Page 12 - NUMERICAL METHOD
P. 12
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STEP 4 :
Second Elimination
For new matrix, eliminate the value of −14 at element 32 by
referring R2 to replace with 0.
( × ) − × 32 = 3
2
22
3
0 0 0
−14 −7 − −7 −14 = 0
−16 −7 14
−22 −7 56
New Augmented Matrix for second elimination:
1 2 3 5
0 −7 −7 −7
0 0 14 56
STEP 5 :
Solve the problem.
+ 2 + 3 = 5
− 7 − 7 = −7
14 = 56
∴ = − ; = − ; = 
