Mixing
Obtain an expression for the inverse of a partitioned matrix.
$begingroup$
Consider the $m$x$m$ partitioned matrix
$
begin{bmatrix}
A_{11} & (0) \
A_{21} & A_{22}
end{bmatrix}
=A$
where the $m_1$x$m_1$ matrix $A_{11}$ and the $m_2$x$m_2$ matrix $A_{22}$ are nonsingular. Obtain an expression for $A^{-1}$ in terms of $A_{11},A_{22},$ and $A_{21}$.
I do not even know where to start with this problem. There are many similar problems in the textbook with no available solution sets. I would greatly appreciate any help towards this problem. Thank you!
matrices
asked Jan 25 at 1:40
MatthewMatthew
687
$endgroup$
add a comment |
$begingroup$
Consider the $m$x$m$ partitioned matrix
$
begin{bmatrix}
A_{11} & (0) \
A_{21} & A_{22}
end{bmatrix}
=A$
where the $m_1$x$m_1$ matrix $A_{11}$ and the $m_2$x$m_2$ matrix $A_{22}$ are nonsingular. Obtain an expression for $A^{-1}$ in terms of $A_{11},A_{22},$ and $A_{21}$.
I do not even know where to start with this problem. There are many similar problems in the textbook with no available solution sets. I would greatly appreciate any help towards this problem. Thank you!
matrices
asked Jan 25 at 1:40
MatthewMatthew
687
$endgroup$
add a comment |
$begingroup$
Consider the $m$x$m$ partitioned matrix
$
begin{bmatrix}
A_{11} & (0) \
A_{21} & A_{22}
end{bmatrix}
=A$
where the $m_1$x$m_1$ matrix $A_{11}$ and the $m_2$x$m_2$ matrix $A_{22}$ are nonsingular. Obtain an expression for $A^{-1}$ in terms of $A_{11},A_{22},$ and $A_{21}$.
I do not even know where to start with this problem. There are many similar problems in the textbook with no available solution sets. I would greatly appreciate any help towards this problem. Thank you!
matrices
asked Jan 25 at 1:40
MatthewMatthew
687
$endgroup$
Consider the $m$x$m$ partitioned matrix
$
begin{bmatrix}
A_{11} & (0) \
A_{21} & A_{22}
end{bmatrix}
=A$
where the $m_1$x$m_1$ matrix $A_{11}$ and the $m_2$x$m_2$ matrix $A_{22}$ are nonsingular. Obtain an expression for $A^{-1}$ in terms of $A_{11},A_{22},$ and $A_{21}$.
I do not even know where to start with this problem. There are many similar problems in the textbook with no available solution sets. I would greatly appreciate any help towards this problem. Thank you!
matrices
matrices
asked Jan 25 at 1:40
MatthewMatthew
687
asked Jan 25 at 1:40
MatthewMatthew
687
asked Jan 25 at 1:40
MatthewMatthew
687
asked Jan 25 at 1:40
MatthewMatthew
687
asked Jan 25 at 1:40
MatthewMatthew
687
687
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The way to do this problem is the same way that you would find an inverse for a numeric $2times 2$ matrix. The only caveat is that we need to remember that there is no such things as division with matrices, but division is equivalent to multiplying by an inverse.
The way that I was always taught to find inverses was to append the first matrix with the identity matrix and reduce the first matrix to the identity matrix; the matrix on the right is now the inverse.
For example, we begin with
$$begin{bmatrix}
A_{11} & 0 & I & 0 \
A_{21} & A_{22} & 0 & I end{bmatrix}$$
If this were a numeric matrix, we would divide the top row by the first entry, which is now the same as multiplying it by the inverse, so we have
$$begin{bmatrix}
A_{11} & 0 & I & 0 \
A_{21} & A_{22} & 0 & I end{bmatrix}simbegin{bmatrix}
I & 0 & A_{11}^{-1} & 0 \
A_{21} & A_{22} & 0 & I end{bmatrix}$$
We continue this process until the left hand side becomes the identity matrix, and you should get something like
$$A^{-1}=begin{bmatrix}
A^{-1}_{11} & 0 \
-A^{-1}_{22}A_{21}A_{11}^{-1} & A_{22}^{-1}end{bmatrix}$$
You can check for yourself that this matrix is indeed the inverse by multiplying by the original matrix.
answered Jan 25 at 1:54
Josh B.Josh B.
1,0298
$endgroup$
$begingroup$
This was an excellent explanation and answer. Thank you for your input!
$endgroup$
– Matthew
Jan 25 at 3:17
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3086599%2fobtain-an-expression-for-the-inverse-of-a-partitioned-matrix%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The way to do this problem is the same way that you would find an inverse for a numeric $2times 2$ matrix. The only caveat is that we need to remember that there is no such things as division with matrices, but division is equivalent to multiplying by an inverse.
The way that I was always taught to find inverses was to append the first matrix with the identity matrix and reduce the first matrix to the identity matrix; the matrix on the right is now the inverse.
For example, we begin with
$$begin{bmatrix}
A_{11} & 0 & I & 0 \
A_{21} & A_{22} & 0 & I end{bmatrix}$$
If this were a numeric matrix, we would divide the top row by the first entry, which is now the same as multiplying it by the inverse, so we have
$$begin{bmatrix}
A_{11} & 0 & I & 0 \
A_{21} & A_{22} & 0 & I end{bmatrix}simbegin{bmatrix}
I & 0 & A_{11}^{-1} & 0 \
A_{21} & A_{22} & 0 & I end{bmatrix}$$
We continue this process until the left hand side becomes the identity matrix, and you should get something like
$$A^{-1}=begin{bmatrix}
A^{-1}_{11} & 0 \
-A^{-1}_{22}A_{21}A_{11}^{-1} & A_{22}^{-1}end{bmatrix}$$
You can check for yourself that this matrix is indeed the inverse by multiplying by the original matrix.
answered Jan 25 at 1:54
Josh B.Josh B.
1,0298
$endgroup$
$begingroup$
This was an excellent explanation and answer. Thank you for your input!
$endgroup$
– Matthew
Jan 25 at 3:17
add a comment |
$begingroup$
The way to do this problem is the same way that you would find an inverse for a numeric $2times 2$ matrix. The only caveat is that we need to remember that there is no such things as division with matrices, but division is equivalent to multiplying by an inverse.
The way that I was always taught to find inverses was to append the first matrix with the identity matrix and reduce the first matrix to the identity matrix; the matrix on the right is now the inverse.
For example, we begin with
$$begin{bmatrix}
A_{11} & 0 & I & 0 \
A_{21} & A_{22} & 0 & I end{bmatrix}$$
If this were a numeric matrix, we would divide the top row by the first entry, which is now the same as multiplying it by the inverse, so we have
$$begin{bmatrix}
A_{11} & 0 & I & 0 \
A_{21} & A_{22} & 0 & I end{bmatrix}simbegin{bmatrix}
I & 0 & A_{11}^{-1} & 0 \
A_{21} & A_{22} & 0 & I end{bmatrix}$$
We continue this process until the left hand side becomes the identity matrix, and you should get something like
$$A^{-1}=begin{bmatrix}
A^{-1}_{11} & 0 \
-A^{-1}_{22}A_{21}A_{11}^{-1} & A_{22}^{-1}end{bmatrix}$$
You can check for yourself that this matrix is indeed the inverse by multiplying by the original matrix.
answered Jan 25 at 1:54
Josh B.Josh B.
1,0298
$endgroup$
$begingroup$
This was an excellent explanation and answer. Thank you for your input!
$endgroup$
– Matthew
Jan 25 at 3:17
add a comment |
$begingroup$
The way to do this problem is the same way that you would find an inverse for a numeric $2times 2$ matrix. The only caveat is that we need to remember that there is no such things as division with matrices, but division is equivalent to multiplying by an inverse.
The way that I was always taught to find inverses was to append the first matrix with the identity matrix and reduce the first matrix to the identity matrix; the matrix on the right is now the inverse.
For example, we begin with
$$begin{bmatrix}
A_{11} & 0 & I & 0 \
A_{21} & A_{22} & 0 & I end{bmatrix}$$
If this were a numeric matrix, we would divide the top row by the first entry, which is now the same as multiplying it by the inverse, so we have
$$begin{bmatrix}
A_{11} & 0 & I & 0 \
A_{21} & A_{22} & 0 & I end{bmatrix}simbegin{bmatrix}
I & 0 & A_{11}^{-1} & 0 \
A_{21} & A_{22} & 0 & I end{bmatrix}$$
We continue this process until the left hand side becomes the identity matrix, and you should get something like
$$A^{-1}=begin{bmatrix}
A^{-1}_{11} & 0 \
-A^{-1}_{22}A_{21}A_{11}^{-1} & A_{22}^{-1}end{bmatrix}$$
You can check for yourself that this matrix is indeed the inverse by multiplying by the original matrix.
answered Jan 25 at 1:54
Josh B.Josh B.
1,0298
$endgroup$
The way to do this problem is the same way that you would find an inverse for a numeric $2times 2$ matrix. The only caveat is that we need to remember that there is no such things as division with matrices, but division is equivalent to multiplying by an inverse.
The way that I was always taught to find inverses was to append the first matrix with the identity matrix and reduce the first matrix to the identity matrix; the matrix on the right is now the inverse.
For example, we begin with
$$begin{bmatrix}
A_{11} & 0 & I & 0 \
A_{21} & A_{22} & 0 & I end{bmatrix}$$
If this were a numeric matrix, we would divide the top row by the first entry, which is now the same as multiplying it by the inverse, so we have
$$begin{bmatrix}
A_{11} & 0 & I & 0 \
A_{21} & A_{22} & 0 & I end{bmatrix}simbegin{bmatrix}
I & 0 & A_{11}^{-1} & 0 \
A_{21} & A_{22} & 0 & I end{bmatrix}$$
We continue this process until the left hand side becomes the identity matrix, and you should get something like
$$A^{-1}=begin{bmatrix}
A^{-1}_{11} & 0 \
-A^{-1}_{22}A_{21}A_{11}^{-1} & A_{22}^{-1}end{bmatrix}$$
You can check for yourself that this matrix is indeed the inverse by multiplying by the original matrix.
answered Jan 25 at 1:54
Josh B.Josh B.
1,0298
answered Jan 25 at 1:54
Josh B.Josh B.
1,0298
answered Jan 25 at 1:54
Josh B.Josh B.
1,0298
answered Jan 25 at 1:54
Josh B.Josh B.
1,0298
1,0298
$begingroup$
This was an excellent explanation and answer. Thank you for your input!
$endgroup$
– Matthew
Jan 25 at 3:17
add a comment |
$begingroup$
This was an excellent explanation and answer. Thank you for your input!
$endgroup$
– Matthew
Jan 25 at 3:17
$begingroup$
This was an excellent explanation and answer. Thank you for your input!
$endgroup$
– Matthew
Jan 25 at 3:17
$begingroup$
This was an excellent explanation and answer. Thank you for your input!
$endgroup$
– Matthew
Jan 25 at 3:17
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3086599%2fobtain-an-expression-for-the-inverse-of-a-partitioned-matrix%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
