Question

Answer the following question:

Solve the following linear equations by Cramer’s Rule:

2x+ 3y + 3z = 5 , x − 2y + z = – 4 , 3x – y – 2z = 3

Answer 1

Given equations are
2x+ 3y + 3z = 5 ,

x − 2y + z = – 4 ,

3x – y – 2z = 3.

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

= 2(4 + 1) – 3(–2 – 3) + 3(– 1 + 6)

= 2(5) –3(– 5) + 3(5)

= 10 + 15 + 15

= 40 ≠ 0

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

= 5(4 + 1) – 3(8 – 3) + 3(4 + 6)

= 5(5) – 3(5) + 3(10)

= 25 – 15 + 30

= 40

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

= 2(8 – 3) – 5(–2 – 3) + 3(3 + 12)

= 2(5) – 5(–5) + 3(15)

= 10 + 25 + 45

= 80

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

= 2(–6 – 4) – 3(3 + 12) + 5(–1 + 6)

= 2(–10) – 3(15) + 5(5)

= –20 – 45 + 25

= – 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.