How to solve for unknown matrix?
How can I solve this?
$$ begin{bmatrix}
1 & 1 \
1 & 2 \
end{bmatrix}
X +
begin{bmatrix}
2 & -1\
-1 & 1\
end{bmatrix}
X
begin{bmatrix}
1 & 5 \
1 & 2 \
end{bmatrix}
=
begin{bmatrix}
1 & 1\
1 & 1\
end{bmatrix}
$$
I know there's similar question like:
Solve for unknown matrix.
But this one is much more complex as there are two separate terms with $X$.
Can I perform something similar like:
$Ax + Bx = C implies (A+B)x = C$? But at the second term, the $X$ is at the middle and that order is important in matrix.
Any help would be appreciated!
matrices matrix-equations
edited Dec 8 at 0:06
Jean-Claude Arbaut
14.7k63464
asked Dec 7 at 22:40
jeispyun
111
add a comment |
How can I solve this?
$$ begin{bmatrix}
1 & 1 \
1 & 2 \
end{bmatrix}
X +
begin{bmatrix}
2 & -1\
-1 & 1\
end{bmatrix}
X
begin{bmatrix}
1 & 5 \
1 & 2 \
end{bmatrix}
=
begin{bmatrix}
1 & 1\
1 & 1\
end{bmatrix}
$$
I know there's similar question like:
Solve for unknown matrix.
But this one is much more complex as there are two separate terms with $X$.
Can I perform something similar like:
$Ax + Bx = C implies (A+B)x = C$? But at the second term, the $X$ is at the middle and that order is important in matrix.
Any help would be appreciated!
matrices matrix-equations
edited Dec 8 at 0:06
Jean-Claude Arbaut
14.7k63464
asked Dec 7 at 22:40
jeispyun
111
Why don't you do exactly the same thing, done in that exercise? Like, do you know, how to multiply and add matrices? Is this the "much more complex" part? I mean, you can also left-multiply both sides by $ begin{bmatrix} 1 & 1 \ 1 & 2 \ end{bmatrix} $ and see what happens
– Makina
Dec 7 at 23:00
If you put $X=begin{bmatrix}x_1 & x_2 \ x_3 & x_4 end{bmatrix}$ and expand all the product, then it is still a set of linear equations.
– Eclipse Sun
Dec 7 at 23:31
add a comment |
How can I solve this?
$$ begin{bmatrix}
1 & 1 \
1 & 2 \
end{bmatrix}
X +
begin{bmatrix}
2 & -1\
-1 & 1\
end{bmatrix}
X
begin{bmatrix}
1 & 5 \
1 & 2 \
end{bmatrix}
=
begin{bmatrix}
1 & 1\
1 & 1\
end{bmatrix}
$$
I know there's similar question like:
Solve for unknown matrix.
But this one is much more complex as there are two separate terms with $X$.
Can I perform something similar like:
$Ax + Bx = C implies (A+B)x = C$? But at the second term, the $X$ is at the middle and that order is important in matrix.
Any help would be appreciated!
matrices matrix-equations
edited Dec 8 at 0:06
Jean-Claude Arbaut
14.7k63464
asked Dec 7 at 22:40
jeispyun
111
How can I solve this?
$$ begin{bmatrix}
1 & 1 \
1 & 2 \
end{bmatrix}
X +
begin{bmatrix}
2 & -1\
-1 & 1\
end{bmatrix}
X
begin{bmatrix}
1 & 5 \
1 & 2 \
end{bmatrix}
=
begin{bmatrix}
1 & 1\
1 & 1\
end{bmatrix}
$$
I know there's similar question like:
Solve for unknown matrix.
But this one is much more complex as there are two separate terms with $X$.
Can I perform something similar like:
$Ax + Bx = C implies (A+B)x = C$? But at the second term, the $X$ is at the middle and that order is important in matrix.
Any help would be appreciated!
matrices matrix-equations
matrices matrix-equations
edited Dec 8 at 0:06
Jean-Claude Arbaut
14.7k63464
asked Dec 7 at 22:40
jeispyun
111
edited Dec 8 at 0:06
Jean-Claude Arbaut
14.7k63464
asked Dec 7 at 22:40
jeispyun
111
edited Dec 8 at 0:06
Jean-Claude Arbaut
14.7k63464
edited Dec 8 at 0:06
Jean-Claude Arbaut
14.7k63464
edited Dec 8 at 0:06
Jean-Claude Arbaut
14.7k63464
14.7k63464
asked Dec 7 at 22:40
jeispyun
111
asked Dec 7 at 22:40
jeispyun
111
asked Dec 7 at 22:40
jeispyun
111
111
Why don't you do exactly the same thing, done in that exercise? Like, do you know, how to multiply and add matrices? Is this the "much more complex" part? I mean, you can also left-multiply both sides by $ begin{bmatrix} 1 & 1 \ 1 & 2 \ end{bmatrix} $ and see what happens
– Makina
Dec 7 at 23:00
If you put $X=begin{bmatrix}x_1 & x_2 \ x_3 & x_4 end{bmatrix}$ and expand all the product, then it is still a set of linear equations.
– Eclipse Sun
Dec 7 at 23:31
add a comment |
Why don't you do exactly the same thing, done in that exercise? Like, do you know, how to multiply and add matrices? Is this the "much more complex" part? I mean, you can also left-multiply both sides by $ begin{bmatrix} 1 & 1 \ 1 & 2 \ end{bmatrix} $ and see what happens
– Makina
Dec 7 at 23:00
If you put $X=begin{bmatrix}x_1 & x_2 \ x_3 & x_4 end{bmatrix}$ and expand all the product, then it is still a set of linear equations.
– Eclipse Sun
Dec 7 at 23:31
Why don't you do exactly the same thing, done in that exercise? Like, do you know, how to multiply and add matrices? Is this the "much more complex" part? I mean, you can also left-multiply both sides by $ begin{bmatrix} 1 & 1 \ 1 & 2 \ end{bmatrix} $ and see what happens
– Makina
Dec 7 at 23:00
Why don't you do exactly the same thing, done in that exercise? Like, do you know, how to multiply and add matrices? Is this the "much more complex" part? I mean, you can also left-multiply both sides by $ begin{bmatrix} 1 & 1 \ 1 & 2 \ end{bmatrix} $ and see what happens
– Makina
Dec 7 at 23:00
If you put $X=begin{bmatrix}x_1 & x_2 \ x_3 & x_4 end{bmatrix}$ and expand all the product, then it is still a set of linear equations.
– Eclipse Sun
Dec 7 at 23:31
If you put $X=begin{bmatrix}x_1 & x_2 \ x_3 & x_4 end{bmatrix}$ and expand all the product, then it is still a set of linear equations.
– Eclipse Sun
Dec 7 at 23:31
add a comment |
1 Answer
1
active
oldest
votes
If we can't see the trick suggested in the comments, by $X=begin{bmatrix}
a & b \
c & d \
end{bmatrix}$ we obtain
$$ begin{bmatrix}
1 & 1 \
1 & 2 \
end{bmatrix}X= begin{bmatrix}
a+b & a+2b \
c+d & c+2d \
end{bmatrix}$$
$$
begin{bmatrix}
2 & -1\
-1 & 1\
end{bmatrix}
X
begin{bmatrix}
1 & 5 \
1 & 2 \
end{bmatrix}
=$$
$$=begin{bmatrix}
2 & -1\
-1 & 1\
end{bmatrix}
begin{bmatrix}
a+b & 5a+2b\
c+d & 5c+2d\
end{bmatrix}
=begin{bmatrix}
2(a+b)-(c+d) & 2(5a+2b)-(5c+2d)\
-(a+b)+(c+d) & -(5a+2b)+(5c+2d)\
end{bmatrix}$$
then the system
$$begin{bmatrix}
3(a+b)-(c+d) & (11a+6b)-(5c+2d)\
-(a+b)+2(c+d) & -(5a+2b)+(6c+4d)\
end{bmatrix}= begin{bmatrix}
1 & 1 \
1 & 1 \
end{bmatrix}$$
answered Dec 7 at 23:54
gimusi
1
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If we can't see the trick suggested in the comments, by $X=begin{bmatrix}
a & b \
c & d \
end{bmatrix}$ we obtain
$$ begin{bmatrix}
1 & 1 \
1 & 2 \
end{bmatrix}X= begin{bmatrix}
a+b & a+2b \
c+d & c+2d \
end{bmatrix}$$
$$
begin{bmatrix}
2 & -1\
-1 & 1\
end{bmatrix}
X
begin{bmatrix}
1 & 5 \
1 & 2 \
end{bmatrix}
=$$
$$=begin{bmatrix}
2 & -1\
-1 & 1\
end{bmatrix}
begin{bmatrix}
a+b & 5a+2b\
c+d & 5c+2d\
end{bmatrix}
=begin{bmatrix}
2(a+b)-(c+d) & 2(5a+2b)-(5c+2d)\
-(a+b)+(c+d) & -(5a+2b)+(5c+2d)\
end{bmatrix}$$
then the system
$$begin{bmatrix}
3(a+b)-(c+d) & (11a+6b)-(5c+2d)\
-(a+b)+2(c+d) & -(5a+2b)+(6c+4d)\
end{bmatrix}= begin{bmatrix}
1 & 1 \
1 & 1 \
end{bmatrix}$$
answered Dec 7 at 23:54
gimusi
1
add a comment |
If we can't see the trick suggested in the comments, by $X=begin{bmatrix}
a & b \
c & d \
end{bmatrix}$ we obtain
$$ begin{bmatrix}
1 & 1 \
1 & 2 \
end{bmatrix}X= begin{bmatrix}
a+b & a+2b \
c+d & c+2d \
end{bmatrix}$$
$$
begin{bmatrix}
2 & -1\
-1 & 1\
end{bmatrix}
X
begin{bmatrix}
1 & 5 \
1 & 2 \
end{bmatrix}
=$$
$$=begin{bmatrix}
2 & -1\
-1 & 1\
end{bmatrix}
begin{bmatrix}
a+b & 5a+2b\
c+d & 5c+2d\
end{bmatrix}
=begin{bmatrix}
2(a+b)-(c+d) & 2(5a+2b)-(5c+2d)\
-(a+b)+(c+d) & -(5a+2b)+(5c+2d)\
end{bmatrix}$$
then the system
$$begin{bmatrix}
3(a+b)-(c+d) & (11a+6b)-(5c+2d)\
-(a+b)+2(c+d) & -(5a+2b)+(6c+4d)\
end{bmatrix}= begin{bmatrix}
1 & 1 \
1 & 1 \
end{bmatrix}$$
answered Dec 7 at 23:54
gimusi
1
add a comment |
If we can't see the trick suggested in the comments, by $X=begin{bmatrix}
a & b \
c & d \
end{bmatrix}$ we obtain
$$ begin{bmatrix}
1 & 1 \
1 & 2 \
end{bmatrix}X= begin{bmatrix}
a+b & a+2b \
c+d & c+2d \
end{bmatrix}$$
$$
begin{bmatrix}
2 & -1\
-1 & 1\
end{bmatrix}
X
begin{bmatrix}
1 & 5 \
1 & 2 \
end{bmatrix}
=$$
$$=begin{bmatrix}
2 & -1\
-1 & 1\
end{bmatrix}
begin{bmatrix}
a+b & 5a+2b\
c+d & 5c+2d\
end{bmatrix}
=begin{bmatrix}
2(a+b)-(c+d) & 2(5a+2b)-(5c+2d)\
-(a+b)+(c+d) & -(5a+2b)+(5c+2d)\
end{bmatrix}$$
then the system
$$begin{bmatrix}
3(a+b)-(c+d) & (11a+6b)-(5c+2d)\
-(a+b)+2(c+d) & -(5a+2b)+(6c+4d)\
end{bmatrix}= begin{bmatrix}
1 & 1 \
1 & 1 \
end{bmatrix}$$
answered Dec 7 at 23:54
gimusi
1
If we can't see the trick suggested in the comments, by $X=begin{bmatrix}
a & b \
c & d \
end{bmatrix}$ we obtain
$$ begin{bmatrix}
1 & 1 \
1 & 2 \
end{bmatrix}X= begin{bmatrix}
a+b & a+2b \
c+d & c+2d \
end{bmatrix}$$
$$
begin{bmatrix}
2 & -1\
-1 & 1\
end{bmatrix}
X
begin{bmatrix}
1 & 5 \
1 & 2 \
end{bmatrix}
=$$
$$=begin{bmatrix}
2 & -1\
-1 & 1\
end{bmatrix}
begin{bmatrix}
a+b & 5a+2b\
c+d & 5c+2d\
end{bmatrix}
=begin{bmatrix}
2(a+b)-(c+d) & 2(5a+2b)-(5c+2d)\
-(a+b)+(c+d) & -(5a+2b)+(5c+2d)\
end{bmatrix}$$
then the system
$$begin{bmatrix}
3(a+b)-(c+d) & (11a+6b)-(5c+2d)\
-(a+b)+2(c+d) & -(5a+2b)+(6c+4d)\
end{bmatrix}= begin{bmatrix}
1 & 1 \
1 & 1 \
end{bmatrix}$$
answered Dec 7 at 23:54
gimusi
1
edited Dec 8 at 0:02
answered Dec 7 at 23:54
gimusi
1
answered Dec 7 at 23:54
gimusi
1
answered Dec 7 at 23:54
gimusi
1
1
add a comment |
add a comment |
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Why don't you do exactly the same thing, done in that exercise? Like, do you know, how to multiply and add matrices? Is this the "much more complex" part? I mean, you can also left-multiply both sides by $ begin{bmatrix} 1 & 1 \ 1 & 2 \ end{bmatrix} $ and see what happens
– Makina
Dec 7 at 23:00
If you put $X=begin{bmatrix}x_1 & x_2 \ x_3 & x_4 end{bmatrix}$ and expand all the product, then it is still a set of linear equations.
– Eclipse Sun
Dec 7 at 23:31