How do you solve 2r + 3s - 4t = 20, 4r - s + 5t = 13, 3r + 2s + 4 = 15?

Question

2634 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-06T15:17:19

$(5, 2, -1)$

You meant $[(2, 3, -4), (4, -1, 5), (3, 2, 4)] * [(r), (s), (t)] = [(20), (13), (15)]$

$L_3 := (L_3 + L_1)/5$

$L_2 := 4 * L_2 + 5 * L_1$

$[(2, 3, -4), (4 * 4 + 5 * 2, 4 * (-1) +5 * 3, 4 * 5 +5 *(-4)), ((3+2)/5, (2+3)/5, (4-4)/5)]$

$* [(r), (s), (t)] = [(20), (4 * 13 + 5 * 20), ((15+20)/5)]$

$[(2, 3, -4), (26, 11, 0), (1, 1, 0)] * [(r), (s), (t)] = [(20), (152), (7)]$

$L_2 := L_2 - 11 * L_3$

$[(2, 3, -4), (26 - 11, 11 - 11, 0), (1, 1, 0)] * [(r), (s), (t)] = [(20), (152-11*7), (7)]$

$[(2, 3, -4), (15, 0, 0), (1, 1, 0)] * [(r), (s), (t)] = [(20), (75), (7)]$

$15 r = 75 Rightarrow r = 5$

$r + s = 7 Rightarrow s = 7 - 5 = 2$

$2r +3s - 4t = 20$

$10 +6 - 4t = 20 Rightarrow t = -1$