Question

Solve the following linear equations by Cramer’s Rule:

x – y + 2z = 7, 3x + 4y – 5z = 5, 2x – y + 3z = 12

Answer 1

Given equations are
x – y + 2z = 7
3x + 4y – 5z = 5
2x – y + 3z = 12

D = `|(1, -1, 2),(3, 4, -5),(2, -1, 3)|`

= 1(12 – 5) – (– 1)(9 + 10) + 2(–3– 8)
= 1(7) + 1(19) + 2(– 11)
= 7 + 19 – 22
= 4 ≠ 0

D_x = `|(7, -1, 2),(5, 4, -5),(12, -1, 3)|`

= 7(12 - 5) – (– 1)(15 + 60) + 2(–5 – 48)
= 7(7) + 1(75) + 2(– 53)
= 49 + 75 – 106
= 18

D_y = `|(1, 7, 2),(3, 5, -5),(2, 12, 3)|`

= 1(15 + 60) –7(9 + 10) + 2(36 – 10)
= 1(75) – 7(19) + 2(26)
= 75 – 133 + 52
= – 6

D_z = `|(1, -1, 7),(3, 4, 5),(2, -1, 12)|`

= 1(48 + 5) – (– 1)(36 – 10) + 7(– 3 – 8)
= 1(53) + 1(26) + 7(– 11)
= 53 + 26 – 77
= 2
By Cramer's Rule,

x = `"D"_x/"D" = 18/4 = 9/2`,

y = `"D"_y/"D" = (-6)/4 = (-3)/2`,

z = `"D"_z/"D" = 2/4 = 1/2`

∴ x = `9/2, y = (-3)/2 and z = 1/2` are the solutions of the given equations.