Solve I2 using cramer's rule. I1+I2+I3=0, 8I1-10I3=8 and 6I1-3I2=12

Question

3.1 solve I2 only using cramer's rule.

3.2 solve for I3 using substitution and elimination

Answer 1

Make determinant Δ from the coefficients of I₁, I₂, I₃:

| 1  1    1 |

| 8  0 -10⎜= 1(0-30)-1(0+60)+1(-24-0)=-30-60-24=-114

| 6 -3    0 |

3.1) Now make the determinant for I₂, using the column of constants in place of the variable's column:

Δ₂=

| 1   0    1 |

| 8   8 -10 | = 1(0+120)+0+1(96-48)=120+48=168

| 6 12    0 |

I₂=Δ₂/Δ=168/-114=-28/19.

3.2) 8I₁-10I₃=8, 6I₁-3I₂=12⇒①4I₁-5I₃=4, ②2I₁-I₂=4; ③I₁+I₂+I₃=0;

②+③=3I₁+I₃=4⇒④15I₁+5I₃=20;

①+④=19I₁=24, I₁=24/19⇒I₃=4-3I₁=4-72/19=4/19.

(We can deduce that I₂=-(I₁+I₃)=-28/19 which confirms 3.1) I₂=-28/19.)