(line1) i1+8i2+3i3=-3 (Line2) 3i1-2i2+i3=-5 (Line 3) 2i1-3i2+2i3=6

Question

electrical circuits

Answer 1

Cramer's Method.

Let determinant Δ=

| 1  8 3 |

| 3 -2 1 | = 1(-4+3)-8(6-2)+3(-9+4)=-1-32-15=-48.

| 2 -3 2 |

This determinant is derived from the coefficients of the 3 variables.

Δ₁=

| -3  8 3 |

| -5 -2 1 | = -3(-4+3)-8(-10-6)+3(15+12)=3+128+81=212.

|  6 -3 2 |

I₁=Δ₁/Δ=212/-48=-53/12.

Δ₂=

| 1 -3 3 |

| 3 -5 1 | =1(-10-6)+3(6-2)+3(18+10)=-16+12+84=80.

| 2  6 2 |

I₂=Δ₂/Δ=80/-48=-5/3.

Δ₃=

| 1  8 -3 |

| 3 -2 -5 | = 1(-12-15)-8(18+10)-3(-9+4)=-27-224+15=-236.

| 2 -3  6 |

I₃=Δ₃/Δ=-236/-48=59/12.

SOLUTION

I₁=-53/12, I₂=-5/3, I₃=59/12.

CHECK

I₁+8I₂+3I₃=-3, 3I₁-2I₂+I₃=-5, 2I₁-3I₂+2I₃=6✔️