Application of the identity matrix II. v05
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xhunga
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You have seen :
a) You must multiply m by four identity matrix,
to eliminate all the values below the pivots.
11/1 14/1 -16/1 19/1
m = 12/1 13/1 13/1 11/1
0/1 0/1 17/1 11/1
12/1 18/1 1/1 -16/1
mID1 : to eliminate all the values under the first pivot.
mID2 : to eliminate all the values under the second pivot.
mID3 : to eliminate all the values under the third pivot.
mID4 : to eliminate the pivot.
Can you do that in one step ?
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Multiply mID4 by mID3 by mID2 by mID1, in this order.
mID = ((mID4 * mID3) * mID2)* mID1.
And then mutiply mID by m.
1/1 14/11 -16/11 19/11
mID * m = 0/1 1/1 -67/5 107/25
0/1 0/1 1/1 11/17
0/1 0/1 0/1 1/1
All the pivots equal 1 (1/1).
All the values below the pivot equal 0 (0/1).
The work must be do without swap two rows.
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