Matrix equations, simplify them.
$begingroup$
I have some equations and I don't know am I doing the simplification right. Can someone check it?
For example, we have an equation $AX = 4X + B$, where $A$ and $B$ are matrices. So, what I have done with it is
- $AX - 4X = B $
- $X(A - 4) = B$
- $X = B(A-4)^{-1}$
And the second example. $(A+4X)^{-1} = B$
- $A^{-1} + frac{1}{4}X^{-1} = B$
- $frac{1}{4}X^{-1} = B - A^{-1}$
- $X^{-1} = 4(B - A^{-1})$
- $X = (4(B - A^{-1}))^{-1}$
I didn't find a lot of information about that, so maybe someone can share any links? Or just help me here pls.
linear-algebra matrices inverse matrix-equations
edited Jan 16 at 17:50
pwerth
3,320417
asked Jan 16 at 17:44
Aliaksei KlimovichAliaksei Klimovich
536
$endgroup$
add a comment |
$begingroup$
I have some equations and I don't know am I doing the simplification right. Can someone check it?
For example, we have an equation $AX = 4X + B$, where $A$ and $B$ are matrices. So, what I have done with it is
- $AX - 4X = B $
- $X(A - 4) = B$
- $X = B(A-4)^{-1}$
And the second example. $(A+4X)^{-1} = B$
- $A^{-1} + frac{1}{4}X^{-1} = B$
- $frac{1}{4}X^{-1} = B - A^{-1}$
- $X^{-1} = 4(B - A^{-1})$
- $X = (4(B - A^{-1}))^{-1}$
I didn't find a lot of information about that, so maybe someone can share any links? Or just help me here pls.
linear-algebra matrices inverse matrix-equations
edited Jan 16 at 17:50
pwerth
3,320417
asked Jan 16 at 17:44
Aliaksei KlimovichAliaksei Klimovich
536
$endgroup$
1
$begingroup$
The expression $A-4$ is undefined: it’s the difference of a matrix and scalar. Get into the habit of writing $A-4I$ (or $4E$, if that’s what the identity matrix is called in your materials.)
$endgroup$
– amd
Jan 16 at 17:53
1
$begingroup$
$(A+4X)^{-1}$ is no more equal to $A^{-1}+frac14X^{-1}$ than $1/(a+4x)$ is equal to $1/a+1/(4x)$.
$endgroup$
– amd
Jan 16 at 17:54
add a comment |
$begingroup$
I have some equations and I don't know am I doing the simplification right. Can someone check it?
For example, we have an equation $AX = 4X + B$, where $A$ and $B$ are matrices. So, what I have done with it is
- $AX - 4X = B $
- $X(A - 4) = B$
- $X = B(A-4)^{-1}$
And the second example. $(A+4X)^{-1} = B$
- $A^{-1} + frac{1}{4}X^{-1} = B$
- $frac{1}{4}X^{-1} = B - A^{-1}$
- $X^{-1} = 4(B - A^{-1})$
- $X = (4(B - A^{-1}))^{-1}$
I didn't find a lot of information about that, so maybe someone can share any links? Or just help me here pls.
linear-algebra matrices inverse matrix-equations
edited Jan 16 at 17:50
pwerth
3,320417
asked Jan 16 at 17:44
Aliaksei KlimovichAliaksei Klimovich
536
$endgroup$
I have some equations and I don't know am I doing the simplification right. Can someone check it?
For example, we have an equation $AX = 4X + B$, where $A$ and $B$ are matrices. So, what I have done with it is
- $AX - 4X = B $
- $X(A - 4) = B$
- $X = B(A-4)^{-1}$
And the second example. $(A+4X)^{-1} = B$
- $A^{-1} + frac{1}{4}X^{-1} = B$
- $frac{1}{4}X^{-1} = B - A^{-1}$
- $X^{-1} = 4(B - A^{-1})$
- $X = (4(B - A^{-1}))^{-1}$
I didn't find a lot of information about that, so maybe someone can share any links? Or just help me here pls.
linear-algebra matrices inverse matrix-equations
linear-algebra matrices inverse matrix-equations
edited Jan 16 at 17:50
pwerth
3,320417
asked Jan 16 at 17:44
Aliaksei KlimovichAliaksei Klimovich
536
edited Jan 16 at 17:50
pwerth
3,320417
asked Jan 16 at 17:44
Aliaksei KlimovichAliaksei Klimovich
536
edited Jan 16 at 17:50
pwerth
3,320417
edited Jan 16 at 17:50
pwerth
3,320417
edited Jan 16 at 17:50
pwerth
3,320417
3,320417
asked Jan 16 at 17:44
Aliaksei KlimovichAliaksei Klimovich
536
asked Jan 16 at 17:44
Aliaksei KlimovichAliaksei Klimovich
536
asked Jan 16 at 17:44
Aliaksei KlimovichAliaksei Klimovich
536
536
1
$begingroup$
The expression $A-4$ is undefined: it’s the difference of a matrix and scalar. Get into the habit of writing $A-4I$ (or $4E$, if that’s what the identity matrix is called in your materials.)
$endgroup$
– amd
Jan 16 at 17:53
1
$begingroup$
$(A+4X)^{-1}$ is no more equal to $A^{-1}+frac14X^{-1}$ than $1/(a+4x)$ is equal to $1/a+1/(4x)$.
$endgroup$
– amd
Jan 16 at 17:54
add a comment |
1
$begingroup$
The expression $A-4$ is undefined: it’s the difference of a matrix and scalar. Get into the habit of writing $A-4I$ (or $4E$, if that’s what the identity matrix is called in your materials.)
$endgroup$
– amd
Jan 16 at 17:53
1
$begingroup$
$(A+4X)^{-1}$ is no more equal to $A^{-1}+frac14X^{-1}$ than $1/(a+4x)$ is equal to $1/a+1/(4x)$.
$endgroup$
– amd
Jan 16 at 17:54
1
1
$begingroup$
The expression $A-4$ is undefined: it’s the difference of a matrix and scalar. Get into the habit of writing $A-4I$ (or $4E$, if that’s what the identity matrix is called in your materials.)
$endgroup$
– amd
Jan 16 at 17:53
$begingroup$
The expression $A-4$ is undefined: it’s the difference of a matrix and scalar. Get into the habit of writing $A-4I$ (or $4E$, if that’s what the identity matrix is called in your materials.)
$endgroup$
– amd
Jan 16 at 17:53
1
1
$begingroup$
$(A+4X)^{-1}$ is no more equal to $A^{-1}+frac14X^{-1}$ than $1/(a+4x)$ is equal to $1/a+1/(4x)$.
$endgroup$
– amd
Jan 16 at 17:54
$begingroup$
$(A+4X)^{-1}$ is no more equal to $A^{-1}+frac14X^{-1}$ than $1/(a+4x)$ is equal to $1/a+1/(4x)$.
$endgroup$
– amd
Jan 16 at 17:54
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Both examples contain some errors. For the first one, you have the right idea subtracting to get
$$AX-4X=B$$
however, you have to be careful about factoring matrices. Observe that
$$AX-4X=(A-4I)X$$
where $I$ is the identity matrix. Keep in mind that "$4$" does not denote a matrix (unless we are talking about $1times1$ matrices). To complete this, we have
$$X=(A-4I)^{-1}B$$
For the second example, your first step is wrong. In general, whether for real numbers or matrices $a$ and $b$ and integers $n$, it is not true that
$$(a+b)^{n}=a^{n}+b^{n}$$
What we can do is the following:
$$(A+4X)^{-1}=B$$
$$I=B(A+4X)=BA+4BX$$
$$I-BA=4BX$$
$$frac{1}{4}(I-BA)=BX$$
$$X=frac{1}{4}B^{-1}(I-BA)$$
I should mention that with all of the above manipulations, things only make sense if the corresponding matrices are invertible. For example, if $(A-4I)$ is not invertible then we cannot simplify the first example as shown.
answered Jan 16 at 17:56
pwerthpwerth
3,320417
$endgroup$
$begingroup$
It should be $AX-4X=(A-4I)X$ else matrix multiplication need not be compatible.
$endgroup$
– Yadati Kiran
Jan 16 at 18:05
$begingroup$
Good point. Updated my answer
$endgroup$
– pwerth
Jan 16 at 18:12
add a comment |
$begingroup$
Your first example is almost correct. You just change your $4$ to $ 4I$
Your second example needs serious adjustments.
First take inverse of both sides , then try to isolate $X$ as you did in the first example.
answered Jan 16 at 17:59
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
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2 Answers
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2 Answers
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votes
$begingroup$
Both examples contain some errors. For the first one, you have the right idea subtracting to get
$$AX-4X=B$$
however, you have to be careful about factoring matrices. Observe that
$$AX-4X=(A-4I)X$$
where $I$ is the identity matrix. Keep in mind that "$4$" does not denote a matrix (unless we are talking about $1times1$ matrices). To complete this, we have
$$X=(A-4I)^{-1}B$$
For the second example, your first step is wrong. In general, whether for real numbers or matrices $a$ and $b$ and integers $n$, it is not true that
$$(a+b)^{n}=a^{n}+b^{n}$$
What we can do is the following:
$$(A+4X)^{-1}=B$$
$$I=B(A+4X)=BA+4BX$$
$$I-BA=4BX$$
$$frac{1}{4}(I-BA)=BX$$
$$X=frac{1}{4}B^{-1}(I-BA)$$
I should mention that with all of the above manipulations, things only make sense if the corresponding matrices are invertible. For example, if $(A-4I)$ is not invertible then we cannot simplify the first example as shown.
answered Jan 16 at 17:56
pwerthpwerth
3,320417
$endgroup$
$begingroup$
It should be $AX-4X=(A-4I)X$ else matrix multiplication need not be compatible.
$endgroup$
– Yadati Kiran
Jan 16 at 18:05
$begingroup$
Good point. Updated my answer
$endgroup$
– pwerth
Jan 16 at 18:12
add a comment |
$begingroup$
Both examples contain some errors. For the first one, you have the right idea subtracting to get
$$AX-4X=B$$
however, you have to be careful about factoring matrices. Observe that
$$AX-4X=(A-4I)X$$
where $I$ is the identity matrix. Keep in mind that "$4$" does not denote a matrix (unless we are talking about $1times1$ matrices). To complete this, we have
$$X=(A-4I)^{-1}B$$
For the second example, your first step is wrong. In general, whether for real numbers or matrices $a$ and $b$ and integers $n$, it is not true that
$$(a+b)^{n}=a^{n}+b^{n}$$
What we can do is the following:
$$(A+4X)^{-1}=B$$
$$I=B(A+4X)=BA+4BX$$
$$I-BA=4BX$$
$$frac{1}{4}(I-BA)=BX$$
$$X=frac{1}{4}B^{-1}(I-BA)$$
I should mention that with all of the above manipulations, things only make sense if the corresponding matrices are invertible. For example, if $(A-4I)$ is not invertible then we cannot simplify the first example as shown.
answered Jan 16 at 17:56
pwerthpwerth
3,320417
$endgroup$
$begingroup$
It should be $AX-4X=(A-4I)X$ else matrix multiplication need not be compatible.
$endgroup$
– Yadati Kiran
Jan 16 at 18:05
$begingroup$
Good point. Updated my answer
$endgroup$
– pwerth
Jan 16 at 18:12
add a comment |
$begingroup$
Both examples contain some errors. For the first one, you have the right idea subtracting to get
$$AX-4X=B$$
however, you have to be careful about factoring matrices. Observe that
$$AX-4X=(A-4I)X$$
where $I$ is the identity matrix. Keep in mind that "$4$" does not denote a matrix (unless we are talking about $1times1$ matrices). To complete this, we have
$$X=(A-4I)^{-1}B$$
For the second example, your first step is wrong. In general, whether for real numbers or matrices $a$ and $b$ and integers $n$, it is not true that
$$(a+b)^{n}=a^{n}+b^{n}$$
What we can do is the following:
$$(A+4X)^{-1}=B$$
$$I=B(A+4X)=BA+4BX$$
$$I-BA=4BX$$
$$frac{1}{4}(I-BA)=BX$$
$$X=frac{1}{4}B^{-1}(I-BA)$$
I should mention that with all of the above manipulations, things only make sense if the corresponding matrices are invertible. For example, if $(A-4I)$ is not invertible then we cannot simplify the first example as shown.
answered Jan 16 at 17:56
pwerthpwerth
3,320417
$endgroup$
Both examples contain some errors. For the first one, you have the right idea subtracting to get
$$AX-4X=B$$
however, you have to be careful about factoring matrices. Observe that
$$AX-4X=(A-4I)X$$
where $I$ is the identity matrix. Keep in mind that "$4$" does not denote a matrix (unless we are talking about $1times1$ matrices). To complete this, we have
$$X=(A-4I)^{-1}B$$
For the second example, your first step is wrong. In general, whether for real numbers or matrices $a$ and $b$ and integers $n$, it is not true that
$$(a+b)^{n}=a^{n}+b^{n}$$
What we can do is the following:
$$(A+4X)^{-1}=B$$
$$I=B(A+4X)=BA+4BX$$
$$I-BA=4BX$$
$$frac{1}{4}(I-BA)=BX$$
$$X=frac{1}{4}B^{-1}(I-BA)$$
I should mention that with all of the above manipulations, things only make sense if the corresponding matrices are invertible. For example, if $(A-4I)$ is not invertible then we cannot simplify the first example as shown.
answered Jan 16 at 17:56
pwerthpwerth
3,320417
edited Jan 16 at 18:12
answered Jan 16 at 17:56
pwerthpwerth
3,320417
answered Jan 16 at 17:56
pwerthpwerth
3,320417
answered Jan 16 at 17:56
pwerthpwerth
3,320417
3,320417
$begingroup$
It should be $AX-4X=(A-4I)X$ else matrix multiplication need not be compatible.
$endgroup$
– Yadati Kiran
Jan 16 at 18:05
$begingroup$
Good point. Updated my answer
$endgroup$
– pwerth
Jan 16 at 18:12
add a comment |
$begingroup$
It should be $AX-4X=(A-4I)X$ else matrix multiplication need not be compatible.
$endgroup$
– Yadati Kiran
Jan 16 at 18:05
$begingroup$
Good point. Updated my answer
$endgroup$
– pwerth
Jan 16 at 18:12
$begingroup$
It should be $AX-4X=(A-4I)X$ else matrix multiplication need not be compatible.
$endgroup$
– Yadati Kiran
Jan 16 at 18:05
$begingroup$
It should be $AX-4X=(A-4I)X$ else matrix multiplication need not be compatible.
$endgroup$
– Yadati Kiran
Jan 16 at 18:05
$begingroup$
Good point. Updated my answer
$endgroup$
– pwerth
Jan 16 at 18:12
$begingroup$
Good point. Updated my answer
$endgroup$
– pwerth
Jan 16 at 18:12
add a comment |
$begingroup$
Your first example is almost correct. You just change your $4$ to $ 4I$
Your second example needs serious adjustments.
First take inverse of both sides , then try to isolate $X$ as you did in the first example.
answered Jan 16 at 17:59
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
Your first example is almost correct. You just change your $4$ to $ 4I$
Your second example needs serious adjustments.
First take inverse of both sides , then try to isolate $X$ as you did in the first example.
answered Jan 16 at 17:59
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
Your first example is almost correct. You just change your $4$ to $ 4I$
Your second example needs serious adjustments.
First take inverse of both sides , then try to isolate $X$ as you did in the first example.
answered Jan 16 at 17:59
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
Your first example is almost correct. You just change your $4$ to $ 4I$
Your second example needs serious adjustments.
First take inverse of both sides , then try to isolate $X$ as you did in the first example.
answered Jan 16 at 17:59
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 16 at 17:59
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 16 at 17:59
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 16 at 17:59
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
41.6k42061
add a comment |
add a comment |
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$begingroup$
The expression $A-4$ is undefined: it’s the difference of a matrix and scalar. Get into the habit of writing $A-4I$ (or $4E$, if that’s what the identity matrix is called in your materials.)
$endgroup$
– amd
Jan 16 at 17:53
1
$begingroup$
$(A+4X)^{-1}$ is no more equal to $A^{-1}+frac14X^{-1}$ than $1/(a+4x)$ is equal to $1/a+1/(4x)$.
$endgroup$
– amd
Jan 16 at 17:54