Largest value of determinantProving the determinant on the L.H.S = determinant on the R.H.S.Prove that the...
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Largest value of determinant
Proving the determinant on the L.H.S = determinant on the R.H.S.Prove that the determinant is $0$ by expressing as a productLeast value of a determinantWrite the given Determinant as the product of two Determinantsdeterminant diagonal zero symmetric matrixSolving $n$th order determinantMatrix Determinant CalculationProve determinant of a matrix with trigonometry functions.Proving that two determinants are equal without expanding themShow the following matrix has determinant = 0
$begingroup$
If $alpha,beta,gamma in [-3,10].$ Then largest value of the determinant
$$begin{vmatrix}3alpha^2&beta^2+alphabeta+alpha^2&gamma^2+alphagamma+alpha^2\\
alpha^2+alphabeta+beta^2& 3beta^2&gamma^2+betagamma+beta^2\\
alpha^2+alphagamma+gamma^2& beta^2+betagamma+gamma^2&3gamma^2end{vmatrix}$$
Try: I am trying to break that determinant into product of 2 determinants but not able to break it.
Could someone help me in this question? Thanks.
linear-algebra determinant
edited 1 hour ago
StubbornAtom
6,74931442
asked 1 hour ago
DXTDXT
5,9812733
$endgroup$
add a comment |
$begingroup$
If $alpha,beta,gamma in [-3,10].$ Then largest value of the determinant
$$begin{vmatrix}3alpha^2&beta^2+alphabeta+alpha^2&gamma^2+alphagamma+alpha^2\\
alpha^2+alphabeta+beta^2& 3beta^2&gamma^2+betagamma+beta^2\\
alpha^2+alphagamma+gamma^2& beta^2+betagamma+gamma^2&3gamma^2end{vmatrix}$$
Try: I am trying to break that determinant into product of 2 determinants but not able to break it.
Could someone help me in this question? Thanks.
linear-algebra determinant
edited 1 hour ago
StubbornAtom
6,74931442
asked 1 hour ago
DXTDXT
5,9812733
$endgroup$
add a comment |
$begingroup$
If $alpha,beta,gamma in [-3,10].$ Then largest value of the determinant
$$begin{vmatrix}3alpha^2&beta^2+alphabeta+alpha^2&gamma^2+alphagamma+alpha^2\\
alpha^2+alphabeta+beta^2& 3beta^2&gamma^2+betagamma+beta^2\\
alpha^2+alphagamma+gamma^2& beta^2+betagamma+gamma^2&3gamma^2end{vmatrix}$$
Try: I am trying to break that determinant into product of 2 determinants but not able to break it.
Could someone help me in this question? Thanks.
linear-algebra determinant
edited 1 hour ago
StubbornAtom
6,74931442
asked 1 hour ago
DXTDXT
5,9812733
$endgroup$
If $alpha,beta,gamma in [-3,10].$ Then largest value of the determinant
$$begin{vmatrix}3alpha^2&beta^2+alphabeta+alpha^2&gamma^2+alphagamma+alpha^2\\
alpha^2+alphabeta+beta^2& 3beta^2&gamma^2+betagamma+beta^2\\
alpha^2+alphagamma+gamma^2& beta^2+betagamma+gamma^2&3gamma^2end{vmatrix}$$
Try: I am trying to break that determinant into product of 2 determinants but not able to break it.
Could someone help me in this question? Thanks.
linear-algebra determinant
linear-algebra determinant
edited 1 hour ago
StubbornAtom
6,74931442
asked 1 hour ago
DXTDXT
5,9812733
edited 1 hour ago
StubbornAtom
6,74931442
asked 1 hour ago
DXTDXT
5,9812733
edited 1 hour ago
StubbornAtom
6,74931442
edited 1 hour ago
StubbornAtom
6,74931442
edited 1 hour ago
StubbornAtom
6,74931442
6,74931442
asked 1 hour ago
DXTDXT
5,9812733
asked 1 hour ago
DXTDXT
5,9812733
asked 1 hour ago
DXTDXT
5,9812733
5,9812733
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: Using the rules of SARRUS we get after simplifying $$- left( alpha-gamma right) ^{2} left( beta-gamma right) ^{2}
left( beta-alpha right) ^{2}
$$
answered 1 hour ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.9k42867
$endgroup$
$begingroup$
I think so! What result do you get?
$endgroup$
– Dr. Sonnhard Graubner
58 mins ago
$begingroup$
This is nice to hear!
$endgroup$
– Dr. Sonnhard Graubner
56 mins ago
$begingroup$
A big typo is only a typo, big or not!
$endgroup$
– Dr. Sonnhard Graubner
55 mins ago
add a comment |
$begingroup$
Hint: The matrix is symmetric. You can use the Cholesky decomposition to rewrite the matrix $A$ as $A=LL^T$, in which $L$ is a lower triangular matrix. Then the determinant is $det A = det L det L^T = det^2 L$. Then $det L$ is given by the product of the diagonal elements of $L$ as it is a lower triangular matrix.
answered 1 hour ago
MachineLearnerMachineLearner
1,906213
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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oldest
votes
$begingroup$
Hint: Using the rules of SARRUS we get after simplifying $$- left( alpha-gamma right) ^{2} left( beta-gamma right) ^{2}
left( beta-alpha right) ^{2}
$$
answered 1 hour ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.9k42867
$endgroup$
$begingroup$
I think so! What result do you get?
$endgroup$
– Dr. Sonnhard Graubner
58 mins ago
$begingroup$
This is nice to hear!
$endgroup$
– Dr. Sonnhard Graubner
56 mins ago
$begingroup$
A big typo is only a typo, big or not!
$endgroup$
– Dr. Sonnhard Graubner
55 mins ago
add a comment |
$begingroup$
Hint: Using the rules of SARRUS we get after simplifying $$- left( alpha-gamma right) ^{2} left( beta-gamma right) ^{2}
left( beta-alpha right) ^{2}
$$
answered 1 hour ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.9k42867
$endgroup$
$begingroup$
I think so! What result do you get?
$endgroup$
– Dr. Sonnhard Graubner
58 mins ago
$begingroup$
This is nice to hear!
$endgroup$
– Dr. Sonnhard Graubner
56 mins ago
$begingroup$
A big typo is only a typo, big or not!
$endgroup$
– Dr. Sonnhard Graubner
55 mins ago
add a comment |
$begingroup$
Hint: Using the rules of SARRUS we get after simplifying $$- left( alpha-gamma right) ^{2} left( beta-gamma right) ^{2}
left( beta-alpha right) ^{2}
$$
answered 1 hour ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.9k42867
$endgroup$
Hint: Using the rules of SARRUS we get after simplifying $$- left( alpha-gamma right) ^{2} left( beta-gamma right) ^{2}
left( beta-alpha right) ^{2}
$$
answered 1 hour ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.9k42867
answered 1 hour ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.9k42867
answered 1 hour ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.9k42867
answered 1 hour ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.9k42867
79.9k42867
$begingroup$
I think so! What result do you get?
$endgroup$
– Dr. Sonnhard Graubner
58 mins ago
$begingroup$
This is nice to hear!
$endgroup$
– Dr. Sonnhard Graubner
56 mins ago
$begingroup$
A big typo is only a typo, big or not!
$endgroup$
– Dr. Sonnhard Graubner
55 mins ago
add a comment |
$begingroup$
I think so! What result do you get?
$endgroup$
– Dr. Sonnhard Graubner
58 mins ago
$begingroup$
This is nice to hear!
$endgroup$
– Dr. Sonnhard Graubner
56 mins ago
$begingroup$
A big typo is only a typo, big or not!
$endgroup$
– Dr. Sonnhard Graubner
55 mins ago
$begingroup$
I think so! What result do you get?
$endgroup$
– Dr. Sonnhard Graubner
58 mins ago
$begingroup$
I think so! What result do you get?
$endgroup$
– Dr. Sonnhard Graubner
58 mins ago
$begingroup$
This is nice to hear!
$endgroup$
– Dr. Sonnhard Graubner
56 mins ago
$begingroup$
This is nice to hear!
$endgroup$
– Dr. Sonnhard Graubner
56 mins ago
$begingroup$
A big typo is only a typo, big or not!
$endgroup$
– Dr. Sonnhard Graubner
55 mins ago
$begingroup$
A big typo is only a typo, big or not!
$endgroup$
– Dr. Sonnhard Graubner
55 mins ago
add a comment |
$begingroup$
Hint: The matrix is symmetric. You can use the Cholesky decomposition to rewrite the matrix $A$ as $A=LL^T$, in which $L$ is a lower triangular matrix. Then the determinant is $det A = det L det L^T = det^2 L$. Then $det L$ is given by the product of the diagonal elements of $L$ as it is a lower triangular matrix.
answered 1 hour ago
MachineLearnerMachineLearner
1,906213
$endgroup$
add a comment |
$begingroup$
Hint: The matrix is symmetric. You can use the Cholesky decomposition to rewrite the matrix $A$ as $A=LL^T$, in which $L$ is a lower triangular matrix. Then the determinant is $det A = det L det L^T = det^2 L$. Then $det L$ is given by the product of the diagonal elements of $L$ as it is a lower triangular matrix.
answered 1 hour ago
MachineLearnerMachineLearner
1,906213
$endgroup$
add a comment |
$begingroup$
Hint: The matrix is symmetric. You can use the Cholesky decomposition to rewrite the matrix $A$ as $A=LL^T$, in which $L$ is a lower triangular matrix. Then the determinant is $det A = det L det L^T = det^2 L$. Then $det L$ is given by the product of the diagonal elements of $L$ as it is a lower triangular matrix.
answered 1 hour ago
MachineLearnerMachineLearner
1,906213
$endgroup$
Hint: The matrix is symmetric. You can use the Cholesky decomposition to rewrite the matrix $A$ as $A=LL^T$, in which $L$ is a lower triangular matrix. Then the determinant is $det A = det L det L^T = det^2 L$. Then $det L$ is given by the product of the diagonal elements of $L$ as it is a lower triangular matrix.
answered 1 hour ago
MachineLearnerMachineLearner
1,906213
answered 1 hour ago
MachineLearnerMachineLearner
1,906213
answered 1 hour ago
MachineLearnerMachineLearner
1,906213
answered 1 hour ago
MachineLearnerMachineLearner
1,906213
1,906213
add a comment |
add a comment |
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