Question

Solve the following linear equations by Cramer’s Rule:  

2x –  y + z = 1, x + 2y + 3z = 8, 3x + y – 4z = 1

Answer 1

Given equations are
2x –  y + z = 1
x + 2y + 3z = 8 
3x + y – 4z = 1

D = `|(2, -1, 1),(1, 2, 3),(3, 1, -4)|`

= 2(– 8 – 3) – (– 1)(– 4 – 9) + 1(1 – 6)
= 2(– 11) + 1(– 13) + 1(– 5)
= – 22 – 13 – 5
= – 40 ≠ 0

D_x = `|(1, -1, 1),(8, 2, 3),(1, 1, -4)|`

= 1(– 8 – 3) – (– 1)(– 32 – 3) + 1(8 – 2)
= 1(– 11) + 1(– 35) + 1(6)
= – 11 – 35 + 6
= – 40

D_y = `|(2, 1, 1),(1, 8, 3),(3, 1, -4)|`

= 2(– 32 – 3) – 1(– 4 – 9) + 1(1 – 24)
= 2( - 35) – 1(– 13) + 1(– 23)
= – 70 + 13 – 23
= – 80

D_z = `|(2, -1, 1),(1, 2, 8),(3, 1, 1)|`

= 2(2 – 8) – (– 1)(1 – 24) + 1(1 – 6)
= 2(– 6) + 1(– 23) + 1(– 5)
= – 12 – 23– 5
= – 40
By Cramer's Rule,

x = `"D"_x/"D" = (-40)/(-40)` = 1

y = `"D"_y/"D" = (-80)/(-40)` = 2,

z = `"D"_z/"D" = (-40)/(-40)` = 1

∴ x = 1, y = 2 and z = 1 are the solutions of the given equations.