I have a problem that I can't seem to solve.
Basically, I have 4 columns of data points and about 200 rows.
So, for Row 1 I have numbers like 1,2,3,4 and the next Row(2) like 5,6,7,8...
The problem is that when I try to solver for the determinate, Excel wants
the data format to be in a 2x2 format, i.e. 1,2 on first row, and 3,4 on
second row. OR, It will take the data in one row (1,2,3,4) provided that
numbers are entered instead of cell designator, as descibed in the help
section:

=MDETERM(A2:D5) Determinant of the matrix above (88)

=MDETERM({3,6,1;1,1,0;3,10,2}) Determinant of the matrix as an array
constant (1)

Does anyone know how to get around this?

Thanks,

Not sure if I follow the question.
The determinant of a matrix is defined only for a square matrix (number of
rows = number of columns)
best wishes
--
Bernard V Liengme
www.stfx.ca/people/bliengme
remove caps from email

"Gary B" wrote in message
...
I have a problem that I can't seem to solve.
Basically, I have 4 columns of data points and about 200 rows.
So, for Row 1 I have numbers like 1,2,3,4 and the next Row(2) like
5,6,7,8...
The problem is that when I try to solver for the determinate, Excel wants
the data format to be in a 2x2 format, i.e. 1,2 on first row, and 3,4 on
second row. OR, It will take the data in one row (1,2,3,4) provided that
numbers are entered instead of cell designator, as descibed in the help
section:

=MDETERM(A2:D5) Determinant of the matrix above (88)

=MDETERM({3,6,1;1,1,0;3,10,2}) Determinant of the matrix as an array
constant (1)

Does anyone know how to get around this?

Thanks,

Hi Bernard:

Thank you for replying. I understand the matrix needs to be square. My
question is how do you make a square matrix from a data set that consists of
2 colums (X,Y) and 200 rows of data? From this data set I need to create the
following square matix for all 200 rows.

a1,1 = x2-x1
a1,2 = y2-y1
a2,1 = x3-x2
a2,2 = y3-y2


"Bernard Liengme" wrote:

Not sure if I follow the question.
The determinant of a matrix is defined only for a square matrix (number of
rows = number of columns)
best wishes
--
Bernard V Liengme
www.stfx.ca/people/bliengme
remove caps from email

"Gary B" wrote in message
...
I have a problem that I can't seem to solve.
Basically, I have 4 columns of data points and about 200 rows.
So, for Row 1 I have numbers like 1,2,3,4 and the next Row(2) like
5,6,7,8...
The problem is that when I try to solver for the determinate, Excel wants
the data format to be in a 2x2 format, i.e. 1,2 on first row, and 3,4 on
second row. OR, It will take the data in one row (1,2,3,4) provided that
numbers are entered instead of cell designator, as descibed in the help
section:

=MDETERM(A2:D5) Determinant of the matrix above (88)

=MDETERM({3,6,1;1,1,0;3,10,2}) Determinant of the matrix as an array
constant (1)

Does anyone know how to get around this?

Thanks,

I do not follow. Give me an example (not formulas but numbers) of a 2 by 4
matrix made square the way you want it.
all the best
--
Bernard V Liengme
www.stfx.ca/people/bliengme
remove caps from email

"Gary B" wrote in message
...
Hi Bernard:

Thank you for replying. I understand the matrix needs to be square. My
question is how do you make a square matrix from a data set that consists
of
2 colums (X,Y) and 200 rows of data? From this data set I need to create
the
following square matix for all 200 rows.

a1,1 = x2-x1
a1,2 = y2-y1
a2,1 = x3-x2
a2,2 = y3-y2


"Bernard Liengme" wrote:

Not sure if I follow the question.
The determinant of a matrix is defined only for a square matrix (number
of
rows = number of columns)
best wishes
--
Bernard V Liengme
www.stfx.ca/people/bliengme
remove caps from email

"Gary B" wrote in message
...
I have a problem that I can't seem to solve.
Basically, I have 4 columns of data points and about 200 rows.
So, for Row 1 I have numbers like 1,2,3,4 and the next Row(2) like
5,6,7,8...
The problem is that when I try to solver for the determinate, Excel
wants
the data format to be in a 2x2 format, i.e. 1,2 on first row, and 3,4
on
second row. OR, It will take the data in one row (1,2,3,4) provided
that
numbers are entered instead of cell designator, as descibed in the help
section:

=MDETERM(A2:D5) Determinant of the matrix above (88)

=MDETERM({3,6,1;1,1,0;3,10,2}) Determinant of the matrix as an array
constant (1)

Does anyone know how to get around this?

Thanks,

Gary, you still do not make much sense, even though a pattern starts
revealing itself.

1. 200 is not a perfect square. Nearest perfect square is 196 (14*14).
Do you want to build a 14 by 14 matrix?
2. In your example you only say what will happen to the first two
columns of the square matrix you want to build. How is, for example,
a3,4 defined? Or a1,3?

HTH
Kostis Vezerides


Gary B wrote:
Hi Bernard:

Thank you for replying. I understand the matrix needs to be square. My
question is how do you make a square matrix from a data set that consists of
2 colums (X,Y) and 200 rows of data? From this data set I need to create the
following square matix for all 200 rows.

a1,1 = x2-x1
a1,2 = y2-y1
a2,1 = x3-x2
a2,2 = y3-y2


"Bernard Liengme" wrote:

Not sure if I follow the question.
The determinant of a matrix is defined only for a square matrix (number of
rows = number of columns)
best wishes
--
Bernard V Liengme
www.stfx.ca/people/bliengme
remove caps from email

"Gary B" wrote in message
...
I have a problem that I can't seem to solve.
Basically, I have 4 columns of data points and about 200 rows.
So, for Row 1 I have numbers like 1,2,3,4 and the next Row(2) like
5,6,7,8...
The problem is that when I try to solver for the determinate, Excel wants
the data format to be in a 2x2 format, i.e. 1,2 on first row, and 3,4 on
second row. OR, It will take the data in one row (1,2,3,4) provided that
numbers are entered instead of cell designator, as descibed in the help
section:

=MDETERM(A2:D5) Determinant of the matrix above (88)

=MDETERM({3,6,1;1,1,0;3,10,2}) Determinant of the matrix as an array
constant (1)

Does anyone know how to get around this?

Thanks,

OK all:

Thanks for responding and I apologize for my unclear messages, I will try to
be as exact as possible. Basically, I have been given a data set in an x,y
coordinate system that essentially gives me a length in the x-direction and
height in the y-direction. The data set consists of about 200 points, so I
have two data columns, i.e.:
X Y
1. 0 1
2. 1 2
3. 2 7
4. 5 3
..
..
200. 210 1

I want to create a 2x2 matrix set, using the following command
=MDETERM({0,1;1,2})...example for the first determinant, where the actual
matrix calculation for simplicity sake would be:

[X(2)-X(1)] [Y(2) - Y(1)]

[X(3)-X(2)] [Y(3) - Y(2)]

Therefore, my table should then look like this:

X Y Determinant
1. 0 1 -1
2. 1 2 3
3. 2 7 -29
4. 5 3
..
..
200. 210 1

I don't want to perform each of these by hand as it gets cumbersome. I am
looking for a way to set up the data into a 2x2 array, such that I can get to
the end result of calculating the determinant and listing it into the third
column.

Hopefully, this is a bit clearer and many thanks for all your help in advance.

Cheers,

Gary


"vezerid" wrote:

Gary, you still do not make much sense, even though a pattern starts
revealing itself.

1. 200 is not a perfect square. Nearest perfect square is 196 (14*14).
Do you want to build a 14 by 14 matrix?
2. In your example you only say what will happen to the first two
columns of the square matrix you want to build. How is, for example,
a3,4 defined? Or a1,3?

HTH
Kostis Vezerides


Gary B wrote:
Hi Bernard:

Thank you for replying. I understand the matrix needs to be square. My
question is how do you make a square matrix from a data set that consists of
2 colums (X,Y) and 200 rows of data? From this data set I need to create the
following square matix for all 200 rows.

a1,1 = x2-x1
a1,2 = y2-y1
a2,1 = x3-x2
a2,2 = y3-y2


"Bernard Liengme" wrote:

Not sure if I follow the question.
The determinant of a matrix is defined only for a square matrix (number of
rows = number of columns)
best wishes
--
Bernard V Liengme
www.stfx.ca/people/bliengme
remove caps from email

"Gary B" wrote in message
...
I have a problem that I can't seem to solve.
Basically, I have 4 columns of data points and about 200 rows.
So, for Row 1 I have numbers like 1,2,3,4 and the next Row(2) like
5,6,7,8...
The problem is that when I try to solver for the determinate, Excel wants
the data format to be in a 2x2 format, i.e. 1,2 on first row, and 3,4 on
second row. OR, It will take the data in one row (1,2,3,4) provided that
numbers are entered instead of cell designator, as descibed in the help
section:

=MDETERM(A2:D5) Determinant of the matrix above (88)

=MDETERM({3,6,1;1,1,0;3,10,2}) Determinant of the matrix as an array
constant (1)

Does anyone know how to get around this?

Thanks,

I don;t know how you manage it, but you are still confusing. The actual
examples that you posted indicate that you want a "sliding" 2*2 matrix
to statrt from A1:B2 and move on to A2:B3 etc. In fact the values that
you have in the 3rd column are exactly the determinants of the sliding
matrices. So these [x2-x1] are still misleading...

But to give a solution which will reproduce the numbers in the third
column as I have understood it,

=MDETERM(A1:B2)

And copy down.

At any rate, if you need to subtract by rows, you can use the
following:

=MDETERM(A2:B3-A1:B2)

But this is now an array formula so it should be committed with
Shift+Ctrl+Enter.

Does this help?
Kostis Vezerides


Gary B wrote:
OK all:

Thanks for responding and I apologize for my unclear messages, I will try to
be as exact as possible. Basically, I have been given a data set in an x,y
coordinate system that essentially gives me a length in the x-direction and
height in the y-direction. The data set consists of about 200 points, so I
have two data columns, i.e.:
X Y
1. 0 1
2. 1 2
3. 2 7
4. 5 3
.
.
200. 210 1

I want to create a 2x2 matrix set, using the following command
=MDETERM({0,1;1,2})...example for the first determinant, where the actual
matrix calculation for simplicity sake would be:

[X(2)-X(1)] [Y(2) - Y(1)]

[X(3)-X(2)] [Y(3) - Y(2)]

Therefore, my table should then look like this:

X Y Determinant
1. 0 1 -1
2. 1 2 3
3. 2 7 -29
4. 5 3
.
.
200. 210 1

I don't want to perform each of these by hand as it gets cumbersome. I am
looking for a way to set up the data into a 2x2 array, such that I can get to
the end result of calculating the determinant and listing it into the third
column.

Hopefully, this is a bit clearer and many thanks for all your help in advance.

Cheers,

Gary


"vezerid" wrote:

Gary, you still do not make much sense, even though a pattern starts
revealing itself.

1. 200 is not a perfect square. Nearest perfect square is 196 (14*14).
Do you want to build a 14 by 14 matrix?
2. In your example you only say what will happen to the first two
columns of the square matrix you want to build. How is, for example,
a3,4 defined? Or a1,3?

HTH
Kostis Vezerides


Gary B wrote:
Hi Bernard:

Thank you for replying. I understand the matrix needs to be square. My
question is how do you make a square matrix from a data set that consists of
2 colums (X,Y) and 200 rows of data? From this data set I need to create the
following square matix for all 200 rows.

a1,1 = x2-x1
a1,2 = y2-y1
a2,1 = x3-x2
a2,2 = y3-y2


"Bernard Liengme" wrote:

Not sure if I follow the question.
The determinant of a matrix is defined only for a square matrix (number of
rows = number of columns)
best wishes
--
Bernard V Liengme
www.stfx.ca/people/bliengme
remove caps from email

"Gary B" wrote in message
...
I have a problem that I can't seem to solve.
Basically, I have 4 columns of data points and about 200 rows.
So, for Row 1 I have numbers like 1,2,3,4 and the next Row(2) like
5,6,7,8...
The problem is that when I try to solver for the determinate, Excel wants
the data format to be in a 2x2 format, i.e. 1,2 on first row, and 3,4 on
second row. OR, It will take the data in one row (1,2,3,4) provided that
numbers are entered instead of cell designator, as descibed in the help
section:

=MDETERM(A2:D5) Determinant of the matrix above (88)

=MDETERM({3,6,1;1,1,0;3,10,2}) Determinant of the matrix as an array
constant (1)

Does anyone know how to get around this?

Thanks,

Sorry Vezerid:

You are very fast...you are correct that I was still confusing and I meant:

I
have two data columns, i.e.:
X Y
1. 0 1
2. 1 2
3. 2 7
4. 5 3
.
.
200. 210 1

I want to create a 2x2 matrix set, using the following command
=MDETERM({0,1;1,2})...example for the first determinant, where the actual
matrix calculation for simplicity sake would be for the first data set:

[X(2)-X(1)] [Y(2) - Y(1)]

[X(3)-X(2)] [Y(3) - Y(2)]

For the second data set, I would increment the matrix array by 1:

[X(3)-X(2)] [Y(3)-Y(2)]
[X(4)-X(3)] [Y(4)-Y(3)]

and so on down to the 198th point. As I am two values ahead in the data
sheet due to the equations.


Therefore, my table should then look like this:

X Y Determinant
1. 0 1 4
2. 1 2 -19
3. 2 7
4. 5 3
.
.
200. 210 1



"vezerid" wrote:

I don;t know how you manage it, but you are still confusing. The actual
examples that you posted indicate that you want a "sliding" 2*2 matrix
to statrt from A1:B2 and move on to A2:B3 etc. In fact the values that
you have in the 3rd column are exactly the determinants of the sliding
matrices. So these [x2-x1] are still misleading...

But to give a solution which will reproduce the numbers in the third
column as I have understood it,

=MDETERM(A1:B2)

And copy down.

At any rate, if you need to subtract by rows, you can use the
following:

=MDETERM(A2:B3-A1:B2)

But this is now an array formula so it should be committed with
Shift+Ctrl+Enter.

Does this help?
Kostis Vezerides


Gary B wrote:
OK all:

Thanks for responding and I apologize for my unclear messages, I will try to
be as exact as possible. Basically, I have been given a data set in an x,y
coordinate system that essentially gives me a length in the x-direction and
height in the y-direction. The data set consists of about 200 points, so I
have two data columns, i.e.:
X Y
1. 0 1
2. 1 2
3. 2 7
4. 5 3
.
.
200. 210 1

I want to create a 2x2 matrix set, using the following command
=MDETERM({0,1;1,2})...example for the first determinant, where the actual
matrix calculation for simplicity sake would be:

[X(2)-X(1)] [Y(2) - Y(1)]

[X(3)-X(2)] [Y(3) - Y(2)]

Therefore, my table should then look like this:

X Y Determinant
1. 0 1 -1
2. 1 2 3
3. 2 7 -29
4. 5 3
.
.
200. 210 1

I don't want to perform each of these by hand as it gets cumbersome. I am
looking for a way to set up the data into a 2x2 array, such that I can get to
the end result of calculating the determinant and listing it into the third
column.

Hopefully, this is a bit clearer and many thanks for all your help in advance.

Cheers,

Gary


"vezerid" wrote:

Gary, you still do not make much sense, even though a pattern starts
revealing itself.

1. 200 is not a perfect square. Nearest perfect square is 196 (14*14).
Do you want to build a 14 by 14 matrix?
2. In your example you only say what will happen to the first two
columns of the square matrix you want to build. How is, for example,
a3,4 defined? Or a1,3?

HTH
Kostis Vezerides


Gary B wrote:
Hi Bernard:

Thank you for replying. I understand the matrix needs to be square. My
question is how do you make a square matrix from a data set that consists of
2 colums (X,Y) and 200 rows of data? From this data set I need to create the
following square matix for all 200 rows.

a1,1 = x2-x1
a1,2 = y2-y1
a2,1 = x3-x2
a2,2 = y3-y2


"Bernard Liengme" wrote:

Not sure if I follow the question.
The determinant of a matrix is defined only for a square matrix (number of
rows = number of columns)
best wishes
--
Bernard V Liengme
www.stfx.ca/people/bliengme
remove caps from email

"Gary B" wrote in message
...
I have a problem that I can't seem to solve.
Basically, I have 4 columns of data points and about 200 rows.
So, for Row 1 I have numbers like 1,2,3,4 and the next Row(2) like
5,6,7,8...
The problem is that when I try to solver for the determinate, Excel wants
the data format to be in a 2x2 format, i.e. 1,2 on first row, and 3,4 on
second row. OR, It will take the data in one row (1,2,3,4) provided that
numbers are entered instead of cell designator, as descibed in the help
section:

=MDETERM(A2:D5) Determinant of the matrix above (88)

=MDETERM({3,6,1;1,1,0;3,10,2}) Determinant of the matrix as an array
constant (1)

Does anyone know how to get around this?

Thanks,

Again...another mistake

The first determinant would be =MDETERM({1,1;1,5}) = 4

"Gary B" wrote:

Sorry Vezerid:

You are very fast...you are correct that I was still confusing and I meant:

I
have two data columns, i.e.:
X Y
1. 0 1
2. 1 2
3. 2 7
4. 5 3
.
.
200. 210 1

I want to create a 2x2 matrix set, using the following command
=MDETERM({0,1;1,2})...example for the first determinant, where the actual
matrix calculation for simplicity sake would be for the first data set:

[X(2)-X(1)] [Y(2) - Y(1)]

[X(3)-X(2)] [Y(3) - Y(2)]

For the second data set, I would increment the matrix array by 1:

[X(3)-X(2)] [Y(3)-Y(2)]
[X(4)-X(3)] [Y(4)-Y(3)]

and so on down to the 198th point. As I am two values ahead in the data
sheet due to the equations.


Therefore, my table should then look like this:

X Y Determinant
1. 0 1 4
2. 1 2 -19
3. 2 7
4. 5 3
.
.
200. 210 1



"vezerid" wrote:

I don;t know how you manage it, but you are still confusing. The actual
examples that you posted indicate that you want a "sliding" 2*2 matrix
to statrt from A1:B2 and move on to A2:B3 etc. In fact the values that
you have in the 3rd column are exactly the determinants of the sliding
matrices. So these [x2-x1] are still misleading...

But to give a solution which will reproduce the numbers in the third
column as I have understood it,

=MDETERM(A1:B2)

And copy down.

At any rate, if you need to subtract by rows, you can use the
following:

=MDETERM(A2:B3-A1:B2)

But this is now an array formula so it should be committed with
Shift+Ctrl+Enter.

Does this help?
Kostis Vezerides


Gary B wrote:
OK all:

Thanks for responding and I apologize for my unclear messages, I will try to
be as exact as possible. Basically, I have been given a data set in an x,y
coordinate system that essentially gives me a length in the x-direction and
height in the y-direction. The data set consists of about 200 points, so I
have two data columns, i.e.:
X Y
1. 0 1
2. 1 2
3. 2 7
4. 5 3
.
.
200. 210 1

I want to create a 2x2 matrix set, using the following command
=MDETERM({0,1;1,2})...example for the first determinant, where the actual
matrix calculation for simplicity sake would be:

[X(2)-X(1)] [Y(2) - Y(1)]

[X(3)-X(2)] [Y(3) - Y(2)]

Therefore, my table should then look like this:

X Y Determinant
1. 0 1 -1
2. 1 2 3
3. 2 7 -29
4. 5 3
.
.
200. 210 1

I don't want to perform each of these by hand as it gets cumbersome. I am
looking for a way to set up the data into a 2x2 array, such that I can get to
the end result of calculating the determinant and listing it into the third
column.

Hopefully, this is a bit clearer and many thanks for all your help in advance.

Cheers,

Gary


"vezerid" wrote:

Gary, you still do not make much sense, even though a pattern starts
revealing itself.

1. 200 is not a perfect square. Nearest perfect square is 196 (14*14).
Do you want to build a 14 by 14 matrix?
2. In your example you only say what will happen to the first two
columns of the square matrix you want to build. How is, for example,
a3,4 defined? Or a1,3?

HTH
Kostis Vezerides


Gary B wrote:
Hi Bernard:

Thank you for replying. I understand the matrix needs to be square. My
question is how do you make a square matrix from a data set that consists of
2 colums (X,Y) and 200 rows of data? From this data set I need to create the
following square matix for all 200 rows.

a1,1 = x2-x1
a1,2 = y2-y1
a2,1 = x3-x2
a2,2 = y3-y2


"Bernard Liengme" wrote:

Not sure if I follow the question.
The determinant of a matrix is defined only for a square matrix (number of
rows = number of columns)
best wishes
--
Bernard V Liengme
www.stfx.ca/people/bliengme
remove caps from email

"Gary B" wrote in message
...
I have a problem that I can't seem to solve.
Basically, I have 4 columns of data points and about 200 rows.
So, for Row 1 I have numbers like 1,2,3,4 and the next Row(2) like
5,6,7,8...
The problem is that when I try to solver for the determinate, Excel wants
the data format to be in a 2x2 format, i.e. 1,2 on first row, and 3,4 on
second row. OR, It will take the data in one row (1,2,3,4) provided that
numbers are entered instead of cell designator, as descibed in the help
section:

=MDETERM(A2:D5) Determinant of the matrix above (88)

=MDETERM({3,6,1;1,1,0;3,10,2}) Determinant of the matrix as an array
constant (1)

Does anyone know how to get around this?

Thanks,

Dear Kostis,

I tried your suggestion, i.e. =MDETERM(A2:B3-A1:B2)

But this is now an array formula so it should be committed with
Shift+Ctrl+Enter.

However, I get the following message, #value! Do you have any other
suggestions, based on all my corrections to my messages?

Many thanks,

"Gary B" wrote:

Again...another mistake

The first determinant would be =MDETERM({1,1;1,5}) = 4

"Gary B" wrote:

Sorry Vezerid:

You are very fast...you are correct that I was still confusing and I meant:

I
have two data columns, i.e.:
X Y
1. 0 1
2. 1 2
3. 2 7
4. 5 3
.
.
200. 210 1

I want to create a 2x2 matrix set, using the following command
=MDETERM({0,1;1,2})...example for the first determinant, where the actual
matrix calculation for simplicity sake would be for the first data set:

[X(2)-X(1)] [Y(2) - Y(1)]

[X(3)-X(2)] [Y(3) - Y(2)]

For the second data set, I would increment the matrix array by 1:

[X(3)-X(2)] [Y(3)-Y(2)]
[X(4)-X(3)] [Y(4)-Y(3)]

and so on down to the 198th point. As I am two values ahead in the data
sheet due to the equations.


Therefore, my table should then look like this:

X Y Determinant
1. 0 1 4
2. 1 2 -19
3. 2 7
4. 5 3
.
.
200. 210 1



"vezerid" wrote:

I don;t know how you manage it, but you are still confusing. The actual
examples that you posted indicate that you want a "sliding" 2*2 matrix
to statrt from A1:B2 and move on to A2:B3 etc. In fact the values that
you have in the 3rd column are exactly the determinants of the sliding
matrices. So these [x2-x1] are still misleading...

But to give a solution which will reproduce the numbers in the third
column as I have understood it,

=MDETERM(A1:B2)

And copy down.

At any rate, if you need to subtract by rows, you can use the
following:

=MDETERM(A2:B3-A1:B2)

But this is now an array formula so it should be committed with
Shift+Ctrl+Enter.

Does this help?
Kostis Vezerides


Gary B wrote:
OK all:

Thanks for responding and I apologize for my unclear messages, I will try to
be as exact as possible. Basically, I have been given a data set in an x,y
coordinate system that essentially gives me a length in the x-direction and
height in the y-direction. The data set consists of about 200 points, so I
have two data columns, i.e.:
X Y
1. 0 1
2. 1 2
3. 2 7
4. 5 3
.
.
200. 210 1

I want to create a 2x2 matrix set, using the following command
=MDETERM({0,1;1,2})...example for the first determinant, where the actual
matrix calculation for simplicity sake would be:

[X(2)-X(1)] [Y(2) - Y(1)]

[X(3)-X(2)] [Y(3) - Y(2)]

Therefore, my table should then look like this:

X Y Determinant
1. 0 1 -1
2. 1 2 3
3. 2 7 -29
4. 5 3
.
.
200. 210 1

I don't want to perform each of these by hand as it gets cumbersome. I am
looking for a way to set up the data into a 2x2 array, such that I can get to
the end result of calculating the determinant and listing it into the third
column.

Hopefully, this is a bit clearer and many thanks for all your help in advance.

Cheers,

Gary


"vezerid" wrote:

Gary, you still do not make much sense, even though a pattern starts
revealing itself.

1. 200 is not a perfect square. Nearest perfect square is 196 (14*14).
Do you want to build a 14 by 14 matrix?
2. In your example you only say what will happen to the first two
columns of the square matrix you want to build. How is, for example,
a3,4 defined? Or a1,3?

HTH
Kostis Vezerides


Gary B wrote:
Hi Bernard:

Thank you for replying. I understand the matrix needs to be square. My
question is how do you make a square matrix from a data set that consists of
2 colums (X,Y) and 200 rows of data? From this data set I need to create the
following square matix for all 200 rows.

a1,1 = x2-x1
a1,2 = y2-y1
a2,1 = x3-x2
a2,2 = y3-y2


"Bernard Liengme" wrote:

Not sure if I follow the question.
The determinant of a matrix is defined only for a square matrix (number of
rows = number of columns)
best wishes
--
Bernard V Liengme
www.stfx.ca/people/bliengme
remove caps from email

"Gary B" wrote in message
...
I have a problem that I can't seem to solve.
Basically, I have 4 columns of data points and about 200 rows.
So, for Row 1 I have numbers like 1,2,3,4 and the next Row(2) like
5,6,7,8...
The problem is that when I try to solver for the determinate, Excel wants
the data format to be in a 2x2 format, i.e. 1,2 on first row, and 3,4 on
second row. OR, It will take the data in one row (1,2,3,4) provided that
numbers are entered instead of cell designator, as descibed in the help
section:

=MDETERM(A2:D5) Determinant of the matrix above (88)

=MDETERM({3,6,1;1,1,0;3,10,2}) Determinant of the matrix as an array
constant (1)

Does anyone know how to get around this?

Thanks,

Gary,

by now I think it is pretty clear that you need my second formula:

=MDETERM(A2:B3-A1:B2)

Remember to commit with Shift+Ctrl+Enter. You can then copy down.

HTH
Kostis

Gary B wrote:
Again...another mistake

The first determinant would be =MDETERM({1,1;1,5}) = 4

Many many thanks for hanging in there with me.

Cheers,

Gary

"vezerid" wrote:

Gary,

by now I think it is pretty clear that you need my second formula:

=MDETERM(A2:B3-A1:B2)

Remember to commit with Shift+Ctrl+Enter. You can then copy down.

HTH
Kostis

Gary B wrote:
Again...another mistake

The first determinant would be =MDETERM({1,1;1,5}) = 4