What is the maximum determinant in 8 by 8 matrix of zeros and ones

(9) and is the formed by eliminating row and column from . This process is called A determinant can also be computed by writing down all , taking each permutation as the subscripts of the letters , , ..., and summing with signs determined by , where is the number of (Muir 1960, p.16), and is the , the permutations and the number of inversions they contain are 123 (0), 132 (1), 213 (1), 231 (2), 312 (2), and 321 (3), so the determinant is given by (28) Important properties of the determinant include the following, which include invariance under 1. Switching two rows or columns changes the sign. 2. Scalars can be factored out from rows and columns. 3. Multiples of rows and columns can be added together without changing the determinant's value. 4. Scalar multiplication of a row by a constant multiplies the determinant by . 5. A determinant with a row or column of zeros has value 0. 6. Any determinant with two rows or columns equal has value 0. Property 1 can be established by induction. For a (38) where is a . Several accounts state that Lewis Carroll (Charles Dodgson) sent Queen Victoria a copy of one of his mathematical works, in one account, An Elementary Treatise on Determinants. Heath (1974) states, "A well-known story tells how Queen Victoria, charmed by Alice in Wonderland, expressed a desire to receive the author's next work, and was presented, in due course, with a loyally inscribed copy of An Elementary Treatise on Determinants," while Gattegno (1974) asserts "Queen...

Hello Sal. I know I'm wrong and the answer is probably staring me in the face! Where's the fallacy in my thinking: As I understand it, a square matrix whose determinant is not zero is invertible. Therefore, using row operations, it can be reduced to having all its column vectors as pivot vectors. That's equvialent to an upper triangular matrix, with the main diagonal elements equal to 1. If normal row operations do not change the determinant, the determinant will be -1. HELP! :) Sal was a bit unfortunate when he said that "row operations do not change the determinant". The proof he gave was that row operations of the form [Ri -c*Rj -> Ri] (that is, replacing row i with row i minus row j times a scalar c) do not change the determinant. But there are row operations of different kind, such as k*Ri -c*Rj -> Ri (That is, replacing row i with row i times a scalar k minus row j times a scalar c). What can be proved is that operations of this kind do change the determinant. In fact, they multiply the determinant by k. And when you put an invertible matrix in RREF (that is, you turn it into an identity matrix), you must do these kinds of operations that scale the determinant. And they always end up scaling the determinant back to one, which is the determinant of any identity matrix. It depends on the size of A and B. Multiplying a matrix by a scalar, is the same as multiplying every row of that matrix by that scalar, and note, that multiplying a single row by a scalar is equivalent...

Vectors & Matrices • • • • • • • • • • • • More than just an online determinant calculator Wolfram|Alpha is the perfect resource to use for computing determinants of matrices. It can also calculate matrix products, rank, nullity, row reduction, diagonalization, eigenvalues, eigenvectors and much more. Learn more about: • Determinants » Tips for entering queries Use plain English or common mathematical syntax to enter your queries. To enter a matrix, separate elements with commas and rows with curly braces, brackets or parentheses. • • • • • • View more examples » Access instant learning tools Get immediate feedback and guidance with step-by-step solutions and Wolfram Problem Generator Learn more about: • Step-by-step solutions » • Wolfram Problem Generator » VIEW ALL CALCULATORS • BMI Calculator • Dilution Calculator • Mortgage Calculator • Interest Calculator • Loan Calculator • Present Value Calculator • Car Payment Calculator • Future Value Calculator • Limit Calculator • Derivative Calculator • Integral Calculator • Double Integral Calculator • Triple Integral Calculator • Series Expansion Calculator • Discontinuity Calculator • Domain and Range Calculator • Factoring Calculator • Quadratic Formula Calculator • Equation Solver Calculator • Partial Fraction Decomposition Calculator • System of Equations Calculator • Eigenvalue Calculator • Matrix Inverse Calculator Knowledgebase about determinants A determinant is a property of a square matrix. The value of the determin...

I shouldn't expect there to be exact results; compare the similar problem with matrices with entries $\pm1$. For an $n$-by-$n$ matrix with entries $\pm1$ one gets an upper bound for the determinant of $n^$. Taking some extra care one can ensure the first column of $A$ has a $1$ atop a column of zeros so that deleting the first row and column gives a smaller matrix with the same determinant. It's not clear to me whether this is the biggest possible.... As pointed out in Robin Chapman's answer and Frederio Poloni's comment thereto, there is a one-to-one map between normalized $n$-by- $n$ $(-1,1)$ matrices and $(n-1)$-by- $(n-1)$ $(0,1)$ matrices under which the determinant gets multiplied by $2^$: The optimal $\tilde G$ is $-J+4B+(n-3)I$ where $B$ is a block-diagonal matrix with all-ones blocks along the diagonal. These all-ones blocks should be as nearly equal in size as possible and their number should be five if $n=7$, five or six if $n=11$, six if $15 \le n \le 59$, and seven if $n \ge 63$. The bound on $\det \tilde G$ one gets from this is seldom a perfect square, and even when it is, Tamura showed that the existence of an $R$ such that $RR^T$ equals the optimal $\tilde G$ is ruled out by the Hasse-Minkowski theorem in many cases. The smallest $n$ for which the bound might be attainable is 511. Nevertheless, there exist matrices with determinant quite close to the bound. For example, when $n=19$, a matrix whose determinant attains 97.5% of the bound is known. Using a lo...

Vectors & Matrices • • • • • • • • • • • • More than just an online determinant calculator Wolfram|Alpha is the perfect resource to use for computing determinants of matrices. It can also calculate matrix products, rank, nullity, row reduction, diagonalization, eigenvalues, eigenvectors and much more. Learn more about: • Determinants » Tips for entering queries Use plain English or common mathematical syntax to enter your queries. To enter a matrix, separate elements with commas and rows with curly braces, brackets or parentheses. • • • • • • View more examples » Access instant learning tools Get immediate feedback and guidance with step-by-step solutions and Wolfram Problem Generator Learn more about: • Step-by-step solutions » • Wolfram Problem Generator » VIEW ALL CALCULATORS • BMI Calculator • Dilution Calculator • Mortgage Calculator • Interest Calculator • Loan Calculator • Present Value Calculator • Car Payment Calculator • Future Value Calculator • Limit Calculator • Derivative Calculator • Integral Calculator • Double Integral Calculator • Triple Integral Calculator • Series Expansion Calculator • Discontinuity Calculator • Domain and Range Calculator • Factoring Calculator • Quadratic Formula Calculator • Equation Solver Calculator • Partial Fraction Decomposition Calculator • System of Equations Calculator • Eigenvalue Calculator • Matrix Inverse Calculator Knowledgebase about determinants A determinant is a property of a square matrix. The value of the determin...

Hi guys I am working with this and I am trying to prove to myself that n by n matrices of the type zero on the diagonal and 1 everywhere else are invertible. I ran some cases and looked at the determinant and came to the conclusion that we can easily find the determinant by using the following $\det(A)=(-1)^(n-1)$ Now to show for a matrix B of size $n+1 \times n+1$. I am not sure I was thinking to take the determinant of the $n \times n$ minors but I am maybe someone can help me. Also is there an easier way to see this is invertible other than the determinant? I am curious. If you have studied eigenvalues and eigenvectors there is a very easy proof. Let $A$ be your $n\times n$ matrix, with $n\ge2$. Then $A+I$ is the matrix consisting entirely of $1$s, which clearly has $n-1$ zero rows after row-reduction. Therefore $A$ has eigenvalue $-1$, repeated (at least) $n-1$ times, and since $(A)=0$, the other eigenvalue is $n-1$. Since every eigenvalue of $A$ is non-zero, the determinant of $A$ is non-zero, so $A$ is invertible. $\begingroup$ @AkashGaur The argument of the answer claimed to prove that the $n$ eigenvalues of $A$ are $-1,-1,\ldots,-1$ ($n-1$ times), and $n-1$ (once). That was not my statement. I just commented that if those $n$ numbers are really the eigenvalues, then their product is the determinant. $\endgroup$ This is easy to calculate by row reduction: Add all rows to first: $$\det(A) =\det \begin$$ Here's an alternative approach. The • all the matrices in the fo...

(9) and is the formed by eliminating row and column from . This process is called A determinant can also be computed by writing down all , taking each permutation as the subscripts of the letters , , ..., and summing with signs determined by , where is the number of (Muir 1960, p.16), and is the , the permutations and the number of inversions they contain are 123 (0), 132 (1), 213 (1), 231 (2), 312 (2), and 321 (3), so the determinant is given by (28) Important properties of the determinant include the following, which include invariance under 1. Switching two rows or columns changes the sign. 2. Scalars can be factored out from rows and columns. 3. Multiples of rows and columns can be added together without changing the determinant's value. 4. Scalar multiplication of a row by a constant multiplies the determinant by . 5. A determinant with a row or column of zeros has value 0. 6. Any determinant with two rows or columns equal has value 0. Property 1 can be established by induction. For a (38) where is a . Several accounts state that Lewis Carroll (Charles Dodgson) sent Queen Victoria a copy of one of his mathematical works, in one account, An Elementary Treatise on Determinants. Heath (1974) states, "A well-known story tells how Queen Victoria, charmed by Alice in Wonderland, expressed a desire to receive the author's next work, and was presented, in due course, with a loyally inscribed copy of An Elementary Treatise on Determinants," while Gattegno (1974) asserts "Queen...

Hello Sal. I know I'm wrong and the answer is probably staring me in the face! Where's the fallacy in my thinking: As I understand it, a square matrix whose determinant is not zero is invertible. Therefore, using row operations, it can be reduced to having all its column vectors as pivot vectors. That's equvialent to an upper triangular matrix, with the main diagonal elements equal to 1. If normal row operations do not change the determinant, the determinant will be -1. HELP! :) Sal was a bit unfortunate when he said that "row operations do not change the determinant". The proof he gave was that row operations of the form [Ri -c*Rj -> Ri] (that is, replacing row i with row i minus row j times a scalar c) do not change the determinant. But there are row operations of different kind, such as k*Ri -c*Rj -> Ri (That is, replacing row i with row i times a scalar k minus row j times a scalar c). What can be proved is that operations of this kind do change the determinant. In fact, they multiply the determinant by k. And when you put an invertible matrix in RREF (that is, you turn it into an identity matrix), you must do these kinds of operations that scale the determinant. And they always end up scaling the determinant back to one, which is the determinant of any identity matrix. It depends on the size of A and B. Multiplying a matrix by a scalar, is the same as multiplying every row of that matrix by that scalar, and note, that multiplying a single row by a scalar is equivalent...

Hello Sal. I know I'm wrong and the answer is probably staring me in the face! Where's the fallacy in my thinking: As I understand it, a square matrix whose determinant is not zero is invertible. Therefore, using row operations, it can be reduced to having all its column vectors as pivot vectors. That's equvialent to an upper triangular matrix, with the main diagonal elements equal to 1. If normal row operations do not change the determinant, the determinant will be -1. HELP! :) Sal was a bit unfortunate when he said that "row operations do not change the determinant". The proof he gave was that row operations of the form [Ri -c*Rj -> Ri] (that is, replacing row i with row i minus row j times a scalar c) do not change the determinant. But there are row operations of different kind, such as k*Ri -c*Rj -> Ri (That is, replacing row i with row i times a scalar k minus row j times a scalar c). What can be proved is that operations of this kind do change the determinant. In fact, they multiply the determinant by k. And when you put an invertible matrix in RREF (that is, you turn it into an identity matrix), you must do these kinds of operations that scale the determinant. And they always end up scaling the determinant back to one, which is the determinant of any identity matrix. It depends on the size of A and B. Multiplying a matrix by a scalar, is the same as multiplying every row of that matrix by that scalar, and note, that multiplying a single row by a scalar is equivalent...

I shouldn't expect there to be exact results; compare the similar problem with matrices with entries $\pm1$. For an $n$-by-$n$ matrix with entries $\pm1$ one gets an upper bound for the determinant of $n^$. Taking some extra care one can ensure the first column of $A$ has a $1$ atop a column of zeros so that deleting the first row and column gives a smaller matrix with the same determinant. It's not clear to me whether this is the biggest possible.... As pointed out in Robin Chapman's answer and Frederio Poloni's comment thereto, there is a one-to-one map between normalized $n$-by- $n$ $(-1,1)$ matrices and $(n-1)$-by- $(n-1)$ $(0,1)$ matrices under which the determinant gets multiplied by $2^$: The optimal $\tilde G$ is $-J+4B+(n-3)I$ where $B$ is a block-diagonal matrix with all-ones blocks along the diagonal. These all-ones blocks should be as nearly equal in size as possible and their number should be five if $n=7$, five or six if $n=11$, six if $15 \le n \le 59$, and seven if $n \ge 63$. The bound on $\det \tilde G$ one gets from this is seldom a perfect square, and even when it is, Tamura showed that the existence of an $R$ such that $RR^T$ equals the optimal $\tilde G$ is ruled out by the Hasse-Minkowski theorem in many cases. The smallest $n$ for which the bound might be attainable is 511. Nevertheless, there exist matrices with determinant quite close to the bound. For example, when $n=19$, a matrix whose determinant attains 97.5% of the bound is known. Using a lo...