Mixing
What is the probability of the matrix being Singular
$begingroup$
Consider the set of all boolean square matrices of order $3 times 3$ as shown below where a,b,c,d,e,f can be either 0 or 1.
$begin{bmatrix}
a&b&c\
0&d&e\
0&0&f
end{bmatrix}$
Out of all possible boolean matrices of above type, a matrix is chosen at random.The probability that the matrix is singular is?
My Work
The above matrix is a triangular matrix and it's determinant can be 0, if either one or more from a,d and f are 0 in which case 0 will be an eigen value of the matrix and hence determinant 0.
Number of ways in which I can set $a,d,f$ to zero are: $binom{3}{1}+binom{3}{2}+binom{3}{3}=7,ways$
Now, total given boolean matrices possible are
$2^6=64$
So, the required probability must be $frac{7}{64}$
Is my answer correct?
probability
asked Jan 22 at 8:12
user3767495user3767495
4078
$endgroup$
add a comment |
$begingroup$
Consider the set of all boolean square matrices of order $3 times 3$ as shown below where a,b,c,d,e,f can be either 0 or 1.
$begin{bmatrix}
a&b&c\
0&d&e\
0&0&f
end{bmatrix}$
Out of all possible boolean matrices of above type, a matrix is chosen at random.The probability that the matrix is singular is?
My Work
The above matrix is a triangular matrix and it's determinant can be 0, if either one or more from a,d and f are 0 in which case 0 will be an eigen value of the matrix and hence determinant 0.
Number of ways in which I can set $a,d,f$ to zero are: $binom{3}{1}+binom{3}{2}+binom{3}{3}=7,ways$
Now, total given boolean matrices possible are
$2^6=64$
So, the required probability must be $frac{7}{64}$
Is my answer correct?
probability
asked Jan 22 at 8:12
user3767495user3767495
4078
$endgroup$
add a comment |
$begingroup$
Consider the set of all boolean square matrices of order $3 times 3$ as shown below where a,b,c,d,e,f can be either 0 or 1.
$begin{bmatrix}
a&b&c\
0&d&e\
0&0&f
end{bmatrix}$
Out of all possible boolean matrices of above type, a matrix is chosen at random.The probability that the matrix is singular is?
My Work
The above matrix is a triangular matrix and it's determinant can be 0, if either one or more from a,d and f are 0 in which case 0 will be an eigen value of the matrix and hence determinant 0.
Number of ways in which I can set $a,d,f$ to zero are: $binom{3}{1}+binom{3}{2}+binom{3}{3}=7,ways$
Now, total given boolean matrices possible are
$2^6=64$
So, the required probability must be $frac{7}{64}$
Is my answer correct?
probability
asked Jan 22 at 8:12
user3767495user3767495
4078
$endgroup$
Consider the set of all boolean square matrices of order $3 times 3$ as shown below where a,b,c,d,e,f can be either 0 or 1.
$begin{bmatrix}
a&b&c\
0&d&e\
0&0&f
end{bmatrix}$
Out of all possible boolean matrices of above type, a matrix is chosen at random.The probability that the matrix is singular is?
My Work
The above matrix is a triangular matrix and it's determinant can be 0, if either one or more from a,d and f are 0 in which case 0 will be an eigen value of the matrix and hence determinant 0.
Number of ways in which I can set $a,d,f$ to zero are: $binom{3}{1}+binom{3}{2}+binom{3}{3}=7,ways$
Now, total given boolean matrices possible are
$2^6=64$
So, the required probability must be $frac{7}{64}$
Is my answer correct?
probability
probability
asked Jan 22 at 8:12
user3767495user3767495
4078
asked Jan 22 at 8:12
user3767495user3767495
4078
asked Jan 22 at 8:12
user3767495user3767495
4078
asked Jan 22 at 8:12
user3767495user3767495
4078
asked Jan 22 at 8:12
user3767495user3767495
4078
4078
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
To be singular, we need $a=0$ or $d=0$ or $f=0$.
To be non-singular, we need $a=d=f=1$.
Hence, the probaility is $$1-frac{1}{2^3}=frac{7}{8}.$$
Note that the values that $b,c,e$ takes are irrelevant. Number of ways to get a singular matrix is $7 times 8=56$.
answered Jan 22 at 8:22
Siong Thye GohSiong Thye Goh
102k1468119
$endgroup$
$begingroup$
So, where I went wrong?
$endgroup$
– user3767495
Jan 22 at 8:25
$begingroup$
you forgot to multiply by $8$ for the values of $b,c,e$.
$endgroup$
– Siong Thye Goh
Jan 22 at 8:26
add a comment |
$begingroup$
It looks basically correct but I'd word it (fully) as follows:
Any given square matrix (over a field) is singular if and only if its determinant is zero. The given matrix is a triangular one and thus its determinant is simply the product of the elements in the main diagonal. Then the matrix is singular iff
$$;adf=0iff a=0,vee,d=0;vee, f=0$$
and etc.
answered Jan 22 at 8:17
DonAntonioDonAntonio
179k1494233
$endgroup$
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%2f3082873%2fwhat-is-the-probability-of-the-matrix-being-singular%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To be singular, we need $a=0$ or $d=0$ or $f=0$.
To be non-singular, we need $a=d=f=1$.
Hence, the probaility is $$1-frac{1}{2^3}=frac{7}{8}.$$
Note that the values that $b,c,e$ takes are irrelevant. Number of ways to get a singular matrix is $7 times 8=56$.
answered Jan 22 at 8:22
Siong Thye GohSiong Thye Goh
102k1468119
$endgroup$
$begingroup$
So, where I went wrong?
$endgroup$
– user3767495
Jan 22 at 8:25
$begingroup$
you forgot to multiply by $8$ for the values of $b,c,e$.
$endgroup$
– Siong Thye Goh
Jan 22 at 8:26
add a comment |
$begingroup$
To be singular, we need $a=0$ or $d=0$ or $f=0$.
To be non-singular, we need $a=d=f=1$.
Hence, the probaility is $$1-frac{1}{2^3}=frac{7}{8}.$$
Note that the values that $b,c,e$ takes are irrelevant. Number of ways to get a singular matrix is $7 times 8=56$.
answered Jan 22 at 8:22
Siong Thye GohSiong Thye Goh
102k1468119
$endgroup$
$begingroup$
So, where I went wrong?
$endgroup$
– user3767495
Jan 22 at 8:25
$begingroup$
you forgot to multiply by $8$ for the values of $b,c,e$.
$endgroup$
– Siong Thye Goh
Jan 22 at 8:26
add a comment |
$begingroup$
To be singular, we need $a=0$ or $d=0$ or $f=0$.
To be non-singular, we need $a=d=f=1$.
Hence, the probaility is $$1-frac{1}{2^3}=frac{7}{8}.$$
Note that the values that $b,c,e$ takes are irrelevant. Number of ways to get a singular matrix is $7 times 8=56$.
answered Jan 22 at 8:22
Siong Thye GohSiong Thye Goh
102k1468119
$endgroup$
To be singular, we need $a=0$ or $d=0$ or $f=0$.
To be non-singular, we need $a=d=f=1$.
Hence, the probaility is $$1-frac{1}{2^3}=frac{7}{8}.$$
Note that the values that $b,c,e$ takes are irrelevant. Number of ways to get a singular matrix is $7 times 8=56$.
answered Jan 22 at 8:22
Siong Thye GohSiong Thye Goh
102k1468119
answered Jan 22 at 8:22
Siong Thye GohSiong Thye Goh
102k1468119
answered Jan 22 at 8:22
Siong Thye GohSiong Thye Goh
102k1468119
answered Jan 22 at 8:22
Siong Thye GohSiong Thye Goh
102k1468119
102k1468119
$begingroup$
So, where I went wrong?
$endgroup$
– user3767495
Jan 22 at 8:25
$begingroup$
you forgot to multiply by $8$ for the values of $b,c,e$.
$endgroup$
– Siong Thye Goh
Jan 22 at 8:26
add a comment |
$begingroup$
So, where I went wrong?
$endgroup$
– user3767495
Jan 22 at 8:25
$begingroup$
you forgot to multiply by $8$ for the values of $b,c,e$.
$endgroup$
– Siong Thye Goh
Jan 22 at 8:26
$begingroup$
So, where I went wrong?
$endgroup$
– user3767495
Jan 22 at 8:25
$begingroup$
So, where I went wrong?
$endgroup$
– user3767495
Jan 22 at 8:25
$begingroup$
you forgot to multiply by $8$ for the values of $b,c,e$.
$endgroup$
– Siong Thye Goh
Jan 22 at 8:26
$begingroup$
you forgot to multiply by $8$ for the values of $b,c,e$.
$endgroup$
– Siong Thye Goh
Jan 22 at 8:26
add a comment |
$begingroup$
It looks basically correct but I'd word it (fully) as follows:
Any given square matrix (over a field) is singular if and only if its determinant is zero. The given matrix is a triangular one and thus its determinant is simply the product of the elements in the main diagonal. Then the matrix is singular iff
$$;adf=0iff a=0,vee,d=0;vee, f=0$$
and etc.
answered Jan 22 at 8:17
DonAntonioDonAntonio
179k1494233
$endgroup$
add a comment |
$begingroup$
It looks basically correct but I'd word it (fully) as follows:
Any given square matrix (over a field) is singular if and only if its determinant is zero. The given matrix is a triangular one and thus its determinant is simply the product of the elements in the main diagonal. Then the matrix is singular iff
$$;adf=0iff a=0,vee,d=0;vee, f=0$$
and etc.
answered Jan 22 at 8:17
DonAntonioDonAntonio
179k1494233
$endgroup$
add a comment |
$begingroup$
It looks basically correct but I'd word it (fully) as follows:
Any given square matrix (over a field) is singular if and only if its determinant is zero. The given matrix is a triangular one and thus its determinant is simply the product of the elements in the main diagonal. Then the matrix is singular iff
$$;adf=0iff a=0,vee,d=0;vee, f=0$$
and etc.
answered Jan 22 at 8:17
DonAntonioDonAntonio
179k1494233
$endgroup$
It looks basically correct but I'd word it (fully) as follows:
Any given square matrix (over a field) is singular if and only if its determinant is zero. The given matrix is a triangular one and thus its determinant is simply the product of the elements in the main diagonal. Then the matrix is singular iff
$$;adf=0iff a=0,vee,d=0;vee, f=0$$
and etc.
answered Jan 22 at 8:17
DonAntonioDonAntonio
179k1494233
edited Jan 22 at 18:27
answered Jan 22 at 8:17
DonAntonioDonAntonio
179k1494233
answered Jan 22 at 8:17
DonAntonioDonAntonio
179k1494233
answered Jan 22 at 8:17
DonAntonioDonAntonio
179k1494233
179k1494233
add a comment |
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%2f3082873%2fwhat-is-the-probability-of-the-matrix-being-singular%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
