Find constant $k$ in matrix so that matrix $A$ is orthogonal
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Problem
Find constant $kin mathbb{R}$ in matrix so that matrix $A$ is orthogonal when:
$$ A = begin{bmatrix} 1 & -1 & -7 \ 1 & 3 & -1 \ 2 & -1 & k end{bmatrix} $$
Attempt to solve
A matrix is orthogonal when $AA^T=A^TA=I$, where $I$ is identity matrix. By utilizing this fact we should be able to solve constant k
$$ begin{bmatrix} 1 & -1 & -7 \ 1 & 3 & -1 \ 2 & -1 & k end{bmatrix}begin{bmatrix} 1 & 1 & 2 \ -1 & 3 & -1 \ -7 & -1 & k end{bmatrix}=begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 end{bmatrix} $$
$$ begin{bmatrix} 51 & 5 & 3-7k \ 5 & 11 & -k -1 \ 3-7k & -k-1 & k^2+5 end{bmatrix} = begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 end{bmatrix} $$
It looks like something that could not be solved, but am i simply wrong ?
linear-algebra matrices orthogonal-matrices
asked Nov 19 at 21:43
Tuki
976316
add a comment |
up vote
0
down vote
favorite
Problem
Find constant $kin mathbb{R}$ in matrix so that matrix $A$ is orthogonal when:
$$ A = begin{bmatrix} 1 & -1 & -7 \ 1 & 3 & -1 \ 2 & -1 & k end{bmatrix} $$
Attempt to solve
A matrix is orthogonal when $AA^T=A^TA=I$, where $I$ is identity matrix. By utilizing this fact we should be able to solve constant k
$$ begin{bmatrix} 1 & -1 & -7 \ 1 & 3 & -1 \ 2 & -1 & k end{bmatrix}begin{bmatrix} 1 & 1 & 2 \ -1 & 3 & -1 \ -7 & -1 & k end{bmatrix}=begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 end{bmatrix} $$
$$ begin{bmatrix} 51 & 5 & 3-7k \ 5 & 11 & -k -1 \ 3-7k & -k-1 & k^2+5 end{bmatrix} = begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 end{bmatrix} $$
It looks like something that could not be solved, but am i simply wrong ?
linear-algebra matrices orthogonal-matrices
asked Nov 19 at 21:43
Tuki
976316
Check your question!
– gimusi
Nov 19 at 21:45
1
I don't think i understand what you mean ? @gimusi
– Tuki
Nov 19 at 21:47
1
It seems that your question can't be "Find constant k in matrix so that matrix A is orthogonal". Are you sure that the question is not different?
– gimusi
Nov 19 at 21:52
2
Tuki, you are not using the word "orthogonal " consistently. There is a value of $k$ that makes the COLUMNS of your matrix pairwise orthogonal. In order to get an orthogonal matrix, you must then divide the columns by, let's see, $sqrt 6,$ then $sqrt {11}$ for the middle column, and a different square root for the thrid column
– Will Jagy
Nov 19 at 21:53
short version: each column in an orthogonal matrix, considered as a vector, has length exactly one.
– Will Jagy
Nov 19 at 21:57
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Problem
Find constant $kin mathbb{R}$ in matrix so that matrix $A$ is orthogonal when:
$$ A = begin{bmatrix} 1 & -1 & -7 \ 1 & 3 & -1 \ 2 & -1 & k end{bmatrix} $$
Attempt to solve
A matrix is orthogonal when $AA^T=A^TA=I$, where $I$ is identity matrix. By utilizing this fact we should be able to solve constant k
$$ begin{bmatrix} 1 & -1 & -7 \ 1 & 3 & -1 \ 2 & -1 & k end{bmatrix}begin{bmatrix} 1 & 1 & 2 \ -1 & 3 & -1 \ -7 & -1 & k end{bmatrix}=begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 end{bmatrix} $$
$$ begin{bmatrix} 51 & 5 & 3-7k \ 5 & 11 & -k -1 \ 3-7k & -k-1 & k^2+5 end{bmatrix} = begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 end{bmatrix} $$
It looks like something that could not be solved, but am i simply wrong ?
linear-algebra matrices orthogonal-matrices
asked Nov 19 at 21:43
Tuki
976316
Problem
Find constant $kin mathbb{R}$ in matrix so that matrix $A$ is orthogonal when:
$$ A = begin{bmatrix} 1 & -1 & -7 \ 1 & 3 & -1 \ 2 & -1 & k end{bmatrix} $$
Attempt to solve
A matrix is orthogonal when $AA^T=A^TA=I$, where $I$ is identity matrix. By utilizing this fact we should be able to solve constant k
$$ begin{bmatrix} 1 & -1 & -7 \ 1 & 3 & -1 \ 2 & -1 & k end{bmatrix}begin{bmatrix} 1 & 1 & 2 \ -1 & 3 & -1 \ -7 & -1 & k end{bmatrix}=begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 end{bmatrix} $$
$$ begin{bmatrix} 51 & 5 & 3-7k \ 5 & 11 & -k -1 \ 3-7k & -k-1 & k^2+5 end{bmatrix} = begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 end{bmatrix} $$
It looks like something that could not be solved, but am i simply wrong ?
linear-algebra matrices orthogonal-matrices
linear-algebra matrices orthogonal-matrices
asked Nov 19 at 21:43
Tuki
976316
asked Nov 19 at 21:43
Tuki
976316
asked Nov 19 at 21:43
Tuki
976316
asked Nov 19 at 21:43
Tuki
976316
asked Nov 19 at 21:43
Tuki
976316
976316
Check your question!
– gimusi
Nov 19 at 21:45
1
I don't think i understand what you mean ? @gimusi
– Tuki
Nov 19 at 21:47
1
It seems that your question can't be "Find constant k in matrix so that matrix A is orthogonal". Are you sure that the question is not different?
– gimusi
Nov 19 at 21:52
2
Tuki, you are not using the word "orthogonal " consistently. There is a value of $k$ that makes the COLUMNS of your matrix pairwise orthogonal. In order to get an orthogonal matrix, you must then divide the columns by, let's see, $sqrt 6,$ then $sqrt {11}$ for the middle column, and a different square root for the thrid column
– Will Jagy
Nov 19 at 21:53
short version: each column in an orthogonal matrix, considered as a vector, has length exactly one.
– Will Jagy
Nov 19 at 21:57
add a comment |
Check your question!
– gimusi
Nov 19 at 21:45
1
I don't think i understand what you mean ? @gimusi
– Tuki
Nov 19 at 21:47
1
It seems that your question can't be "Find constant k in matrix so that matrix A is orthogonal". Are you sure that the question is not different?
– gimusi
Nov 19 at 21:52
2
Tuki, you are not using the word "orthogonal " consistently. There is a value of $k$ that makes the COLUMNS of your matrix pairwise orthogonal. In order to get an orthogonal matrix, you must then divide the columns by, let's see, $sqrt 6,$ then $sqrt {11}$ for the middle column, and a different square root for the thrid column
– Will Jagy
Nov 19 at 21:53
short version: each column in an orthogonal matrix, considered as a vector, has length exactly one.
– Will Jagy
Nov 19 at 21:57
Check your question!
– gimusi
Nov 19 at 21:45
Check your question!
– gimusi
Nov 19 at 21:45
1
1
I don't think i understand what you mean ? @gimusi
– Tuki
Nov 19 at 21:47
I don't think i understand what you mean ? @gimusi
– Tuki
Nov 19 at 21:47
1
1
It seems that your question can't be "Find constant k in matrix so that matrix A is orthogonal". Are you sure that the question is not different?
– gimusi
Nov 19 at 21:52
It seems that your question can't be "Find constant k in matrix so that matrix A is orthogonal". Are you sure that the question is not different?
– gimusi
Nov 19 at 21:52
2
2
Tuki, you are not using the word "orthogonal " consistently. There is a value of $k$ that makes the COLUMNS of your matrix pairwise orthogonal. In order to get an orthogonal matrix, you must then divide the columns by, let's see, $sqrt 6,$ then $sqrt {11}$ for the middle column, and a different square root for the thrid column
– Will Jagy
Nov 19 at 21:53
Tuki, you are not using the word "orthogonal " consistently. There is a value of $k$ that makes the COLUMNS of your matrix pairwise orthogonal. In order to get an orthogonal matrix, you must then divide the columns by, let's see, $sqrt 6,$ then $sqrt {11}$ for the middle column, and a different square root for the thrid column
– Will Jagy
Nov 19 at 21:53
short version: each column in an orthogonal matrix, considered as a vector, has length exactly one.
– Will Jagy
Nov 19 at 21:57
short version: each column in an orthogonal matrix, considered as a vector, has length exactly one.
– Will Jagy
Nov 19 at 21:57
add a comment |
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Check your question!
– gimusi
Nov 19 at 21:45
1
I don't think i understand what you mean ? @gimusi
– Tuki
Nov 19 at 21:47
1
It seems that your question can't be "Find constant k in matrix so that matrix A is orthogonal". Are you sure that the question is not different?
– gimusi
Nov 19 at 21:52
2
Tuki, you are not using the word "orthogonal " consistently. There is a value of $k$ that makes the COLUMNS of your matrix pairwise orthogonal. In order to get an orthogonal matrix, you must then divide the columns by, let's see, $sqrt 6,$ then $sqrt {11}$ for the middle column, and a different square root for the thrid column
– Will Jagy
Nov 19 at 21:53
short version: each column in an orthogonal matrix, considered as a vector, has length exactly one.
– Will Jagy
Nov 19 at 21:57