Question

Answer the following 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.