Page 330 - 35Linear Algebra
P. 330
330 Sample First Midterm
1 0 0 0 1 1 −1 2
1 1 0 0 0 2 3 0
=
−1 −1 1 0 0 0 −2 8
0 2 0 1 0 0 1 −2
1 0 0 0 1 1 −1 2
1 1 0 0 0 2 3 0
= .
−1 −1 1 0 0 0 −2 8
0 2 − 1 1 0 0 0 2
2
To solve MX = V using M = LU we first solve LW = V whose augmented
matrix reads
1 0 0 0 7 1 0 0 0 7
1 1 0 0 6 0 1 0 0 −1
∼
−1 −1 1 0 12 0 0 1 0 18
0 2 − 1 1 −7 0 2 − 1 1 −7
2 2
1 0 0 0 7
0 1 0 0 −1
∼ ,
0 0 1 0 18
0 0 0 1 4
from which we can read off W. Now we compute X by solving UX = W
with the augmented matrix
1 1 −1 2 7 1 1 −1 2 7
0 2 3 0 −1 0 2 3 0 −1
∼
0 0 −2 8 18 0 0 −2 0 2
0 0 0 2 4 0 0 0 1 2
1 1 −1 2 7 1 0 0 0 1
0 2 0 0 2 0 1 0 0 1
∼ ∼ .
0 0 1 0 −1 0 0 1 0 −1
0 0 0 1 2 0 0 0 1 2
So x = 1, y = 1, z = −1 and w = 2.
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