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Determinant of a matrix with cofactor expansion
Determinant of 4x4 Matrix by Expansion MethodDeterminant of block $n times n$ matrixDeterminant of sum of orthogonal matrix with rank-$1$ matrixBlock matrix determinant with symmetriclly placed blocksdeterminant of matrix $X$How to compute determinant (or eigenvalues) of this matrix?Determinant of MDeterminant of $P_n$Determinant of Partitioned Upper Triangular MatrixA interesting question on Skew-symmetric matrix…finding the determinant.
$begingroup$
I have this question here:
Suppose that $A$ is an $(n,n)$-matrices, $n geq 3$, with $det(A)=5$ and whose $(2,3)$-minor $M_{2,3}(A)$ equals $12$. Let $S_{2,3}$ be the $(n,n)$-matrix all whose entries are $0$, except the entry
in position $(2,3)$ equals $2$. With $B:=(A+S_{2,3})$, find $det(B)$.
The problem is that I am not sure how to go about this. I know for a fact that if $C=A+B$ in general, then $det(C) neq det(A) + det(B)$ so thus:
$det(B) = det(A+S_{2,3})$
I tried a sample matrix:
If $A=begin{bmatrix}4 & 1 & 2 \ 3 & 1 & 1 \ 4 & 4 & 1end{bmatrix}$
The determinant of this matrix is $5$ and the minor $M_{2,3}(A)$ is $A=begin{bmatrix}4 & 1\ 4 & 4end{bmatrix}$. The determinant of this minor is $12$.
$S_{2,3} = A=begin{bmatrix}0 & 0 & 0 \ 0 & 0 & 2 \ 0 & 0 & 0end{bmatrix} $
Thus, $A + S_{2,3} = B =begin{bmatrix}4 & 1 & 2 \ 3 & 1 & 3 \ 4 & 4 & 1end{bmatrix}$. The determiant of this matrix gives $-19$. However, is there a better way of doing this in general? I don't think using specific matrices is the best idea.
linear-algebra matrices determinant
asked Mar 13 at 7:15
Future Math personFuture Math person
993817
$endgroup$
add a comment |
$begingroup$
I have this question here:
Suppose that $A$ is an $(n,n)$-matrices, $n geq 3$, with $det(A)=5$ and whose $(2,3)$-minor $M_{2,3}(A)$ equals $12$. Let $S_{2,3}$ be the $(n,n)$-matrix all whose entries are $0$, except the entry
in position $(2,3)$ equals $2$. With $B:=(A+S_{2,3})$, find $det(B)$.
The problem is that I am not sure how to go about this. I know for a fact that if $C=A+B$ in general, then $det(C) neq det(A) + det(B)$ so thus:
$det(B) = det(A+S_{2,3})$
I tried a sample matrix:
If $A=begin{bmatrix}4 & 1 & 2 \ 3 & 1 & 1 \ 4 & 4 & 1end{bmatrix}$
The determinant of this matrix is $5$ and the minor $M_{2,3}(A)$ is $A=begin{bmatrix}4 & 1\ 4 & 4end{bmatrix}$. The determinant of this minor is $12$.
$S_{2,3} = A=begin{bmatrix}0 & 0 & 0 \ 0 & 0 & 2 \ 0 & 0 & 0end{bmatrix} $
Thus, $A + S_{2,3} = B =begin{bmatrix}4 & 1 & 2 \ 3 & 1 & 3 \ 4 & 4 & 1end{bmatrix}$. The determiant of this matrix gives $-19$. However, is there a better way of doing this in general? I don't think using specific matrices is the best idea.
linear-algebra matrices determinant
asked Mar 13 at 7:15
Future Math personFuture Math person
993817
$endgroup$
$begingroup$
You could check out LU decomposition, where te determinant is the product of the L or U diagonal. This is a fast in general, the Laplace expansion is heavely computational for general cases.
$endgroup$
– Max
Mar 13 at 7:42
$begingroup$
We didn't learn that. EDIT: We covered vector motion, complex numbers, matrix addition/subtraction, multiplication, RREF, and determinants along with their properties.
$endgroup$
– Future Math person
Mar 13 at 7:43
$begingroup$
Did you arrive at a conclusion?
$endgroup$
– Max
Mar 14 at 17:54
$begingroup$
Hi yes. I did it a little differently but I got it.
$endgroup$
– Future Math person
Mar 15 at 5:13
add a comment |
$begingroup$
I have this question here:
Suppose that $A$ is an $(n,n)$-matrices, $n geq 3$, with $det(A)=5$ and whose $(2,3)$-minor $M_{2,3}(A)$ equals $12$. Let $S_{2,3}$ be the $(n,n)$-matrix all whose entries are $0$, except the entry
in position $(2,3)$ equals $2$. With $B:=(A+S_{2,3})$, find $det(B)$.
The problem is that I am not sure how to go about this. I know for a fact that if $C=A+B$ in general, then $det(C) neq det(A) + det(B)$ so thus:
$det(B) = det(A+S_{2,3})$
I tried a sample matrix:
If $A=begin{bmatrix}4 & 1 & 2 \ 3 & 1 & 1 \ 4 & 4 & 1end{bmatrix}$
The determinant of this matrix is $5$ and the minor $M_{2,3}(A)$ is $A=begin{bmatrix}4 & 1\ 4 & 4end{bmatrix}$. The determinant of this minor is $12$.
$S_{2,3} = A=begin{bmatrix}0 & 0 & 0 \ 0 & 0 & 2 \ 0 & 0 & 0end{bmatrix} $
Thus, $A + S_{2,3} = B =begin{bmatrix}4 & 1 & 2 \ 3 & 1 & 3 \ 4 & 4 & 1end{bmatrix}$. The determiant of this matrix gives $-19$. However, is there a better way of doing this in general? I don't think using specific matrices is the best idea.
linear-algebra matrices determinant
asked Mar 13 at 7:15
Future Math personFuture Math person
993817
$endgroup$
I have this question here:
Suppose that $A$ is an $(n,n)$-matrices, $n geq 3$, with $det(A)=5$ and whose $(2,3)$-minor $M_{2,3}(A)$ equals $12$. Let $S_{2,3}$ be the $(n,n)$-matrix all whose entries are $0$, except the entry
in position $(2,3)$ equals $2$. With $B:=(A+S_{2,3})$, find $det(B)$.
The problem is that I am not sure how to go about this. I know for a fact that if $C=A+B$ in general, then $det(C) neq det(A) + det(B)$ so thus:
$det(B) = det(A+S_{2,3})$
I tried a sample matrix:
If $A=begin{bmatrix}4 & 1 & 2 \ 3 & 1 & 1 \ 4 & 4 & 1end{bmatrix}$
The determinant of this matrix is $5$ and the minor $M_{2,3}(A)$ is $A=begin{bmatrix}4 & 1\ 4 & 4end{bmatrix}$. The determinant of this minor is $12$.
$S_{2,3} = A=begin{bmatrix}0 & 0 & 0 \ 0 & 0 & 2 \ 0 & 0 & 0end{bmatrix} $
Thus, $A + S_{2,3} = B =begin{bmatrix}4 & 1 & 2 \ 3 & 1 & 3 \ 4 & 4 & 1end{bmatrix}$. The determiant of this matrix gives $-19$. However, is there a better way of doing this in general? I don't think using specific matrices is the best idea.
linear-algebra matrices determinant
linear-algebra matrices determinant
asked Mar 13 at 7:15
Future Math personFuture Math person
993817
asked Mar 13 at 7:15
Future Math personFuture Math person
993817
asked Mar 13 at 7:15
Future Math personFuture Math person
993817
asked Mar 13 at 7:15
Future Math personFuture Math person
993817
asked Mar 13 at 7:15
Future Math personFuture Math person
993817
993817
$begingroup$
You could check out LU decomposition, where te determinant is the product of the L or U diagonal. This is a fast in general, the Laplace expansion is heavely computational for general cases.
$endgroup$
– Max
Mar 13 at 7:42
$begingroup$
We didn't learn that. EDIT: We covered vector motion, complex numbers, matrix addition/subtraction, multiplication, RREF, and determinants along with their properties.
$endgroup$
– Future Math person
Mar 13 at 7:43
$begingroup$
Did you arrive at a conclusion?
$endgroup$
– Max
Mar 14 at 17:54
$begingroup$
Hi yes. I did it a little differently but I got it.
$endgroup$
– Future Math person
Mar 15 at 5:13
add a comment |
$begingroup$
You could check out LU decomposition, where te determinant is the product of the L or U diagonal. This is a fast in general, the Laplace expansion is heavely computational for general cases.
$endgroup$
– Max
Mar 13 at 7:42
$begingroup$
We didn't learn that. EDIT: We covered vector motion, complex numbers, matrix addition/subtraction, multiplication, RREF, and determinants along with their properties.
$endgroup$
– Future Math person
Mar 13 at 7:43
$begingroup$
Did you arrive at a conclusion?
$endgroup$
– Max
Mar 14 at 17:54
$begingroup$
Hi yes. I did it a little differently but I got it.
$endgroup$
– Future Math person
Mar 15 at 5:13
$begingroup$
You could check out LU decomposition, where te determinant is the product of the L or U diagonal. This is a fast in general, the Laplace expansion is heavely computational for general cases.
$endgroup$
– Max
Mar 13 at 7:42
$begingroup$
You could check out LU decomposition, where te determinant is the product of the L or U diagonal. This is a fast in general, the Laplace expansion is heavely computational for general cases.
$endgroup$
– Max
Mar 13 at 7:42
$begingroup$
We didn't learn that. EDIT: We covered vector motion, complex numbers, matrix addition/subtraction, multiplication, RREF, and determinants along with their properties.
$endgroup$
– Future Math person
Mar 13 at 7:43
$begingroup$
We didn't learn that. EDIT: We covered vector motion, complex numbers, matrix addition/subtraction, multiplication, RREF, and determinants along with their properties.
$endgroup$
– Future Math person
Mar 13 at 7:43
$begingroup$
Did you arrive at a conclusion?
$endgroup$
– Max
Mar 14 at 17:54
$begingroup$
Did you arrive at a conclusion?
$endgroup$
– Max
Mar 14 at 17:54
$begingroup$
Hi yes. I did it a little differently but I got it.
$endgroup$
– Future Math person
Mar 15 at 5:13
$begingroup$
Hi yes. I did it a little differently but I got it.
$endgroup$
– Future Math person
Mar 15 at 5:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A faster way then using the Laplace expansion (cofacter expansion), would be first using row operations and then take the determinant in REF. A determinant of a triangular matrix is just the product of the diagonal. Hope this may help.
answered Mar 13 at 7:59
MaxMax
9071318
$endgroup$
add a comment |
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1 Answer
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$begingroup$
A faster way then using the Laplace expansion (cofacter expansion), would be first using row operations and then take the determinant in REF. A determinant of a triangular matrix is just the product of the diagonal. Hope this may help.
answered Mar 13 at 7:59
MaxMax
9071318
$endgroup$
add a comment |
$begingroup$
A faster way then using the Laplace expansion (cofacter expansion), would be first using row operations and then take the determinant in REF. A determinant of a triangular matrix is just the product of the diagonal. Hope this may help.
answered Mar 13 at 7:59
MaxMax
9071318
$endgroup$
add a comment |
$begingroup$
A faster way then using the Laplace expansion (cofacter expansion), would be first using row operations and then take the determinant in REF. A determinant of a triangular matrix is just the product of the diagonal. Hope this may help.
answered Mar 13 at 7:59
MaxMax
9071318
$endgroup$
A faster way then using the Laplace expansion (cofacter expansion), would be first using row operations and then take the determinant in REF. A determinant of a triangular matrix is just the product of the diagonal. Hope this may help.
answered Mar 13 at 7:59
MaxMax
9071318
answered Mar 13 at 7:59
MaxMax
9071318
answered Mar 13 at 7:59
MaxMax
9071318
answered Mar 13 at 7:59
MaxMax
9071318
9071318
add a comment |
add a comment |
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$begingroup$
You could check out LU decomposition, where te determinant is the product of the L or U diagonal. This is a fast in general, the Laplace expansion is heavely computational for general cases.
$endgroup$
– Max
Mar 13 at 7:42
$begingroup$
We didn't learn that. EDIT: We covered vector motion, complex numbers, matrix addition/subtraction, multiplication, RREF, and determinants along with their properties.
$endgroup$
– Future Math person
Mar 13 at 7:43
$begingroup$
Did you arrive at a conclusion?
$endgroup$
– Max
Mar 14 at 17:54
$begingroup$
Hi yes. I did it a little differently but I got it.
$endgroup$
– Future Math person
Mar 15 at 5:13