Why do the properties of determinants (used to calculate determinants from multiple matrices) apply not only...
A set of rules in my textbook is as follows:
a. If $A$ has a zero row (column), then $det A = 0$
b. If $B$ is obtained by interchanging two rows (columns) of $A$, then $det B = -det A$
c. If $A$ has two identical rows (columns), then $det A = 0$
d. If $B$ is obtained by multiplying a row (column) of $A$ by $k$, then $det B = kcdotdet A$
e. If $A$, $B$, and $C$ are identical except that the $i$-th row (column) of $C$ is the sum of the $i$-th rows (columns) of $A$ and $B$, then $det C = det B + det A$
f. If $B$ is obtained by adding a multiple of one row (column) of $A$ to another row (column), then $det B = det A$
I don't understand why "column" is in parentheses after every instance of row. Is that the same as saying, for example with clause a: "if $A$ has a zero row or a zero column, then $det A = 0$"? As in, it's saying that column and row can be used interchangably in the statement, as the statement holds true either way?
If the above interpretation is correct, then how? Why would these statements that apply to rows also apply to columns. The only situation I could see it applying is if the matrix is symmetrical, but the question doesn't specify that, it only says that the matrix is square.
Any help is appreciated.
linear-algebra matrices determinant
edited 2 hours ago
amWhy
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A set of rules in my textbook is as follows:
a. If $A$ has a zero row (column), then $det A = 0$
b. If $B$ is obtained by interchanging two rows (columns) of $A$, then $det B = -det A$
c. If $A$ has two identical rows (columns), then $det A = 0$
d. If $B$ is obtained by multiplying a row (column) of $A$ by $k$, then $det B = kcdotdet A$
e. If $A$, $B$, and $C$ are identical except that the $i$-th row (column) of $C$ is the sum of the $i$-th rows (columns) of $A$ and $B$, then $det C = det B + det A$
f. If $B$ is obtained by adding a multiple of one row (column) of $A$ to another row (column), then $det B = det A$
I don't understand why "column" is in parentheses after every instance of row. Is that the same as saying, for example with clause a: "if $A$ has a zero row or a zero column, then $det A = 0$"? As in, it's saying that column and row can be used interchangably in the statement, as the statement holds true either way?
If the above interpretation is correct, then how? Why would these statements that apply to rows also apply to columns. The only situation I could see it applying is if the matrix is symmetrical, but the question doesn't specify that, it only says that the matrix is square.
Any help is appreciated.
linear-algebra matrices determinant
edited 2 hours ago
amWhy
191k28224439
asked 2 hours ago
Nest Doberman
161
New contributor
Nest Doberman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Yes, the same rules apply to columns too. This startled me as well when I was taught it. See J. G.'s answer. And welcome to the site.
– timtfj
15 mins ago
add a comment |
A set of rules in my textbook is as follows:
a. If $A$ has a zero row (column), then $det A = 0$
b. If $B$ is obtained by interchanging two rows (columns) of $A$, then $det B = -det A$
c. If $A$ has two identical rows (columns), then $det A = 0$
d. If $B$ is obtained by multiplying a row (column) of $A$ by $k$, then $det B = kcdotdet A$
e. If $A$, $B$, and $C$ are identical except that the $i$-th row (column) of $C$ is the sum of the $i$-th rows (columns) of $A$ and $B$, then $det C = det B + det A$
f. If $B$ is obtained by adding a multiple of one row (column) of $A$ to another row (column), then $det B = det A$
I don't understand why "column" is in parentheses after every instance of row. Is that the same as saying, for example with clause a: "if $A$ has a zero row or a zero column, then $det A = 0$"? As in, it's saying that column and row can be used interchangably in the statement, as the statement holds true either way?
If the above interpretation is correct, then how? Why would these statements that apply to rows also apply to columns. The only situation I could see it applying is if the matrix is symmetrical, but the question doesn't specify that, it only says that the matrix is square.
Any help is appreciated.
linear-algebra matrices determinant
edited 2 hours ago
amWhy
191k28224439
asked 2 hours ago
Nest Doberman
161
New contributor
Nest Doberman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
A set of rules in my textbook is as follows:
a. If $A$ has a zero row (column), then $det A = 0$
b. If $B$ is obtained by interchanging two rows (columns) of $A$, then $det B = -det A$
c. If $A$ has two identical rows (columns), then $det A = 0$
d. If $B$ is obtained by multiplying a row (column) of $A$ by $k$, then $det B = kcdotdet A$
e. If $A$, $B$, and $C$ are identical except that the $i$-th row (column) of $C$ is the sum of the $i$-th rows (columns) of $A$ and $B$, then $det C = det B + det A$
f. If $B$ is obtained by adding a multiple of one row (column) of $A$ to another row (column), then $det B = det A$
I don't understand why "column" is in parentheses after every instance of row. Is that the same as saying, for example with clause a: "if $A$ has a zero row or a zero column, then $det A = 0$"? As in, it's saying that column and row can be used interchangably in the statement, as the statement holds true either way?
If the above interpretation is correct, then how? Why would these statements that apply to rows also apply to columns. The only situation I could see it applying is if the matrix is symmetrical, but the question doesn't specify that, it only says that the matrix is square.
Any help is appreciated.
linear-algebra matrices determinant
linear-algebra matrices determinant
edited 2 hours ago
amWhy
191k28224439
asked 2 hours ago
Nest Doberman
161
New contributor
Nest Doberman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
amWhy
191k28224439
asked 2 hours ago
Nest Doberman
161
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Check out our Code of Conduct.
edited 2 hours ago
amWhy
191k28224439
edited 2 hours ago
amWhy
191k28224439
edited 2 hours ago
amWhy
191k28224439
191k28224439
asked 2 hours ago
Nest Doberman
161
New contributor
Nest Doberman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Nest Doberman
161
asked 2 hours ago
Nest Doberman
161
161
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Nest Doberman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Nest Doberman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Check out our Code of Conduct.
Yes, the same rules apply to columns too. This startled me as well when I was taught it. See J. G.'s answer. And welcome to the site.
– timtfj
15 mins ago
add a comment |
Yes, the same rules apply to columns too. This startled me as well when I was taught it. See J. G.'s answer. And welcome to the site.
– timtfj
15 mins ago
Yes, the same rules apply to columns too. This startled me as well when I was taught it. See J. G.'s answer. And welcome to the site.
– timtfj
15 mins ago
Yes, the same rules apply to columns too. This startled me as well when I was taught it. See J. G.'s answer. And welcome to the site.
– timtfj
15 mins ago
add a comment |
1 Answer
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The trick is that $det A^T=det A$. Therefore, if there's a zero column in $A$ there's a zero row in $A^T$, implying $det A^T=0$. You can check all the other copy-paste-to-column results in that way.
answered 2 hours ago
J.G.
22.7k22136
And, if you define determinants by the formula using signs of permutations, $det A^T=det A$ just amounts to the fact that the sign of a permutation is the same as the sign of its inverse.
– Eric Wofsey
1 hour ago
@EricWolsey Which, since signs multiply on composition of permutations, amounts to the primary-school fact that $(pm 1)^2=1$. (Or, if we work with parities, $1+1=2$ does it.)
– J.G.
1 hour ago
add a comment |
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1 Answer
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1 Answer
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votes
The trick is that $det A^T=det A$. Therefore, if there's a zero column in $A$ there's a zero row in $A^T$, implying $det A^T=0$. You can check all the other copy-paste-to-column results in that way.
answered 2 hours ago
J.G.
22.7k22136
And, if you define determinants by the formula using signs of permutations, $det A^T=det A$ just amounts to the fact that the sign of a permutation is the same as the sign of its inverse.
– Eric Wofsey
1 hour ago
@EricWolsey Which, since signs multiply on composition of permutations, amounts to the primary-school fact that $(pm 1)^2=1$. (Or, if we work with parities, $1+1=2$ does it.)
– J.G.
1 hour ago
add a comment |
The trick is that $det A^T=det A$. Therefore, if there's a zero column in $A$ there's a zero row in $A^T$, implying $det A^T=0$. You can check all the other copy-paste-to-column results in that way.
answered 2 hours ago
J.G.
22.7k22136
And, if you define determinants by the formula using signs of permutations, $det A^T=det A$ just amounts to the fact that the sign of a permutation is the same as the sign of its inverse.
– Eric Wofsey
1 hour ago
@EricWolsey Which, since signs multiply on composition of permutations, amounts to the primary-school fact that $(pm 1)^2=1$. (Or, if we work with parities, $1+1=2$ does it.)
– J.G.
1 hour ago
add a comment |
The trick is that $det A^T=det A$. Therefore, if there's a zero column in $A$ there's a zero row in $A^T$, implying $det A^T=0$. You can check all the other copy-paste-to-column results in that way.
answered 2 hours ago
J.G.
22.7k22136
The trick is that $det A^T=det A$. Therefore, if there's a zero column in $A$ there's a zero row in $A^T$, implying $det A^T=0$. You can check all the other copy-paste-to-column results in that way.
answered 2 hours ago
J.G.
22.7k22136
answered 2 hours ago
J.G.
22.7k22136
answered 2 hours ago
J.G.
22.7k22136
answered 2 hours ago
J.G.
22.7k22136
22.7k22136
And, if you define determinants by the formula using signs of permutations, $det A^T=det A$ just amounts to the fact that the sign of a permutation is the same as the sign of its inverse.
– Eric Wofsey
1 hour ago
@EricWolsey Which, since signs multiply on composition of permutations, amounts to the primary-school fact that $(pm 1)^2=1$. (Or, if we work with parities, $1+1=2$ does it.)
– J.G.
1 hour ago
add a comment |
And, if you define determinants by the formula using signs of permutations, $det A^T=det A$ just amounts to the fact that the sign of a permutation is the same as the sign of its inverse.
– Eric Wofsey
1 hour ago
@EricWolsey Which, since signs multiply on composition of permutations, amounts to the primary-school fact that $(pm 1)^2=1$. (Or, if we work with parities, $1+1=2$ does it.)
– J.G.
1 hour ago
And, if you define determinants by the formula using signs of permutations, $det A^T=det A$ just amounts to the fact that the sign of a permutation is the same as the sign of its inverse.
– Eric Wofsey
1 hour ago
And, if you define determinants by the formula using signs of permutations, $det A^T=det A$ just amounts to the fact that the sign of a permutation is the same as the sign of its inverse.
– Eric Wofsey
1 hour ago
@EricWolsey Which, since signs multiply on composition of permutations, amounts to the primary-school fact that $(pm 1)^2=1$. (Or, if we work with parities, $1+1=2$ does it.)
– J.G.
1 hour ago
@EricWolsey Which, since signs multiply on composition of permutations, amounts to the primary-school fact that $(pm 1)^2=1$. (Or, if we work with parities, $1+1=2$ does it.)
– J.G.
1 hour ago
add a comment |
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Yes, the same rules apply to columns too. This startled me as well when I was taught it. See J. G.'s answer. And welcome to the site.
– timtfj
15 mins ago