Mixing
Stuck with matrix equation
I'm trying to solve a matrix equation problem and I can't work out the correct form for the equation for it to be valid.
The matrices given are:
A= $begin{bmatrix}
1 & -1 & 3\
4 & 1 & 5\
0 & 0 & 0\
end{bmatrix}$, B= $begin{bmatrix}
1 & -1\
3 & 6\
1 & 0\
end{bmatrix}$, C= $begin{bmatrix}
-1 & 0\
5 & 6\
0 & 1\
end{bmatrix}$
The equation goes as follows:
$AX + B = C - X$
I arrange it to: $X= (C - B)*(A+I)^{-1}$ via the following steps:
$$AX + B = C - X$$
$$AX +X = C - B$$
$$X(A+I) = C - B /(A+I)^{-1}$$
$$X = (C - B) (A+I)^{-1}$$
But the problem is that the matrices $(C-B)$ and $(A+I)^{-1}$ can't be multiplied because they're not chained (the number of rows and collumns don't allow multiplication). I've been looking at this for over half an hour and can't figure out a different approach. Any help would be highly appreciated.
matrices matrix-equations
asked Nov 21 '18 at 17:17
Arcturus
475
|
show 1 more comment
I'm trying to solve a matrix equation problem and I can't work out the correct form for the equation for it to be valid.
The matrices given are:
A= $begin{bmatrix}
1 & -1 & 3\
4 & 1 & 5\
0 & 0 & 0\
end{bmatrix}$, B= $begin{bmatrix}
1 & -1\
3 & 6\
1 & 0\
end{bmatrix}$, C= $begin{bmatrix}
-1 & 0\
5 & 6\
0 & 1\
end{bmatrix}$
The equation goes as follows:
$AX + B = C - X$
I arrange it to: $X= (C - B)*(A+I)^{-1}$ via the following steps:
$$AX + B = C - X$$
$$AX +X = C - B$$
$$X(A+I) = C - B /(A+I)^{-1}$$
$$X = (C - B) (A+I)^{-1}$$
But the problem is that the matrices $(C-B)$ and $(A+I)^{-1}$ can't be multiplied because they're not chained (the number of rows and collumns don't allow multiplication). I've been looking at this for over half an hour and can't figure out a different approach. Any help would be highly appreciated.
matrices matrix-equations
asked Nov 21 '18 at 17:17
Arcturus
475
Pre multiply by $(A+I)^{-1}$. Your equation is $(A+I)X=C-B implies X=(A+I)^{-1}(C-B )$.
– Yadati Kiran
Nov 21 '18 at 17:21
I don't understand how you got to $(A+I)X=(C−B)$
– Arcturus
Nov 21 '18 at 17:29
See we have to be careful with matrix multplication. we have $AX+X$ so number of columns of $A$ must be equal to number of rows of $X$.
– Yadati Kiran
Nov 21 '18 at 17:32
Can you guide me step to step through how you got from $AX+B=C−X$ to $(A+I)X=(C−B)$? That's the only thing I don't understand. I can find the inverse of $A+I$ just fine.
– Arcturus
Nov 21 '18 at 17:34
1
$AX+B=C−X $. Adding addtitive inverses of $B$ and $-X$ on both sides we get $AX+(B-B)+X=C+(−X+X)-Bimplies AX+Icdot X=C-Bimplies (A+I)X=C-B$. Assuming $(A+I)$ is invertible we premultiply both sides by $(A+I)^{-1}$ i.e. $(A+I)^{-1}(A+I)X=(A+I)^{-1}(C-B)implies X=(A+I)^{-1}(C-B)$
– Yadati Kiran
Nov 21 '18 at 17:39
|
show 1 more comment
I'm trying to solve a matrix equation problem and I can't work out the correct form for the equation for it to be valid.
The matrices given are:
A= $begin{bmatrix}
1 & -1 & 3\
4 & 1 & 5\
0 & 0 & 0\
end{bmatrix}$, B= $begin{bmatrix}
1 & -1\
3 & 6\
1 & 0\
end{bmatrix}$, C= $begin{bmatrix}
-1 & 0\
5 & 6\
0 & 1\
end{bmatrix}$
The equation goes as follows:
$AX + B = C - X$
I arrange it to: $X= (C - B)*(A+I)^{-1}$ via the following steps:
$$AX + B = C - X$$
$$AX +X = C - B$$
$$X(A+I) = C - B /(A+I)^{-1}$$
$$X = (C - B) (A+I)^{-1}$$
But the problem is that the matrices $(C-B)$ and $(A+I)^{-1}$ can't be multiplied because they're not chained (the number of rows and collumns don't allow multiplication). I've been looking at this for over half an hour and can't figure out a different approach. Any help would be highly appreciated.
matrices matrix-equations
asked Nov 21 '18 at 17:17
Arcturus
475
I'm trying to solve a matrix equation problem and I can't work out the correct form for the equation for it to be valid.
The matrices given are:
A= $begin{bmatrix}
1 & -1 & 3\
4 & 1 & 5\
0 & 0 & 0\
end{bmatrix}$, B= $begin{bmatrix}
1 & -1\
3 & 6\
1 & 0\
end{bmatrix}$, C= $begin{bmatrix}
-1 & 0\
5 & 6\
0 & 1\
end{bmatrix}$
The equation goes as follows:
$AX + B = C - X$
I arrange it to: $X= (C - B)*(A+I)^{-1}$ via the following steps:
$$AX + B = C - X$$
$$AX +X = C - B$$
$$X(A+I) = C - B /(A+I)^{-1}$$
$$X = (C - B) (A+I)^{-1}$$
But the problem is that the matrices $(C-B)$ and $(A+I)^{-1}$ can't be multiplied because they're not chained (the number of rows and collumns don't allow multiplication). I've been looking at this for over half an hour and can't figure out a different approach. Any help would be highly appreciated.
matrices matrix-equations
matrices matrix-equations
asked Nov 21 '18 at 17:17
Arcturus
475
asked Nov 21 '18 at 17:17
Arcturus
475
asked Nov 21 '18 at 17:17
Arcturus
475
asked Nov 21 '18 at 17:17
Arcturus
475
asked Nov 21 '18 at 17:17
Arcturus
475
475
Pre multiply by $(A+I)^{-1}$. Your equation is $(A+I)X=C-B implies X=(A+I)^{-1}(C-B )$.
– Yadati Kiran
Nov 21 '18 at 17:21
I don't understand how you got to $(A+I)X=(C−B)$
– Arcturus
Nov 21 '18 at 17:29
See we have to be careful with matrix multplication. we have $AX+X$ so number of columns of $A$ must be equal to number of rows of $X$.
– Yadati Kiran
Nov 21 '18 at 17:32
Can you guide me step to step through how you got from $AX+B=C−X$ to $(A+I)X=(C−B)$? That's the only thing I don't understand. I can find the inverse of $A+I$ just fine.
– Arcturus
Nov 21 '18 at 17:34
1
$AX+B=C−X $. Adding addtitive inverses of $B$ and $-X$ on both sides we get $AX+(B-B)+X=C+(−X+X)-Bimplies AX+Icdot X=C-Bimplies (A+I)X=C-B$. Assuming $(A+I)$ is invertible we premultiply both sides by $(A+I)^{-1}$ i.e. $(A+I)^{-1}(A+I)X=(A+I)^{-1}(C-B)implies X=(A+I)^{-1}(C-B)$
– Yadati Kiran
Nov 21 '18 at 17:39
|
show 1 more comment
Pre multiply by $(A+I)^{-1}$. Your equation is $(A+I)X=C-B implies X=(A+I)^{-1}(C-B )$.
– Yadati Kiran
Nov 21 '18 at 17:21
I don't understand how you got to $(A+I)X=(C−B)$
– Arcturus
Nov 21 '18 at 17:29
See we have to be careful with matrix multplication. we have $AX+X$ so number of columns of $A$ must be equal to number of rows of $X$.
– Yadati Kiran
Nov 21 '18 at 17:32
Can you guide me step to step through how you got from $AX+B=C−X$ to $(A+I)X=(C−B)$? That's the only thing I don't understand. I can find the inverse of $A+I$ just fine.
– Arcturus
Nov 21 '18 at 17:34
1
$AX+B=C−X $. Adding addtitive inverses of $B$ and $-X$ on both sides we get $AX+(B-B)+X=C+(−X+X)-Bimplies AX+Icdot X=C-Bimplies (A+I)X=C-B$. Assuming $(A+I)$ is invertible we premultiply both sides by $(A+I)^{-1}$ i.e. $(A+I)^{-1}(A+I)X=(A+I)^{-1}(C-B)implies X=(A+I)^{-1}(C-B)$
– Yadati Kiran
Nov 21 '18 at 17:39
Pre multiply by $(A+I)^{-1}$. Your equation is $(A+I)X=C-B implies X=(A+I)^{-1}(C-B )$.
– Yadati Kiran
Nov 21 '18 at 17:21
Pre multiply by $(A+I)^{-1}$. Your equation is $(A+I)X=C-B implies X=(A+I)^{-1}(C-B )$.
– Yadati Kiran
Nov 21 '18 at 17:21
I don't understand how you got to $(A+I)X=(C−B)$
– Arcturus
Nov 21 '18 at 17:29
I don't understand how you got to $(A+I)X=(C−B)$
– Arcturus
Nov 21 '18 at 17:29
See we have to be careful with matrix multplication. we have $AX+X$ so number of columns of $A$ must be equal to number of rows of $X$.
– Yadati Kiran
Nov 21 '18 at 17:32
See we have to be careful with matrix multplication. we have $AX+X$ so number of columns of $A$ must be equal to number of rows of $X$.
– Yadati Kiran
Nov 21 '18 at 17:32
Can you guide me step to step through how you got from $AX+B=C−X$ to $(A+I)X=(C−B)$? That's the only thing I don't understand. I can find the inverse of $A+I$ just fine.
– Arcturus
Nov 21 '18 at 17:34
Can you guide me step to step through how you got from $AX+B=C−X$ to $(A+I)X=(C−B)$? That's the only thing I don't understand. I can find the inverse of $A+I$ just fine.
– Arcturus
Nov 21 '18 at 17:34
1
1
$AX+B=C−X $. Adding addtitive inverses of $B$ and $-X$ on both sides we get $AX+(B-B)+X=C+(−X+X)-Bimplies AX+Icdot X=C-Bimplies (A+I)X=C-B$. Assuming $(A+I)$ is invertible we premultiply both sides by $(A+I)^{-1}$ i.e. $(A+I)^{-1}(A+I)X=(A+I)^{-1}(C-B)implies X=(A+I)^{-1}(C-B)$
– Yadati Kiran
Nov 21 '18 at 17:39
$AX+B=C−X $. Adding addtitive inverses of $B$ and $-X$ on both sides we get $AX+(B-B)+X=C+(−X+X)-Bimplies AX+Icdot X=C-Bimplies (A+I)X=C-B$. Assuming $(A+I)$ is invertible we premultiply both sides by $(A+I)^{-1}$ i.e. $(A+I)^{-1}(A+I)X=(A+I)^{-1}(C-B)implies X=(A+I)^{-1}(C-B)$
– Yadati Kiran
Nov 21 '18 at 17:39
|
show 1 more comment
1 Answer
1
active
oldest
votes
$X=(A+I)^{-1}(C-B )=begin{bmatrix}frac14 &frac18 &frac{-11}{8}\frac{-1}{2} ¼ ¼\0 &0 &1end{bmatrix}begin{bmatrix}-2 &1\2 &0\-1 &1end{bmatrix}=begin{bmatrix}frac98 &frac{-9}{8}\frac54 &frac{-1}{4}\-1 &1end{bmatrix}$
We get two independent solutions for $X$.
answered Nov 21 '18 at 17:32
Yadati Kiran
1,694619
add a comment |
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1 Answer
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$X=(A+I)^{-1}(C-B )=begin{bmatrix}frac14 &frac18 &frac{-11}{8}\frac{-1}{2} ¼ ¼\0 &0 &1end{bmatrix}begin{bmatrix}-2 &1\2 &0\-1 &1end{bmatrix}=begin{bmatrix}frac98 &frac{-9}{8}\frac54 &frac{-1}{4}\-1 &1end{bmatrix}$
We get two independent solutions for $X$.
answered Nov 21 '18 at 17:32
Yadati Kiran
1,694619
add a comment |
$X=(A+I)^{-1}(C-B )=begin{bmatrix}frac14 &frac18 &frac{-11}{8}\frac{-1}{2} ¼ ¼\0 &0 &1end{bmatrix}begin{bmatrix}-2 &1\2 &0\-1 &1end{bmatrix}=begin{bmatrix}frac98 &frac{-9}{8}\frac54 &frac{-1}{4}\-1 &1end{bmatrix}$
We get two independent solutions for $X$.
answered Nov 21 '18 at 17:32
Yadati Kiran
1,694619
add a comment |
$X=(A+I)^{-1}(C-B )=begin{bmatrix}frac14 &frac18 &frac{-11}{8}\frac{-1}{2} ¼ ¼\0 &0 &1end{bmatrix}begin{bmatrix}-2 &1\2 &0\-1 &1end{bmatrix}=begin{bmatrix}frac98 &frac{-9}{8}\frac54 &frac{-1}{4}\-1 &1end{bmatrix}$
We get two independent solutions for $X$.
answered Nov 21 '18 at 17:32
Yadati Kiran
1,694619
$X=(A+I)^{-1}(C-B )=begin{bmatrix}frac14 &frac18 &frac{-11}{8}\frac{-1}{2} ¼ ¼\0 &0 &1end{bmatrix}begin{bmatrix}-2 &1\2 &0\-1 &1end{bmatrix}=begin{bmatrix}frac98 &frac{-9}{8}\frac54 &frac{-1}{4}\-1 &1end{bmatrix}$
We get two independent solutions for $X$.
answered Nov 21 '18 at 17:32
Yadati Kiran
1,694619
answered Nov 21 '18 at 17:32
Yadati Kiran
1,694619
answered Nov 21 '18 at 17:32
Yadati Kiran
1,694619
answered Nov 21 '18 at 17:32
Yadati Kiran
1,694619
1,694619
add a comment |
add a comment |
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Pre multiply by $(A+I)^{-1}$. Your equation is $(A+I)X=C-B implies X=(A+I)^{-1}(C-B )$.
– Yadati Kiran
Nov 21 '18 at 17:21
I don't understand how you got to $(A+I)X=(C−B)$
– Arcturus
Nov 21 '18 at 17:29
See we have to be careful with matrix multplication. we have $AX+X$ so number of columns of $A$ must be equal to number of rows of $X$.
– Yadati Kiran
Nov 21 '18 at 17:32
Can you guide me step to step through how you got from $AX+B=C−X$ to $(A+I)X=(C−B)$? That's the only thing I don't understand. I can find the inverse of $A+I$ just fine.
– Arcturus
Nov 21 '18 at 17:34
1
$AX+B=C−X $. Adding addtitive inverses of $B$ and $-X$ on both sides we get $AX+(B-B)+X=C+(−X+X)-Bimplies AX+Icdot X=C-Bimplies (A+I)X=C-B$. Assuming $(A+I)$ is invertible we premultiply both sides by $(A+I)^{-1}$ i.e. $(A+I)^{-1}(A+I)X=(A+I)^{-1}(C-B)implies X=(A+I)^{-1}(C-B)$
– Yadati Kiran
Nov 21 '18 at 17:39