Mixing
Row swapping not working as expected for determinant
$begingroup$
Suppose I have determinant A such that
$$
A=
begin {vmatrix}
1&0&1 \
5&-1&0 \
1&0&0
end {vmatrix}
$$
I conduct row operation $ R_1 <-> R_3 $
So I should have
$$
A = -
begin {vmatrix}
1&0&0 \
5&-1&0 \
1&0&1
end {vmatrix}
$$
but that does not seem to be the case because in the first case A=-1 and in the other A=1
That isn't working as expected. What am I missing here?
matrices determinant
asked Jan 18 at 7:38
ShashwatShashwat
143
$endgroup$
add a comment |
$begingroup$
Suppose I have determinant A such that
$$
A=
begin {vmatrix}
1&0&1 \
5&-1&0 \
1&0&0
end {vmatrix}
$$
I conduct row operation $ R_1 <-> R_3 $
So I should have
$$
A = -
begin {vmatrix}
1&0&0 \
5&-1&0 \
1&0&1
end {vmatrix}
$$
but that does not seem to be the case because in the first case A=-1 and in the other A=1
That isn't working as expected. What am I missing here?
matrices determinant
asked Jan 18 at 7:38
ShashwatShashwat
143
$endgroup$
$begingroup$
Why did you put a minus sign in the front? Then you're also multiplying the whole matrix by a number ($-1$). Remember: If we multiply a row (column) of A by a number, the determinant of A will be multiplied by the same number. And if we swap two rows (columns) in A, the determinant will change its sign.
$endgroup$
– Matti P.
Jan 18 at 7:43
1
$begingroup$
I put the minus sign because the swapping of row will change the sign, so to keep the value unchanged I put the minus sign.
$endgroup$
– Shashwat
Jan 18 at 7:44
4
$begingroup$
I get that both your $A$s equal $1$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 7:45
$begingroup$
If you would tell us how you got the incorrect value $-1$ for the first determinant, maybe we could tell you what you did wrong. What method did you use to evaluate it?
$endgroup$
– bof
Jan 18 at 7:48
add a comment |
$begingroup$
Suppose I have determinant A such that
$$
A=
begin {vmatrix}
1&0&1 \
5&-1&0 \
1&0&0
end {vmatrix}
$$
I conduct row operation $ R_1 <-> R_3 $
So I should have
$$
A = -
begin {vmatrix}
1&0&0 \
5&-1&0 \
1&0&1
end {vmatrix}
$$
but that does not seem to be the case because in the first case A=-1 and in the other A=1
That isn't working as expected. What am I missing here?
matrices determinant
asked Jan 18 at 7:38
ShashwatShashwat
143
$endgroup$
Suppose I have determinant A such that
$$
A=
begin {vmatrix}
1&0&1 \
5&-1&0 \
1&0&0
end {vmatrix}
$$
I conduct row operation $ R_1 <-> R_3 $
So I should have
$$
A = -
begin {vmatrix}
1&0&0 \
5&-1&0 \
1&0&1
end {vmatrix}
$$
but that does not seem to be the case because in the first case A=-1 and in the other A=1
That isn't working as expected. What am I missing here?
matrices determinant
matrices determinant
asked Jan 18 at 7:38
ShashwatShashwat
143
asked Jan 18 at 7:38
ShashwatShashwat
143
asked Jan 18 at 7:38
ShashwatShashwat
143
asked Jan 18 at 7:38
ShashwatShashwat
143
asked Jan 18 at 7:38
ShashwatShashwat
143
143
$begingroup$
Why did you put a minus sign in the front? Then you're also multiplying the whole matrix by a number ($-1$). Remember: If we multiply a row (column) of A by a number, the determinant of A will be multiplied by the same number. And if we swap two rows (columns) in A, the determinant will change its sign.
$endgroup$
– Matti P.
Jan 18 at 7:43
1
$begingroup$
I put the minus sign because the swapping of row will change the sign, so to keep the value unchanged I put the minus sign.
$endgroup$
– Shashwat
Jan 18 at 7:44
4
$begingroup$
I get that both your $A$s equal $1$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 7:45
$begingroup$
If you would tell us how you got the incorrect value $-1$ for the first determinant, maybe we could tell you what you did wrong. What method did you use to evaluate it?
$endgroup$
– bof
Jan 18 at 7:48
add a comment |
$begingroup$
Why did you put a minus sign in the front? Then you're also multiplying the whole matrix by a number ($-1$). Remember: If we multiply a row (column) of A by a number, the determinant of A will be multiplied by the same number. And if we swap two rows (columns) in A, the determinant will change its sign.
$endgroup$
– Matti P.
Jan 18 at 7:43
1
$begingroup$
I put the minus sign because the swapping of row will change the sign, so to keep the value unchanged I put the minus sign.
$endgroup$
– Shashwat
Jan 18 at 7:44
4
$begingroup$
I get that both your $A$s equal $1$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 7:45
$begingroup$
If you would tell us how you got the incorrect value $-1$ for the first determinant, maybe we could tell you what you did wrong. What method did you use to evaluate it?
$endgroup$
– bof
Jan 18 at 7:48
$begingroup$
Why did you put a minus sign in the front? Then you're also multiplying the whole matrix by a number ($-1$). Remember: If we multiply a row (column) of A by a number, the determinant of A will be multiplied by the same number. And if we swap two rows (columns) in A, the determinant will change its sign.
$endgroup$
– Matti P.
Jan 18 at 7:43
$begingroup$
Why did you put a minus sign in the front? Then you're also multiplying the whole matrix by a number ($-1$). Remember: If we multiply a row (column) of A by a number, the determinant of A will be multiplied by the same number. And if we swap two rows (columns) in A, the determinant will change its sign.
$endgroup$
– Matti P.
Jan 18 at 7:43
1
1
$begingroup$
I put the minus sign because the swapping of row will change the sign, so to keep the value unchanged I put the minus sign.
$endgroup$
– Shashwat
Jan 18 at 7:44
$begingroup$
I put the minus sign because the swapping of row will change the sign, so to keep the value unchanged I put the minus sign.
$endgroup$
– Shashwat
Jan 18 at 7:44
4
4
$begingroup$
I get that both your $A$s equal $1$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 7:45
$begingroup$
I get that both your $A$s equal $1$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 7:45
$begingroup$
If you would tell us how you got the incorrect value $-1$ for the first determinant, maybe we could tell you what you did wrong. What method did you use to evaluate it?
$endgroup$
– bof
Jan 18 at 7:48
$begingroup$
If you would tell us how you got the incorrect value $-1$ for the first determinant, maybe we could tell you what you did wrong. What method did you use to evaluate it?
$endgroup$
– bof
Jan 18 at 7:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is$$begin {vmatrix}
1&0&1 \
5&-1&0 \
1&0&0
end {vmatrix}=1$$ and
$$begin {vmatrix}
1&0&0 \
5&-1&0 \
1&0&1
end {vmatrix}=-1.$$ So
$$begin {vmatrix}
1&0&1 \
5&-1&0 \
1&0&0
end {vmatrix}=-begin {vmatrix}
1&0&0 \
5&-1&0 \
1&0&1
end {vmatrix}.$$
answered Jan 18 at 7:45
mflmfl
26.7k12142
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
It is$$begin {vmatrix}
1&0&1 \
5&-1&0 \
1&0&0
end {vmatrix}=1$$ and
$$begin {vmatrix}
1&0&0 \
5&-1&0 \
1&0&1
end {vmatrix}=-1.$$ So
$$begin {vmatrix}
1&0&1 \
5&-1&0 \
1&0&0
end {vmatrix}=-begin {vmatrix}
1&0&0 \
5&-1&0 \
1&0&1
end {vmatrix}.$$
answered Jan 18 at 7:45
mflmfl
26.7k12142
$endgroup$
add a comment |
$begingroup$
It is$$begin {vmatrix}
1&0&1 \
5&-1&0 \
1&0&0
end {vmatrix}=1$$ and
$$begin {vmatrix}
1&0&0 \
5&-1&0 \
1&0&1
end {vmatrix}=-1.$$ So
$$begin {vmatrix}
1&0&1 \
5&-1&0 \
1&0&0
end {vmatrix}=-begin {vmatrix}
1&0&0 \
5&-1&0 \
1&0&1
end {vmatrix}.$$
answered Jan 18 at 7:45
mflmfl
26.7k12142
$endgroup$
add a comment |
$begingroup$
It is$$begin {vmatrix}
1&0&1 \
5&-1&0 \
1&0&0
end {vmatrix}=1$$ and
$$begin {vmatrix}
1&0&0 \
5&-1&0 \
1&0&1
end {vmatrix}=-1.$$ So
$$begin {vmatrix}
1&0&1 \
5&-1&0 \
1&0&0
end {vmatrix}=-begin {vmatrix}
1&0&0 \
5&-1&0 \
1&0&1
end {vmatrix}.$$
answered Jan 18 at 7:45
mflmfl
26.7k12142
$endgroup$
It is$$begin {vmatrix}
1&0&1 \
5&-1&0 \
1&0&0
end {vmatrix}=1$$ and
$$begin {vmatrix}
1&0&0 \
5&-1&0 \
1&0&1
end {vmatrix}=-1.$$ So
$$begin {vmatrix}
1&0&1 \
5&-1&0 \
1&0&0
end {vmatrix}=-begin {vmatrix}
1&0&0 \
5&-1&0 \
1&0&1
end {vmatrix}.$$
answered Jan 18 at 7:45
mflmfl
26.7k12142
answered Jan 18 at 7:45
mflmfl
26.7k12142
answered Jan 18 at 7:45
mflmfl
26.7k12142
answered Jan 18 at 7:45
mflmfl
26.7k12142
26.7k12142
add a comment |
add a comment |
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$begingroup$
Why did you put a minus sign in the front? Then you're also multiplying the whole matrix by a number ($-1$). Remember: If we multiply a row (column) of A by a number, the determinant of A will be multiplied by the same number. And if we swap two rows (columns) in A, the determinant will change its sign.
$endgroup$
– Matti P.
Jan 18 at 7:43
1
$begingroup$
I put the minus sign because the swapping of row will change the sign, so to keep the value unchanged I put the minus sign.
$endgroup$
– Shashwat
Jan 18 at 7:44
4
$begingroup$
I get that both your $A$s equal $1$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 7:45
$begingroup$
If you would tell us how you got the incorrect value $-1$ for the first determinant, maybe we could tell you what you did wrong. What method did you use to evaluate it?
$endgroup$
– bof
Jan 18 at 7:48