Assuming matrices $A,B,C ;text{and};D;$ are $ ntimes n,$ show two ways that $(A+B)(C+D)=AC+AD+BC+BD$
$begingroup$
My way first of showing this is by letting $A,B,C ;text{and};D;$ equal
begin{bmatrix}
a_{ij}
end{bmatrix}
begin{bmatrix}
b_{ij}
end{bmatrix}
begin{bmatrix}
c_{ij}
end{bmatrix}
begin{bmatrix}
d_{ij}
end{bmatrix}
Respectively. I then used the distributive laws to show that both side are equal, but I cannot think of another way.
matrices matrix-equations matrix-rank
edited Dec 8 '18 at 19:01
user376343
3,3583826
asked Dec 8 '18 at 17:53
David MBDavid MB
267
$endgroup$
add a comment |
$begingroup$
My way first of showing this is by letting $A,B,C ;text{and};D;$ equal
begin{bmatrix}
a_{ij}
end{bmatrix}
begin{bmatrix}
b_{ij}
end{bmatrix}
begin{bmatrix}
c_{ij}
end{bmatrix}
begin{bmatrix}
d_{ij}
end{bmatrix}
Respectively. I then used the distributive laws to show that both side are equal, but I cannot think of another way.
matrices matrix-equations matrix-rank
edited Dec 8 '18 at 19:01
user376343
3,3583826
asked Dec 8 '18 at 17:53
David MBDavid MB
267
$endgroup$
$begingroup$
multiply $[a_{ij}+b_{ij}]$ and $[c_{ij}+d_{ij}]$ ?
$endgroup$
– user113102
Dec 8 '18 at 17:57
$begingroup$
That's a good point, but is that not the same doing it if they were both singular as I did e.g. ([a]+[b])([c]+[d]) ? If not, then I guess that answers my question.
$endgroup$
– David MB
Dec 8 '18 at 18:10
add a comment |
$begingroup$
My way first of showing this is by letting $A,B,C ;text{and};D;$ equal
begin{bmatrix}
a_{ij}
end{bmatrix}
begin{bmatrix}
b_{ij}
end{bmatrix}
begin{bmatrix}
c_{ij}
end{bmatrix}
begin{bmatrix}
d_{ij}
end{bmatrix}
Respectively. I then used the distributive laws to show that both side are equal, but I cannot think of another way.
matrices matrix-equations matrix-rank
edited Dec 8 '18 at 19:01
user376343
3,3583826
asked Dec 8 '18 at 17:53
David MBDavid MB
267
$endgroup$
My way first of showing this is by letting $A,B,C ;text{and};D;$ equal
begin{bmatrix}
a_{ij}
end{bmatrix}
begin{bmatrix}
b_{ij}
end{bmatrix}
begin{bmatrix}
c_{ij}
end{bmatrix}
begin{bmatrix}
d_{ij}
end{bmatrix}
Respectively. I then used the distributive laws to show that both side are equal, but I cannot think of another way.
matrices matrix-equations matrix-rank
matrices matrix-equations matrix-rank
edited Dec 8 '18 at 19:01
user376343
3,3583826
asked Dec 8 '18 at 17:53
David MBDavid MB
267
edited Dec 8 '18 at 19:01
user376343
3,3583826
asked Dec 8 '18 at 17:53
David MBDavid MB
267
edited Dec 8 '18 at 19:01
user376343
3,3583826
edited Dec 8 '18 at 19:01
user376343
3,3583826
edited Dec 8 '18 at 19:01
user376343
3,3583826
3,3583826
asked Dec 8 '18 at 17:53
David MBDavid MB
267
asked Dec 8 '18 at 17:53
David MBDavid MB
267
asked Dec 8 '18 at 17:53
David MBDavid MB
267
267
$begingroup$
multiply $[a_{ij}+b_{ij}]$ and $[c_{ij}+d_{ij}]$ ?
$endgroup$
– user113102
Dec 8 '18 at 17:57
$begingroup$
That's a good point, but is that not the same doing it if they were both singular as I did e.g. ([a]+[b])([c]+[d]) ? If not, then I guess that answers my question.
$endgroup$
– David MB
Dec 8 '18 at 18:10
add a comment |
$begingroup$
multiply $[a_{ij}+b_{ij}]$ and $[c_{ij}+d_{ij}]$ ?
$endgroup$
– user113102
Dec 8 '18 at 17:57
$begingroup$
That's a good point, but is that not the same doing it if they were both singular as I did e.g. ([a]+[b])([c]+[d]) ? If not, then I guess that answers my question.
$endgroup$
– David MB
Dec 8 '18 at 18:10
$begingroup$
multiply $[a_{ij}+b_{ij}]$ and $[c_{ij}+d_{ij}]$ ?
$endgroup$
– user113102
Dec 8 '18 at 17:57
$begingroup$
multiply $[a_{ij}+b_{ij}]$ and $[c_{ij}+d_{ij}]$ ?
$endgroup$
– user113102
Dec 8 '18 at 17:57
$begingroup$
That's a good point, but is that not the same doing it if they were both singular as I did e.g. ([a]+[b])([c]+[d]) ? If not, then I guess that answers my question.
$endgroup$
– David MB
Dec 8 '18 at 18:10
$begingroup$
That's a good point, but is that not the same doing it if they were both singular as I did e.g. ([a]+[b])([c]+[d]) ? If not, then I guess that answers my question.
$endgroup$
– David MB
Dec 8 '18 at 18:10
add a comment |
1 Answer
1
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oldest
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$begingroup$
You can use the distributive property of matrices to show this the second way.
Observe that $(A+B)$ and $(C+D)$ are each themselves $ntimes n$ matrices. Using right distribution, $(A+B)(C+D) = (A+B)C + (A+B)D$. Then distributing to the left of each term, we obtain the desired result.
answered Dec 8 '18 at 19:28
Adam CartisanoAdam Cartisano
1614
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You can use the distributive property of matrices to show this the second way.
Observe that $(A+B)$ and $(C+D)$ are each themselves $ntimes n$ matrices. Using right distribution, $(A+B)(C+D) = (A+B)C + (A+B)D$. Then distributing to the left of each term, we obtain the desired result.
answered Dec 8 '18 at 19:28
Adam CartisanoAdam Cartisano
1614
$endgroup$
add a comment |
$begingroup$
You can use the distributive property of matrices to show this the second way.
Observe that $(A+B)$ and $(C+D)$ are each themselves $ntimes n$ matrices. Using right distribution, $(A+B)(C+D) = (A+B)C + (A+B)D$. Then distributing to the left of each term, we obtain the desired result.
answered Dec 8 '18 at 19:28
Adam CartisanoAdam Cartisano
1614
$endgroup$
add a comment |
$begingroup$
You can use the distributive property of matrices to show this the second way.
Observe that $(A+B)$ and $(C+D)$ are each themselves $ntimes n$ matrices. Using right distribution, $(A+B)(C+D) = (A+B)C + (A+B)D$. Then distributing to the left of each term, we obtain the desired result.
answered Dec 8 '18 at 19:28
Adam CartisanoAdam Cartisano
1614
$endgroup$
You can use the distributive property of matrices to show this the second way.
Observe that $(A+B)$ and $(C+D)$ are each themselves $ntimes n$ matrices. Using right distribution, $(A+B)(C+D) = (A+B)C + (A+B)D$. Then distributing to the left of each term, we obtain the desired result.
answered Dec 8 '18 at 19:28
Adam CartisanoAdam Cartisano
1614
answered Dec 8 '18 at 19:28
Adam CartisanoAdam Cartisano
1614
answered Dec 8 '18 at 19:28
Adam CartisanoAdam Cartisano
1614
answered Dec 8 '18 at 19:28
Adam CartisanoAdam Cartisano
1614
1614
add a comment |
add a comment |
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$begingroup$
multiply $[a_{ij}+b_{ij}]$ and $[c_{ij}+d_{ij}]$ ?
$endgroup$
– user113102
Dec 8 '18 at 17:57
$begingroup$
That's a good point, but is that not the same doing it if they were both singular as I did e.g. ([a]+[b])([c]+[d]) ? If not, then I guess that answers my question.
$endgroup$
– David MB
Dec 8 '18 at 18:10