Mixing
Solving for unknown matrices. Specific example and useful rules.
$begingroup$
I wanted to ask how i could go about finding matrix X in the following equations:
AX-B^T= X + AB
and
AX=BX-2X+I
A and B are given in both examples. A,B and X are not the same in the 2 examples.
I realize that its supposedly quite basic,but i haven't been able to find a source online to pinpoint the rules of matrix arithmetic.
Thank you in advance! :)
matrices matrix-equations
edited Jan 26 at 20:39
Jean Marie
30.9k42155
asked Jan 26 at 19:04
user569685user569685
102
$endgroup$
add a comment |
$begingroup$
I wanted to ask how i could go about finding matrix X in the following equations:
AX-B^T= X + AB
and
AX=BX-2X+I
A and B are given in both examples. A,B and X are not the same in the 2 examples.
I realize that its supposedly quite basic,but i haven't been able to find a source online to pinpoint the rules of matrix arithmetic.
Thank you in advance! :)
matrices matrix-equations
edited Jan 26 at 20:39
Jean Marie
30.9k42155
asked Jan 26 at 19:04
user569685user569685
102
$endgroup$
add a comment |
$begingroup$
I wanted to ask how i could go about finding matrix X in the following equations:
AX-B^T= X + AB
and
AX=BX-2X+I
A and B are given in both examples. A,B and X are not the same in the 2 examples.
I realize that its supposedly quite basic,but i haven't been able to find a source online to pinpoint the rules of matrix arithmetic.
Thank you in advance! :)
matrices matrix-equations
edited Jan 26 at 20:39
Jean Marie
30.9k42155
asked Jan 26 at 19:04
user569685user569685
102
$endgroup$
I wanted to ask how i could go about finding matrix X in the following equations:
AX-B^T= X + AB
and
AX=BX-2X+I
A and B are given in both examples. A,B and X are not the same in the 2 examples.
I realize that its supposedly quite basic,but i haven't been able to find a source online to pinpoint the rules of matrix arithmetic.
Thank you in advance! :)
matrices matrix-equations
matrices matrix-equations
edited Jan 26 at 20:39
Jean Marie
30.9k42155
asked Jan 26 at 19:04
user569685user569685
102
edited Jan 26 at 20:39
Jean Marie
30.9k42155
asked Jan 26 at 19:04
user569685user569685
102
edited Jan 26 at 20:39
Jean Marie
30.9k42155
edited Jan 26 at 20:39
Jean Marie
30.9k42155
edited Jan 26 at 20:39
Jean Marie
30.9k42155
30.9k42155
asked Jan 26 at 19:04
user569685user569685
102
asked Jan 26 at 19:04
user569685user569685
102
asked Jan 26 at 19:04
user569685user569685
102
102
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes, it is easy. For example, rewrite the second equation as $(A-B+2I)X=I$.
This means that $X=(A-B+2I)^{-1}$.
answered Jan 26 at 19:08
Dietrich BurdeDietrich Burde
81.2k648106
$endgroup$
$begingroup$
Could you please explain how you got to that? Im not sure how the rules work in matrix arithmetic.
$endgroup$
– user569685
Jan 26 at 19:11
$begingroup$
It is the same arithmetic as below: $AX-BX=(A-B)X$.
$endgroup$
– Dietrich Burde
Jan 26 at 19:31
add a comment |
$begingroup$
We have that
$$AX-B^T=X+ABimplies AX-X=AB+B^Timplies (A-I)X=AB+B^T.$$ Assuming $A-I$ is invertible (and I bet it is in your case) then
$$X=(A-I)^{-1}(AB+B^T).$$
answered Jan 26 at 19:09
mflmfl
26.9k12142
$endgroup$
$begingroup$
Why does AX-X = (A-I)X ?
$endgroup$
– user569685
Jan 26 at 19:14
$begingroup$
$AX-X=AX-IX=(A-I)X.$
$endgroup$
– mfl
Jan 26 at 19:15
$begingroup$
Thank you so much.
$endgroup$
– user569685
Jan 26 at 19:20
add a comment |
Your Answer
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, it is easy. For example, rewrite the second equation as $(A-B+2I)X=I$.
This means that $X=(A-B+2I)^{-1}$.
answered Jan 26 at 19:08
Dietrich BurdeDietrich Burde
81.2k648106
$endgroup$
$begingroup$
Could you please explain how you got to that? Im not sure how the rules work in matrix arithmetic.
$endgroup$
– user569685
Jan 26 at 19:11
$begingroup$
It is the same arithmetic as below: $AX-BX=(A-B)X$.
$endgroup$
– Dietrich Burde
Jan 26 at 19:31
add a comment |
$begingroup$
Yes, it is easy. For example, rewrite the second equation as $(A-B+2I)X=I$.
This means that $X=(A-B+2I)^{-1}$.
answered Jan 26 at 19:08
Dietrich BurdeDietrich Burde
81.2k648106
$endgroup$
$begingroup$
Could you please explain how you got to that? Im not sure how the rules work in matrix arithmetic.
$endgroup$
– user569685
Jan 26 at 19:11
$begingroup$
It is the same arithmetic as below: $AX-BX=(A-B)X$.
$endgroup$
– Dietrich Burde
Jan 26 at 19:31
add a comment |
$begingroup$
Yes, it is easy. For example, rewrite the second equation as $(A-B+2I)X=I$.
This means that $X=(A-B+2I)^{-1}$.
answered Jan 26 at 19:08
Dietrich BurdeDietrich Burde
81.2k648106
$endgroup$
Yes, it is easy. For example, rewrite the second equation as $(A-B+2I)X=I$.
This means that $X=(A-B+2I)^{-1}$.
answered Jan 26 at 19:08
Dietrich BurdeDietrich Burde
81.2k648106
answered Jan 26 at 19:08
Dietrich BurdeDietrich Burde
81.2k648106
answered Jan 26 at 19:08
Dietrich BurdeDietrich Burde
81.2k648106
answered Jan 26 at 19:08
Dietrich BurdeDietrich Burde
81.2k648106
81.2k648106
$begingroup$
Could you please explain how you got to that? Im not sure how the rules work in matrix arithmetic.
$endgroup$
– user569685
Jan 26 at 19:11
$begingroup$
It is the same arithmetic as below: $AX-BX=(A-B)X$.
$endgroup$
– Dietrich Burde
Jan 26 at 19:31
add a comment |
$begingroup$
Could you please explain how you got to that? Im not sure how the rules work in matrix arithmetic.
$endgroup$
– user569685
Jan 26 at 19:11
$begingroup$
It is the same arithmetic as below: $AX-BX=(A-B)X$.
$endgroup$
– Dietrich Burde
Jan 26 at 19:31
$begingroup$
Could you please explain how you got to that? Im not sure how the rules work in matrix arithmetic.
$endgroup$
– user569685
Jan 26 at 19:11
$begingroup$
Could you please explain how you got to that? Im not sure how the rules work in matrix arithmetic.
$endgroup$
– user569685
Jan 26 at 19:11
$begingroup$
It is the same arithmetic as below: $AX-BX=(A-B)X$.
$endgroup$
– Dietrich Burde
Jan 26 at 19:31
$begingroup$
It is the same arithmetic as below: $AX-BX=(A-B)X$.
$endgroup$
– Dietrich Burde
Jan 26 at 19:31
add a comment |
$begingroup$
We have that
$$AX-B^T=X+ABimplies AX-X=AB+B^Timplies (A-I)X=AB+B^T.$$ Assuming $A-I$ is invertible (and I bet it is in your case) then
$$X=(A-I)^{-1}(AB+B^T).$$
answered Jan 26 at 19:09
mflmfl
26.9k12142
$endgroup$
$begingroup$
Why does AX-X = (A-I)X ?
$endgroup$
– user569685
Jan 26 at 19:14
$begingroup$
$AX-X=AX-IX=(A-I)X.$
$endgroup$
– mfl
Jan 26 at 19:15
$begingroup$
Thank you so much.
$endgroup$
– user569685
Jan 26 at 19:20
add a comment |
$begingroup$
We have that
$$AX-B^T=X+ABimplies AX-X=AB+B^Timplies (A-I)X=AB+B^T.$$ Assuming $A-I$ is invertible (and I bet it is in your case) then
$$X=(A-I)^{-1}(AB+B^T).$$
answered Jan 26 at 19:09
mflmfl
26.9k12142
$endgroup$
$begingroup$
Why does AX-X = (A-I)X ?
$endgroup$
– user569685
Jan 26 at 19:14
$begingroup$
$AX-X=AX-IX=(A-I)X.$
$endgroup$
– mfl
Jan 26 at 19:15
$begingroup$
Thank you so much.
$endgroup$
– user569685
Jan 26 at 19:20
add a comment |
$begingroup$
We have that
$$AX-B^T=X+ABimplies AX-X=AB+B^Timplies (A-I)X=AB+B^T.$$ Assuming $A-I$ is invertible (and I bet it is in your case) then
$$X=(A-I)^{-1}(AB+B^T).$$
answered Jan 26 at 19:09
mflmfl
26.9k12142
$endgroup$
We have that
$$AX-B^T=X+ABimplies AX-X=AB+B^Timplies (A-I)X=AB+B^T.$$ Assuming $A-I$ is invertible (and I bet it is in your case) then
$$X=(A-I)^{-1}(AB+B^T).$$
answered Jan 26 at 19:09
mflmfl
26.9k12142
answered Jan 26 at 19:09
mflmfl
26.9k12142
answered Jan 26 at 19:09
mflmfl
26.9k12142
answered Jan 26 at 19:09
mflmfl
26.9k12142
26.9k12142
$begingroup$
Why does AX-X = (A-I)X ?
$endgroup$
– user569685
Jan 26 at 19:14
$begingroup$
$AX-X=AX-IX=(A-I)X.$
$endgroup$
– mfl
Jan 26 at 19:15
$begingroup$
Thank you so much.
$endgroup$
– user569685
Jan 26 at 19:20
add a comment |
$begingroup$
Why does AX-X = (A-I)X ?
$endgroup$
– user569685
Jan 26 at 19:14
$begingroup$
$AX-X=AX-IX=(A-I)X.$
$endgroup$
– mfl
Jan 26 at 19:15
$begingroup$
Thank you so much.
$endgroup$
– user569685
Jan 26 at 19:20
$begingroup$
Why does AX-X = (A-I)X ?
$endgroup$
– user569685
Jan 26 at 19:14
$begingroup$
Why does AX-X = (A-I)X ?
$endgroup$
– user569685
Jan 26 at 19:14
$begingroup$
$AX-X=AX-IX=(A-I)X.$
$endgroup$
– mfl
Jan 26 at 19:15
$begingroup$
$AX-X=AX-IX=(A-I)X.$
$endgroup$
– mfl
Jan 26 at 19:15
$begingroup$
Thank you so much.
$endgroup$
– user569685
Jan 26 at 19:20
$begingroup$
Thank you so much.
$endgroup$
– user569685
Jan 26 at 19:20
add a comment |
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