linear algebra, matrix

[ 1 1 1 1 | 1
2 1 2 0 | 3
1 2 1 3 | a
3 4 3 5 | d ]

R2 - 2 x R1 to R2
R3 - R1 to R3
R4 - 3 x R1 to R4 we get

[ 1 1 1 1 | 1
0 -1 0 -2 | 1
0 1 0 2 | a - 1
0 1 0 2 | d - 3 ]

R4 - R3 to R4 and then R2 + R3 to R3

[ 1 1 1 1 | 1
0 -1 0 -2 | 1
0 0 0 0 | a
0 0 0 0 | (d - 3) - (a - 1) ]

For the system to be consistent
a = 0 and
(d - 3) - (a - 1) = 0
so d = a +2 = 2

Solution of the system (x,y,z,w) :
Let w = t
Let z = s
-y + (-2t) = 1
y = - 1 - 2t
x + (-1 - 2t) + s + t = 1
x = t - s + 2.

用 Gaussian elimination 計到 (reduced) row echelon form。

如果無人幫你我今晚計畀你睇，而家唔得閒住．．．

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