Distribution property on the internal and external multiplication of matrices
$begingroup$
I have relation as :
A=E*(p1.p2)=270
which
E=[1 2 3];
p1=transpose([4 5 6]);
p2= transpose( [7 8 9]);
How can I withdraw p2 or make a factor from p2 f to make isolate (Ep1)
calculus linear-algebra combinatorics matrices matlab
asked Dec 28 '18 at 21:52
Hamed HatamiHamed Hatami
1
$endgroup$
add a comment |
$begingroup$
I have relation as :
A=E*(p1.p2)=270
which
E=[1 2 3];
p1=transpose([4 5 6]);
p2= transpose( [7 8 9]);
How can I withdraw p2 or make a factor from p2 f to make isolate (Ep1)
calculus linear-algebra combinatorics matrices matlab
asked Dec 28 '18 at 21:52
Hamed HatamiHamed Hatami
1
$endgroup$
$begingroup$
The dot product is symmetric.
$endgroup$
– amd
Dec 28 '18 at 22:58
add a comment |
$begingroup$
I have relation as :
A=E*(p1.p2)=270
which
E=[1 2 3];
p1=transpose([4 5 6]);
p2= transpose( [7 8 9]);
How can I withdraw p2 or make a factor from p2 f to make isolate (Ep1)
calculus linear-algebra combinatorics matrices matlab
asked Dec 28 '18 at 21:52
Hamed HatamiHamed Hatami
1
$endgroup$
I have relation as :
A=E*(p1.p2)=270
which
E=[1 2 3];
p1=transpose([4 5 6]);
p2= transpose( [7 8 9]);
How can I withdraw p2 or make a factor from p2 f to make isolate (Ep1)
calculus linear-algebra combinatorics matrices matlab
calculus linear-algebra combinatorics matrices matlab
asked Dec 28 '18 at 21:52
Hamed HatamiHamed Hatami
1
asked Dec 28 '18 at 21:52
Hamed HatamiHamed Hatami
1
asked Dec 28 '18 at 21:52
Hamed HatamiHamed Hatami
1
asked Dec 28 '18 at 21:52
Hamed HatamiHamed Hatami
1
asked Dec 28 '18 at 21:52
Hamed HatamiHamed Hatami
1
1
$begingroup$
The dot product is symmetric.
$endgroup$
– amd
Dec 28 '18 at 22:58
add a comment |
$begingroup$
The dot product is symmetric.
$endgroup$
– amd
Dec 28 '18 at 22:58
$begingroup$
The dot product is symmetric.
$endgroup$
– amd
Dec 28 '18 at 22:58
$begingroup$
The dot product is symmetric.
$endgroup$
– amd
Dec 28 '18 at 22:58
add a comment |
0
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$begingroup$
The dot product is symmetric.
$endgroup$
– amd
Dec 28 '18 at 22:58