Mixing
For which values of $x$ the matrix is invertible?
$begingroup$
The following matrix has coefficients in $Bbb Z_{11}$:
$left(begin{matrix}
1 & 0 & 3 & 0 & 5 \
0 & 3 & 0 & 5 & 0 \
3 & 0 & x & 0 & 7 \
0 & 5 & 0 & 7 & 0 \
5 & 0 & 7 & 0 & 9
end{matrix}right)$
To determine for which values of $x$ it is invertible, I tried to find the correspondent triangular matrix so I can easily calculate the determinant and then understand for which values $x$ is $0$. I have come to this point:
$left(begin{matrix}
1 & 0 & 3 & 0 & 5 \
0 & 1 & 0 & 7 & 0 \
0 & 0 & 2x & 0 & 3 \
0 & 0 & 0 & 5 & 0 \
0 & 0 & 3 & 0 & 6
end{matrix}right)$
I don't know how to remove the $3$ to make the matrix triangular. Any help?
linear-algebra matrices discrete-mathematics linear-transformations
asked Jan 23 at 16:42
JackJack
344
$endgroup$
|
show 3 more comments
$begingroup$
The following matrix has coefficients in $Bbb Z_{11}$:
$left(begin{matrix}
1 & 0 & 3 & 0 & 5 \
0 & 3 & 0 & 5 & 0 \
3 & 0 & x & 0 & 7 \
0 & 5 & 0 & 7 & 0 \
5 & 0 & 7 & 0 & 9
end{matrix}right)$
To determine for which values of $x$ it is invertible, I tried to find the correspondent triangular matrix so I can easily calculate the determinant and then understand for which values $x$ is $0$. I have come to this point:
$left(begin{matrix}
1 & 0 & 3 & 0 & 5 \
0 & 1 & 0 & 7 & 0 \
0 & 0 & 2x & 0 & 3 \
0 & 0 & 0 & 5 & 0 \
0 & 0 & 3 & 0 & 6
end{matrix}right)$
I don't know how to remove the $3$ to make the matrix triangular. Any help?
linear-algebra matrices discrete-mathematics linear-transformations
asked Jan 23 at 16:42
JackJack
344
$endgroup$
$begingroup$
Did you learned about determinants?
$endgroup$
– gbox
Jan 23 at 16:44
2
$begingroup$
Hint: if $xneq 0$, you can subtract a multiple of the third row from the last row.
$endgroup$
– Wojowu
Jan 23 at 16:44
$begingroup$
@gbox I'm not a veteran of the topic, but something I know. What in particular are you referring to?
$endgroup$
– Jack
Jan 23 at 16:46
$begingroup$
@Wojowu but basically I don't know the values of $x$. Is this operation allowed?
$endgroup$
– Jack
Jan 23 at 16:48
$begingroup$
@Jack: it is when $xne0$. But you may prefer to swap rows 3, 5 and eliminate as usual.
$endgroup$
– Yves Daoust
Jan 23 at 16:54
|
show 3 more comments
$begingroup$
The following matrix has coefficients in $Bbb Z_{11}$:
$left(begin{matrix}
1 & 0 & 3 & 0 & 5 \
0 & 3 & 0 & 5 & 0 \
3 & 0 & x & 0 & 7 \
0 & 5 & 0 & 7 & 0 \
5 & 0 & 7 & 0 & 9
end{matrix}right)$
To determine for which values of $x$ it is invertible, I tried to find the correspondent triangular matrix so I can easily calculate the determinant and then understand for which values $x$ is $0$. I have come to this point:
$left(begin{matrix}
1 & 0 & 3 & 0 & 5 \
0 & 1 & 0 & 7 & 0 \
0 & 0 & 2x & 0 & 3 \
0 & 0 & 0 & 5 & 0 \
0 & 0 & 3 & 0 & 6
end{matrix}right)$
I don't know how to remove the $3$ to make the matrix triangular. Any help?
linear-algebra matrices discrete-mathematics linear-transformations
asked Jan 23 at 16:42
JackJack
344
$endgroup$
The following matrix has coefficients in $Bbb Z_{11}$:
$left(begin{matrix}
1 & 0 & 3 & 0 & 5 \
0 & 3 & 0 & 5 & 0 \
3 & 0 & x & 0 & 7 \
0 & 5 & 0 & 7 & 0 \
5 & 0 & 7 & 0 & 9
end{matrix}right)$
To determine for which values of $x$ it is invertible, I tried to find the correspondent triangular matrix so I can easily calculate the determinant and then understand for which values $x$ is $0$. I have come to this point:
$left(begin{matrix}
1 & 0 & 3 & 0 & 5 \
0 & 1 & 0 & 7 & 0 \
0 & 0 & 2x & 0 & 3 \
0 & 0 & 0 & 5 & 0 \
0 & 0 & 3 & 0 & 6
end{matrix}right)$
I don't know how to remove the $3$ to make the matrix triangular. Any help?
linear-algebra matrices discrete-mathematics linear-transformations
linear-algebra matrices discrete-mathematics linear-transformations
asked Jan 23 at 16:42
JackJack
344
asked Jan 23 at 16:42
JackJack
344
asked Jan 23 at 16:42
JackJack
344
asked Jan 23 at 16:42
JackJack
344
asked Jan 23 at 16:42
JackJack
344
344
$begingroup$
Did you learned about determinants?
$endgroup$
– gbox
Jan 23 at 16:44
2
$begingroup$
Hint: if $xneq 0$, you can subtract a multiple of the third row from the last row.
$endgroup$
– Wojowu
Jan 23 at 16:44
$begingroup$
@gbox I'm not a veteran of the topic, but something I know. What in particular are you referring to?
$endgroup$
– Jack
Jan 23 at 16:46
$begingroup$
@Wojowu but basically I don't know the values of $x$. Is this operation allowed?
$endgroup$
– Jack
Jan 23 at 16:48
$begingroup$
@Jack: it is when $xne0$. But you may prefer to swap rows 3, 5 and eliminate as usual.
$endgroup$
– Yves Daoust
Jan 23 at 16:54
|
show 3 more comments
$begingroup$
Did you learned about determinants?
$endgroup$
– gbox
Jan 23 at 16:44
2
$begingroup$
Hint: if $xneq 0$, you can subtract a multiple of the third row from the last row.
$endgroup$
– Wojowu
Jan 23 at 16:44
$begingroup$
@gbox I'm not a veteran of the topic, but something I know. What in particular are you referring to?
$endgroup$
– Jack
Jan 23 at 16:46
$begingroup$
@Wojowu but basically I don't know the values of $x$. Is this operation allowed?
$endgroup$
– Jack
Jan 23 at 16:48
$begingroup$
@Jack: it is when $xne0$. But you may prefer to swap rows 3, 5 and eliminate as usual.
$endgroup$
– Yves Daoust
Jan 23 at 16:54
$begingroup$
Did you learned about determinants?
$endgroup$
– gbox
Jan 23 at 16:44
$begingroup$
Did you learned about determinants?
$endgroup$
– gbox
Jan 23 at 16:44
2
2
$begingroup$
Hint: if $xneq 0$, you can subtract a multiple of the third row from the last row.
$endgroup$
– Wojowu
Jan 23 at 16:44
$begingroup$
Hint: if $xneq 0$, you can subtract a multiple of the third row from the last row.
$endgroup$
– Wojowu
Jan 23 at 16:44
$begingroup$
@gbox I'm not a veteran of the topic, but something I know. What in particular are you referring to?
$endgroup$
– Jack
Jan 23 at 16:46
$begingroup$
@gbox I'm not a veteran of the topic, but something I know. What in particular are you referring to?
$endgroup$
– Jack
Jan 23 at 16:46
$begingroup$
@Wojowu but basically I don't know the values of $x$. Is this operation allowed?
$endgroup$
– Jack
Jan 23 at 16:48
$begingroup$
@Wojowu but basically I don't know the values of $x$. Is this operation allowed?
$endgroup$
– Jack
Jan 23 at 16:48
$begingroup$
@Jack: it is when $xne0$. But you may prefer to swap rows 3, 5 and eliminate as usual.
$endgroup$
– Yves Daoust
Jan 23 at 16:54
$begingroup$
@Jack: it is when $xne0$. But you may prefer to swap rows 3, 5 and eliminate as usual.
$endgroup$
– Yves Daoust
Jan 23 at 16:54
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
The original matrix $A$ will not be invertible if and only if there is a nonzero vector $v=(v_1,ldots,v_5)^T$ such that $Av=0$.
By the pattern of zeros of $A$ we see that the equations from $Av=0$ for $v_2,v_4$ are independent of those for $v_1,v_3,v_5$. Moreover we have $3v_2+5v_4=0=5v_2+7v_4$, which are independent of each other in $mathbb{Z}_{11}$, so $v_2=0=v_4$.
Now we have to impose that the matrix for $v_1,v_3,v_5$ is not invertible. That matrix is equivalent modulo $11$ to
$$begin{pmatrix}1 & 3 & 5 \ 3 & x & -4\ 5 & -4 & -2end{pmatrix}.$$
Its determinant is equivalent modulo $11$ to $6x+3$, so $det(A)equiv 0pmod{11}$ iff $xequiv (-3)6^{-1}equiv (-3)2equiv 5pmod{11}$.
answered Jan 23 at 17:14
Jose BroxJose Brox
3,15711128
$endgroup$
add a comment |
$begingroup$
The matrix is singular (and not invertible) if any row can be described as a linear combination of the other rows (other than the trivial combination).
Due to the placement of the zeros, it should be obvious that second and fourth rows must be independent from the third row.
But if we take the first plus the fifth minus twice the third we get
$(0,0,10-2x,0,0)$,
and we have a singular matrix if
$10-2x equiv 0pmod {11}$.
edited Jan 23 at 21:11
J. W. Tanner
3,2401320
answered Jan 23 at 17:20
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
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2 Answers
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2 Answers
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votes
$begingroup$
The original matrix $A$ will not be invertible if and only if there is a nonzero vector $v=(v_1,ldots,v_5)^T$ such that $Av=0$.
By the pattern of zeros of $A$ we see that the equations from $Av=0$ for $v_2,v_4$ are independent of those for $v_1,v_3,v_5$. Moreover we have $3v_2+5v_4=0=5v_2+7v_4$, which are independent of each other in $mathbb{Z}_{11}$, so $v_2=0=v_4$.
Now we have to impose that the matrix for $v_1,v_3,v_5$ is not invertible. That matrix is equivalent modulo $11$ to
$$begin{pmatrix}1 & 3 & 5 \ 3 & x & -4\ 5 & -4 & -2end{pmatrix}.$$
Its determinant is equivalent modulo $11$ to $6x+3$, so $det(A)equiv 0pmod{11}$ iff $xequiv (-3)6^{-1}equiv (-3)2equiv 5pmod{11}$.
answered Jan 23 at 17:14
Jose BroxJose Brox
3,15711128
$endgroup$
add a comment |
$begingroup$
The original matrix $A$ will not be invertible if and only if there is a nonzero vector $v=(v_1,ldots,v_5)^T$ such that $Av=0$.
By the pattern of zeros of $A$ we see that the equations from $Av=0$ for $v_2,v_4$ are independent of those for $v_1,v_3,v_5$. Moreover we have $3v_2+5v_4=0=5v_2+7v_4$, which are independent of each other in $mathbb{Z}_{11}$, so $v_2=0=v_4$.
Now we have to impose that the matrix for $v_1,v_3,v_5$ is not invertible. That matrix is equivalent modulo $11$ to
$$begin{pmatrix}1 & 3 & 5 \ 3 & x & -4\ 5 & -4 & -2end{pmatrix}.$$
Its determinant is equivalent modulo $11$ to $6x+3$, so $det(A)equiv 0pmod{11}$ iff $xequiv (-3)6^{-1}equiv (-3)2equiv 5pmod{11}$.
answered Jan 23 at 17:14
Jose BroxJose Brox
3,15711128
$endgroup$
add a comment |
$begingroup$
The original matrix $A$ will not be invertible if and only if there is a nonzero vector $v=(v_1,ldots,v_5)^T$ such that $Av=0$.
By the pattern of zeros of $A$ we see that the equations from $Av=0$ for $v_2,v_4$ are independent of those for $v_1,v_3,v_5$. Moreover we have $3v_2+5v_4=0=5v_2+7v_4$, which are independent of each other in $mathbb{Z}_{11}$, so $v_2=0=v_4$.
Now we have to impose that the matrix for $v_1,v_3,v_5$ is not invertible. That matrix is equivalent modulo $11$ to
$$begin{pmatrix}1 & 3 & 5 \ 3 & x & -4\ 5 & -4 & -2end{pmatrix}.$$
Its determinant is equivalent modulo $11$ to $6x+3$, so $det(A)equiv 0pmod{11}$ iff $xequiv (-3)6^{-1}equiv (-3)2equiv 5pmod{11}$.
answered Jan 23 at 17:14
Jose BroxJose Brox
3,15711128
$endgroup$
The original matrix $A$ will not be invertible if and only if there is a nonzero vector $v=(v_1,ldots,v_5)^T$ such that $Av=0$.
By the pattern of zeros of $A$ we see that the equations from $Av=0$ for $v_2,v_4$ are independent of those for $v_1,v_3,v_5$. Moreover we have $3v_2+5v_4=0=5v_2+7v_4$, which are independent of each other in $mathbb{Z}_{11}$, so $v_2=0=v_4$.
Now we have to impose that the matrix for $v_1,v_3,v_5$ is not invertible. That matrix is equivalent modulo $11$ to
$$begin{pmatrix}1 & 3 & 5 \ 3 & x & -4\ 5 & -4 & -2end{pmatrix}.$$
Its determinant is equivalent modulo $11$ to $6x+3$, so $det(A)equiv 0pmod{11}$ iff $xequiv (-3)6^{-1}equiv (-3)2equiv 5pmod{11}$.
answered Jan 23 at 17:14
Jose BroxJose Brox
3,15711128
answered Jan 23 at 17:14
Jose BroxJose Brox
3,15711128
answered Jan 23 at 17:14
Jose BroxJose Brox
3,15711128
answered Jan 23 at 17:14
Jose BroxJose Brox
3,15711128
3,15711128
add a comment |
add a comment |
$begingroup$
The matrix is singular (and not invertible) if any row can be described as a linear combination of the other rows (other than the trivial combination).
Due to the placement of the zeros, it should be obvious that second and fourth rows must be independent from the third row.
But if we take the first plus the fifth minus twice the third we get
$(0,0,10-2x,0,0)$,
and we have a singular matrix if
$10-2x equiv 0pmod {11}$.
edited Jan 23 at 21:11
J. W. Tanner
3,2401320
answered Jan 23 at 17:20
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
$begingroup$
The matrix is singular (and not invertible) if any row can be described as a linear combination of the other rows (other than the trivial combination).
Due to the placement of the zeros, it should be obvious that second and fourth rows must be independent from the third row.
But if we take the first plus the fifth minus twice the third we get
$(0,0,10-2x,0,0)$,
and we have a singular matrix if
$10-2x equiv 0pmod {11}$.
edited Jan 23 at 21:11
J. W. Tanner
3,2401320
answered Jan 23 at 17:20
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
$begingroup$
The matrix is singular (and not invertible) if any row can be described as a linear combination of the other rows (other than the trivial combination).
Due to the placement of the zeros, it should be obvious that second and fourth rows must be independent from the third row.
But if we take the first plus the fifth minus twice the third we get
$(0,0,10-2x,0,0)$,
and we have a singular matrix if
$10-2x equiv 0pmod {11}$.
edited Jan 23 at 21:11
J. W. Tanner
3,2401320
answered Jan 23 at 17:20
Doug MDoug M
45.3k31954
$endgroup$
The matrix is singular (and not invertible) if any row can be described as a linear combination of the other rows (other than the trivial combination).
Due to the placement of the zeros, it should be obvious that second and fourth rows must be independent from the third row.
But if we take the first plus the fifth minus twice the third we get
$(0,0,10-2x,0,0)$,
and we have a singular matrix if
$10-2x equiv 0pmod {11}$.
edited Jan 23 at 21:11
J. W. Tanner
3,2401320
answered Jan 23 at 17:20
Doug MDoug M
45.3k31954
edited Jan 23 at 21:11
J. W. Tanner
3,2401320
edited Jan 23 at 21:11
J. W. Tanner
3,2401320
edited Jan 23 at 21:11
J. W. Tanner
3,2401320
3,2401320
answered Jan 23 at 17:20
Doug MDoug M
45.3k31954
answered Jan 23 at 17:20
Doug MDoug M
45.3k31954
answered Jan 23 at 17:20
Doug MDoug M
45.3k31954
45.3k31954
add a comment |
add a comment |
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$begingroup$
Did you learned about determinants?
$endgroup$
– gbox
Jan 23 at 16:44
2
$begingroup$
Hint: if $xneq 0$, you can subtract a multiple of the third row from the last row.
$endgroup$
– Wojowu
Jan 23 at 16:44
$begingroup$
@gbox I'm not a veteran of the topic, but something I know. What in particular are you referring to?
$endgroup$
– Jack
Jan 23 at 16:46
$begingroup$
@Wojowu but basically I don't know the values of $x$. Is this operation allowed?
$endgroup$
– Jack
Jan 23 at 16:48
$begingroup$
@Jack: it is when $xne0$. But you may prefer to swap rows 3, 5 and eliminate as usual.
$endgroup$
– Yves Daoust
Jan 23 at 16:54