How to solve for unknown matrix?
$begingroup$
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 '18 at 0:06
Jean-Claude Arbaut
14.8k63464
asked Dec 7 '18 at 22:40
jeispyunjeispyun
111
$endgroup$
add a comment |
$begingroup$
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 '18 at 0:06
Jean-Claude Arbaut
14.8k63464
asked Dec 7 '18 at 22:40
jeispyunjeispyun
111
$endgroup$
$begingroup$
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
$endgroup$
– Makina
Dec 7 '18 at 23:00
$begingroup$
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.
$endgroup$
– Eclipse Sun
Dec 7 '18 at 23:31
add a comment |
$begingroup$
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 '18 at 0:06
Jean-Claude Arbaut
14.8k63464
asked Dec 7 '18 at 22:40
jeispyunjeispyun
111
$endgroup$
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 '18 at 0:06
Jean-Claude Arbaut
14.8k63464
asked Dec 7 '18 at 22:40
jeispyunjeispyun
111
edited Dec 8 '18 at 0:06
Jean-Claude Arbaut
14.8k63464
asked Dec 7 '18 at 22:40
jeispyunjeispyun
111
edited Dec 8 '18 at 0:06
Jean-Claude Arbaut
14.8k63464
edited Dec 8 '18 at 0:06
Jean-Claude Arbaut
14.8k63464
edited Dec 8 '18 at 0:06
Jean-Claude Arbaut
14.8k63464
14.8k63464
asked Dec 7 '18 at 22:40
jeispyunjeispyun
111
asked Dec 7 '18 at 22:40
jeispyunjeispyun
111
asked Dec 7 '18 at 22:40
jeispyunjeispyun
111
111
$begingroup$
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
$endgroup$
– Makina
Dec 7 '18 at 23:00
$begingroup$
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.
$endgroup$
– Eclipse Sun
Dec 7 '18 at 23:31
add a comment |
$begingroup$
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
$endgroup$
– Makina
Dec 7 '18 at 23:00
$begingroup$
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.
$endgroup$
– Eclipse Sun
Dec 7 '18 at 23:31
$begingroup$
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
$endgroup$
– Makina
Dec 7 '18 at 23:00
$begingroup$
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
$endgroup$
– Makina
Dec 7 '18 at 23:00
$begingroup$
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.
$endgroup$
– Eclipse Sun
Dec 7 '18 at 23:31
$begingroup$
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.
$endgroup$
– Eclipse Sun
Dec 7 '18 at 23:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 '18 at 23:54
gimusigimusi
92.8k84494
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
$begingroup$
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 '18 at 23:54
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
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 '18 at 23:54
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
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 '18 at 23:54
gimusigimusi
92.8k84494
$endgroup$
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 '18 at 23:54
gimusigimusi
92.8k84494
edited Dec 8 '18 at 0:02
answered Dec 7 '18 at 23:54
gimusigimusi
92.8k84494
answered Dec 7 '18 at 23:54
gimusigimusi
92.8k84494
answered Dec 7 '18 at 23:54
gimusigimusi
92.8k84494
92.8k84494
add a comment |
add a comment |
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$begingroup$
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
$endgroup$
– Makina
Dec 7 '18 at 23:00
$begingroup$
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.
$endgroup$
– Eclipse Sun
Dec 7 '18 at 23:31