How to derive the general form of determinants through this method.
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If we take the definition of the determinant to be the amount by which areas/volumes are transformed by a linear transformation, a common definition, the determinant of a Matrix $A$ will be the area/volume contained in the parallelogram/parallelepiped formed by the matrix's vector components.
We can thus obtain an expression for the determinant by performing the elementary row operation of ading a multiple of one row to another until the matrix is diagonal, since this operation corresponds to a shear transformation the volume of the parallelepiped and thus the determinant will not change. Once the matrix is diagonal, the volume will simply be the diagonal elements multiplied together. This is easy to show for a $2times2$ matrix:
$$beginbmatrix
a&b \
c&d
end bmatrix$$
This becomes
$$beginbmatrix
a&0 \
0&d-fracbca
end bmatrix$$
So the determinant is
$$beginvmatrix
a&b \
c&d
end vmatrix=ad-bc$$
The same can be done for a $3times3$ matrix
$$beginbmatrix
a&b&c \
d&e&f \
g&h&i
end bmatrix$$
Which can be made into
$$beginbmatrix
a&0&0 \
0&e-fracbda&f -fraccda\
0&h-fracbga&i-fraccga
end bmatrix$$
So the determinant will be
$$beginvmatrix
a&b&c \
d&e&f \
g&h&i
end vmatrix=
abeginvmatrix
e-fracbda&f -fraccda\
h-fracbga&i-fraccga
end vmatrix$$
Which can be rearranged to
$$beginvmatrix
a&b&c \
d&e&f \
g&h&i
end vmatrix
=
abeginvmatrix
e&f\
h&i
end vmatrix-b
beginvmatrix
d&f \
g&i
end vmatrix+c
beginvmatrix
d&e \
g&h
end vmatrix$$
My question is, how can this be generalized to higher dimensional matrices using the same method?
linear-algebra matrices determinant
asked Aug 31 at 3:12
user140323
164
add a comment |Â
up vote
0
down vote
favorite
If we take the definition of the determinant to be the amount by which areas/volumes are transformed by a linear transformation, a common definition, the determinant of a Matrix $A$ will be the area/volume contained in the parallelogram/parallelepiped formed by the matrix's vector components.
We can thus obtain an expression for the determinant by performing the elementary row operation of ading a multiple of one row to another until the matrix is diagonal, since this operation corresponds to a shear transformation the volume of the parallelepiped and thus the determinant will not change. Once the matrix is diagonal, the volume will simply be the diagonal elements multiplied together. This is easy to show for a $2times2$ matrix:
$$beginbmatrix
a&b \
c&d
end bmatrix$$
This becomes
$$beginbmatrix
a&0 \
0&d-fracbca
end bmatrix$$
So the determinant is
$$beginvmatrix
a&b \
c&d
end vmatrix=ad-bc$$
The same can be done for a $3times3$ matrix
$$beginbmatrix
a&b&c \
d&e&f \
g&h&i
end bmatrix$$
Which can be made into
$$beginbmatrix
a&0&0 \
0&e-fracbda&f -fraccda\
0&h-fracbga&i-fraccga
end bmatrix$$
So the determinant will be
$$beginvmatrix
a&b&c \
d&e&f \
g&h&i
end vmatrix=
abeginvmatrix
e-fracbda&f -fraccda\
h-fracbga&i-fraccga
end vmatrix$$
Which can be rearranged to
$$beginvmatrix
a&b&c \
d&e&f \
g&h&i
end vmatrix
=
abeginvmatrix
e&f\
h&i
end vmatrix-b
beginvmatrix
d&f \
g&i
end vmatrix+c
beginvmatrix
d&e \
g&h
end vmatrix$$
My question is, how can this be generalized to higher dimensional matrices using the same method?
linear-algebra matrices determinant
asked Aug 31 at 3:12
user140323
164
Look up any linear algebra text on the computation of a determinant by Gaussian elimination (for this is what you've been doing in the $2 times 2$ and $3 times 3$ cases).
– darij grinberg
Sep 8 at 23:44
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If we take the definition of the determinant to be the amount by which areas/volumes are transformed by a linear transformation, a common definition, the determinant of a Matrix $A$ will be the area/volume contained in the parallelogram/parallelepiped formed by the matrix's vector components.
We can thus obtain an expression for the determinant by performing the elementary row operation of ading a multiple of one row to another until the matrix is diagonal, since this operation corresponds to a shear transformation the volume of the parallelepiped and thus the determinant will not change. Once the matrix is diagonal, the volume will simply be the diagonal elements multiplied together. This is easy to show for a $2times2$ matrix:
$$beginbmatrix
a&b \
c&d
end bmatrix$$
This becomes
$$beginbmatrix
a&0 \
0&d-fracbca
end bmatrix$$
So the determinant is
$$beginvmatrix
a&b \
c&d
end vmatrix=ad-bc$$
The same can be done for a $3times3$ matrix
$$beginbmatrix
a&b&c \
d&e&f \
g&h&i
end bmatrix$$
Which can be made into
$$beginbmatrix
a&0&0 \
0&e-fracbda&f -fraccda\
0&h-fracbga&i-fraccga
end bmatrix$$
So the determinant will be
$$beginvmatrix
a&b&c \
d&e&f \
g&h&i
end vmatrix=
abeginvmatrix
e-fracbda&f -fraccda\
h-fracbga&i-fraccga
end vmatrix$$
Which can be rearranged to
$$beginvmatrix
a&b&c \
d&e&f \
g&h&i
end vmatrix
=
abeginvmatrix
e&f\
h&i
end vmatrix-b
beginvmatrix
d&f \
g&i
end vmatrix+c
beginvmatrix
d&e \
g&h
end vmatrix$$
My question is, how can this be generalized to higher dimensional matrices using the same method?
linear-algebra matrices determinant
asked Aug 31 at 3:12
user140323
164
If we take the definition of the determinant to be the amount by which areas/volumes are transformed by a linear transformation, a common definition, the determinant of a Matrix $A$ will be the area/volume contained in the parallelogram/parallelepiped formed by the matrix's vector components.
We can thus obtain an expression for the determinant by performing the elementary row operation of ading a multiple of one row to another until the matrix is diagonal, since this operation corresponds to a shear transformation the volume of the parallelepiped and thus the determinant will not change. Once the matrix is diagonal, the volume will simply be the diagonal elements multiplied together. This is easy to show for a $2times2$ matrix:
$$beginbmatrix
a&b \
c&d
end bmatrix$$
This becomes
$$beginbmatrix
a&0 \
0&d-fracbca
end bmatrix$$
So the determinant is
$$beginvmatrix
a&b \
c&d
end vmatrix=ad-bc$$
The same can be done for a $3times3$ matrix
$$beginbmatrix
a&b&c \
d&e&f \
g&h&i
end bmatrix$$
Which can be made into
$$beginbmatrix
a&0&0 \
0&e-fracbda&f -fraccda\
0&h-fracbga&i-fraccga
end bmatrix$$
So the determinant will be
$$beginvmatrix
a&b&c \
d&e&f \
g&h&i
end vmatrix=
abeginvmatrix
e-fracbda&f -fraccda\
h-fracbga&i-fraccga
end vmatrix$$
Which can be rearranged to
$$beginvmatrix
a&b&c \
d&e&f \
g&h&i
end vmatrix
=
abeginvmatrix
e&f\
h&i
end vmatrix-b
beginvmatrix
d&f \
g&i
end vmatrix+c
beginvmatrix
d&e \
g&h
end vmatrix$$
My question is, how can this be generalized to higher dimensional matrices using the same method?
linear-algebra matrices determinant
linear-algebra matrices determinant
asked Aug 31 at 3:12
user140323
164
asked Aug 31 at 3:12
user140323
164
asked Aug 31 at 3:12
user140323
164
asked Aug 31 at 3:12
user140323
164
asked Aug 31 at 3:12
user140323
164
164
Look up any linear algebra text on the computation of a determinant by Gaussian elimination (for this is what you've been doing in the $2 times 2$ and $3 times 3$ cases).
– darij grinberg
Sep 8 at 23:44
add a comment |Â
Look up any linear algebra text on the computation of a determinant by Gaussian elimination (for this is what you've been doing in the $2 times 2$ and $3 times 3$ cases).
– darij grinberg
Sep 8 at 23:44
Look up any linear algebra text on the computation of a determinant by Gaussian elimination (for this is what you've been doing in the $2 times 2$ and $3 times 3$ cases).
– darij grinberg
Sep 8 at 23:44
Look up any linear algebra text on the computation of a determinant by Gaussian elimination (for this is what you've been doing in the $2 times 2$ and $3 times 3$ cases).
– darij grinberg
Sep 8 at 23:44
add a comment |Â
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Look up any linear algebra text on the computation of a determinant by Gaussian elimination (for this is what you've been doing in the $2 times 2$ and $3 times 3$ cases).
– darij grinberg
Sep 8 at 23:44