)) - (-31)(3(2) - 1(2)) + 3(3(6) - (-5)(2))$$ $$= 1(-10 -6) + 31(6 - 2) + 3(18 + 10)$$ $$= -16 + 31(4) + 3(28)$$ $$= -16 + 124 + 84 = 192$$ 6. **Find determinant $D_3$ by replacing third column of $A$ with $B$:** $$D_3 = \det\begin{bmatrix}1 & 8 & -31 \\ 3 & -2 & -5 \\ 2 & -3 & 6 \end{bmatrix}$$ Calculate: $$D_3 = 1((-2)(6) - (-5)(-3)) - 8(3(6) - (-5)(2)) + (-31)(3(-3) - (-2)(2))$$ $$= 1(-12 -15) - 8(18 + 10) - 31(-9 + 4)$$ $$= 1(-27) - 8(28) - 31(-5)$$ $$= -27 - 224 + 155 = -96$$ 7. **Calculate the unknown currents using Cramer's rule:** $$I_1 = \frac{D_1}{D} = \frac{240}{-48} = -5$$ $$I_2 = \frac{D_2}{D} = \frac{192}{-48} = -4$$ $$I_3 = \frac{D_3}{D} = \frac{-96}{-48} = 2$$ **Final answer:** $$I_1 = -5, \quad I_2 = -4, \quad I_3 = 2$$