Mixing
Relationship between two given determinants .
$begingroup$
. Let $D1$ = det $begin{pmatrix}
a & b & c \
x & y & z \
p & q & r \
end{pmatrix}$
and $D2$ = det$begin{pmatrix}
-x & a & -p \
y & -b & q \
z & -c & r \
end{pmatrix}
$
$(i)D_1=D_2 $ $(ii)D_1=2D_2 $ $(iii)D_1=-D_2 $ $(iv)D_2=2D_1 $
I am trying to find out the relation between given deteriminants so I am using some properties of determinants on $D_2$
$D2$ = $begin{vmatrix}
-x & a & -p \
y & -b & q \
z & -c & r \
end{vmatrix}
implies (-1)begin{vmatrix}
x & -a & p \
y & -b & q \
z & -c & r \
end{vmatrix} text{Took $(-1)$ common from first row}$
$implies (1)begin{vmatrix}
x & a & p \
y & b & q \
z & c & r \
end{vmatrix} text{Took $(-1)$ common from $2nd$ column}$
$implies(-1)begin{vmatrix}
a & x & p \
b & y & q \
c & z & r \
end{vmatrix} text{interchanged first and 2nd column}$
$implies(-1)D_1$
I don't have an answer available for this question so I want to know if I did it right? Can anyone tell me any other way of solving this? I Just wanna know any other method.
linear-algebra determinant
asked Jan 17 at 4:47
Daman deepDaman deep
731418
$endgroup$
add a comment |
$begingroup$
. Let $D1$ = det $begin{pmatrix}
a & b & c \
x & y & z \
p & q & r \
end{pmatrix}$
and $D2$ = det$begin{pmatrix}
-x & a & -p \
y & -b & q \
z & -c & r \
end{pmatrix}
$
$(i)D_1=D_2 $ $(ii)D_1=2D_2 $ $(iii)D_1=-D_2 $ $(iv)D_2=2D_1 $
I am trying to find out the relation between given deteriminants so I am using some properties of determinants on $D_2$
$D2$ = $begin{vmatrix}
-x & a & -p \
y & -b & q \
z & -c & r \
end{vmatrix}
implies (-1)begin{vmatrix}
x & -a & p \
y & -b & q \
z & -c & r \
end{vmatrix} text{Took $(-1)$ common from first row}$
$implies (1)begin{vmatrix}
x & a & p \
y & b & q \
z & c & r \
end{vmatrix} text{Took $(-1)$ common from $2nd$ column}$
$implies(-1)begin{vmatrix}
a & x & p \
b & y & q \
c & z & r \
end{vmatrix} text{interchanged first and 2nd column}$
$implies(-1)D_1$
I don't have an answer available for this question so I want to know if I did it right? Can anyone tell me any other way of solving this? I Just wanna know any other method.
linear-algebra determinant
asked Jan 17 at 4:47
Daman deepDaman deep
731418
$endgroup$
$begingroup$
Have you tried calculating explicitly the determinant? It's a 3 by 3 matrix, so it should be easy.
$endgroup$
– Andrei
Jan 17 at 4:52
3
$begingroup$
You solved it correctly. In the last step you are using $|A|=|A^T|$. Sanity check: consider $a=y=r=1$ and others are $0$, $|D_1|=1,|D_2|=-1$.
$endgroup$
– farruhota
Jan 17 at 4:53
$begingroup$
math.stackexchange.com/questions/1468064/…
$endgroup$
– lab bhattacharjee
Jan 17 at 4:54
$begingroup$
@Andrei It came in one mark so I tried to find out quick way. Although I didn't try it.
$endgroup$
– Daman deep
Jan 17 at 4:56
add a comment |
$begingroup$
. Let $D1$ = det $begin{pmatrix}
a & b & c \
x & y & z \
p & q & r \
end{pmatrix}$
and $D2$ = det$begin{pmatrix}
-x & a & -p \
y & -b & q \
z & -c & r \
end{pmatrix}
$
$(i)D_1=D_2 $ $(ii)D_1=2D_2 $ $(iii)D_1=-D_2 $ $(iv)D_2=2D_1 $
I am trying to find out the relation between given deteriminants so I am using some properties of determinants on $D_2$
$D2$ = $begin{vmatrix}
-x & a & -p \
y & -b & q \
z & -c & r \
end{vmatrix}
implies (-1)begin{vmatrix}
x & -a & p \
y & -b & q \
z & -c & r \
end{vmatrix} text{Took $(-1)$ common from first row}$
$implies (1)begin{vmatrix}
x & a & p \
y & b & q \
z & c & r \
end{vmatrix} text{Took $(-1)$ common from $2nd$ column}$
$implies(-1)begin{vmatrix}
a & x & p \
b & y & q \
c & z & r \
end{vmatrix} text{interchanged first and 2nd column}$
$implies(-1)D_1$
I don't have an answer available for this question so I want to know if I did it right? Can anyone tell me any other way of solving this? I Just wanna know any other method.
linear-algebra determinant
asked Jan 17 at 4:47
Daman deepDaman deep
731418
$endgroup$
. Let $D1$ = det $begin{pmatrix}
a & b & c \
x & y & z \
p & q & r \
end{pmatrix}$
and $D2$ = det$begin{pmatrix}
-x & a & -p \
y & -b & q \
z & -c & r \
end{pmatrix}
$
$(i)D_1=D_2 $ $(ii)D_1=2D_2 $ $(iii)D_1=-D_2 $ $(iv)D_2=2D_1 $
I am trying to find out the relation between given deteriminants so I am using some properties of determinants on $D_2$
$D2$ = $begin{vmatrix}
-x & a & -p \
y & -b & q \
z & -c & r \
end{vmatrix}
implies (-1)begin{vmatrix}
x & -a & p \
y & -b & q \
z & -c & r \
end{vmatrix} text{Took $(-1)$ common from first row}$
$implies (1)begin{vmatrix}
x & a & p \
y & b & q \
z & c & r \
end{vmatrix} text{Took $(-1)$ common from $2nd$ column}$
$implies(-1)begin{vmatrix}
a & x & p \
b & y & q \
c & z & r \
end{vmatrix} text{interchanged first and 2nd column}$
$implies(-1)D_1$
I don't have an answer available for this question so I want to know if I did it right? Can anyone tell me any other way of solving this? I Just wanna know any other method.
linear-algebra determinant
linear-algebra determinant
asked Jan 17 at 4:47
Daman deepDaman deep
731418
asked Jan 17 at 4:47
Daman deepDaman deep
731418
asked Jan 17 at 4:47
Daman deepDaman deep
731418
asked Jan 17 at 4:47
Daman deepDaman deep
731418
asked Jan 17 at 4:47
Daman deepDaman deep
731418
731418
$begingroup$
Have you tried calculating explicitly the determinant? It's a 3 by 3 matrix, so it should be easy.
$endgroup$
– Andrei
Jan 17 at 4:52
3
$begingroup$
You solved it correctly. In the last step you are using $|A|=|A^T|$. Sanity check: consider $a=y=r=1$ and others are $0$, $|D_1|=1,|D_2|=-1$.
$endgroup$
– farruhota
Jan 17 at 4:53
$begingroup$
math.stackexchange.com/questions/1468064/…
$endgroup$
– lab bhattacharjee
Jan 17 at 4:54
$begingroup$
@Andrei It came in one mark so I tried to find out quick way. Although I didn't try it.
$endgroup$
– Daman deep
Jan 17 at 4:56
add a comment |
$begingroup$
Have you tried calculating explicitly the determinant? It's a 3 by 3 matrix, so it should be easy.
$endgroup$
– Andrei
Jan 17 at 4:52
3
$begingroup$
You solved it correctly. In the last step you are using $|A|=|A^T|$. Sanity check: consider $a=y=r=1$ and others are $0$, $|D_1|=1,|D_2|=-1$.
$endgroup$
– farruhota
Jan 17 at 4:53
$begingroup$
math.stackexchange.com/questions/1468064/…
$endgroup$
– lab bhattacharjee
Jan 17 at 4:54
$begingroup$
@Andrei It came in one mark so I tried to find out quick way. Although I didn't try it.
$endgroup$
– Daman deep
Jan 17 at 4:56
$begingroup$
Have you tried calculating explicitly the determinant? It's a 3 by 3 matrix, so it should be easy.
$endgroup$
– Andrei
Jan 17 at 4:52
$begingroup$
Have you tried calculating explicitly the determinant? It's a 3 by 3 matrix, so it should be easy.
$endgroup$
– Andrei
Jan 17 at 4:52
3
3
$begingroup$
You solved it correctly. In the last step you are using $|A|=|A^T|$. Sanity check: consider $a=y=r=1$ and others are $0$, $|D_1|=1,|D_2|=-1$.
$endgroup$
– farruhota
Jan 17 at 4:53
$begingroup$
You solved it correctly. In the last step you are using $|A|=|A^T|$. Sanity check: consider $a=y=r=1$ and others are $0$, $|D_1|=1,|D_2|=-1$.
$endgroup$
– farruhota
Jan 17 at 4:53
$begingroup$
math.stackexchange.com/questions/1468064/…
$endgroup$
– lab bhattacharjee
Jan 17 at 4:54
$begingroup$
math.stackexchange.com/questions/1468064/…
$endgroup$
– lab bhattacharjee
Jan 17 at 4:54
$begingroup$
@Andrei It came in one mark so I tried to find out quick way. Although I didn't try it.
$endgroup$
– Daman deep
Jan 17 at 4:56
$begingroup$
@Andrei It came in one mark so I tried to find out quick way. Although I didn't try it.
$endgroup$
– Daman deep
Jan 17 at 4:56
add a comment |
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$begingroup$
Have you tried calculating explicitly the determinant? It's a 3 by 3 matrix, so it should be easy.
$endgroup$
– Andrei
Jan 17 at 4:52
3
$begingroup$
You solved it correctly. In the last step you are using $|A|=|A^T|$. Sanity check: consider $a=y=r=1$ and others are $0$, $|D_1|=1,|D_2|=-1$.
$endgroup$
– farruhota
Jan 17 at 4:53
$begingroup$
math.stackexchange.com/questions/1468064/…
$endgroup$
– lab bhattacharjee
Jan 17 at 4:54
$begingroup$
@Andrei It came in one mark so I tried to find out quick way. Although I didn't try it.
$endgroup$
– Daman deep
Jan 17 at 4:56