Name: Sudhan Arumugam
Student #: 1286902
Assignment 1
4.
From x1 − 5x2 = 1 we have x1 = 1 + 5x2 . Substitute into 3x1 − 7x2 = 5:
3(1 + 5x2 ) − 7x2 = 5 ⇒ 3 + 8x2 = 5 ⇒ x2 = 14 .
Then x1 = 1 + 5 · 14 = 94 . Intersection ( 94 , 41 ).
8.




1 1 5 0
1 0 0 −4
Row reduce  0 1 9 0  to  0 1 0 9  . So (x1 , x2 , x3 ) = (−4, 9, −1).
0 0 7 −7
0 0 1 −1
12.
Use E2 − 3E1 : 2x2 − 5x3 = 4 and E3 + 4E1 : −x2 + 3x3 = −2 ⇒ x2 = 2 + 3x3 . Plugging
in gives x3 = 0 so x2 = 2. From E1 get x1 = 2. Final (2, 2, 0).
16.
1
Check (2, 2, 0) in the three equations:
2 − 3 · 2 + 4 · 0 = −4,
3 · 2 − 7 · 2 + 7 · 0 = −8,
−4 · 2 + 6 · 2 + 2 · 0 = 4.
All match, so it is correct.
20.
From 2x2 + 2x3 = 0 get x2 = −x3 . From x3 + 3x4 = 1 get x3 = 1 − 3x4 so x2 = −1 + 3x4 .
From x1 − 2x4 = −3 get x1 = −3 + 2x4 . The last equation is automatically true. Let t = x4 :
(x1 , x2 , x3 , x4 ) = (−3 + 2t, −1 + 3t, 1 − 3t, t), t ∈ R.
22.
From x2 − x3 = 1 get x2 = 1 + x3 . From x1 + 3x2 = 0 get x1 = −3 − 3x3 . Plugging into
x1 + 2x2 + x3 = 4 gives −1 = 4, which is impossible. No common point.
24.
1 h
Coefficient matrix
has det = 4 + 2h. If h ̸= −2 there is a unique solution. If
−2 4
h = −2 the rows are multiples and constants match, so still consistent. Therefore consistent
for every real h.
2
26.
Row 2 coefficients are −2 times row 1, so require 9 = −2h. Hence h = − 92 . Otherwise
inconsistent.
3