Find values of x so that the matrix is invertibleFind values so matrix not invertible?values of $t inmathbb R$ the matrix is not invertibleFind the values of $k$ so that the matrix is not invertibleFind values of $t$ for which a matrix is invertible?For what values of $x_1, x_2, x_3, x_4$ is the matrix $A$ invertible?matrix that fails to be invertible using variablesFind the invertible matrixFind values of x so that the matrix is not invertibleInvertible matrix and spanFinding an invertible matrix P and some matrix C
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Find values of x so that the matrix is invertible
Find values so matrix not invertible?values of $t inmathbb R$ the matrix is not invertibleFind the values of $k$ so that the matrix is not invertibleFind values of $t$ for which a matrix is invertible?For what values of $x_1, x_2, x_3, x_4$ is the matrix $A$ invertible?matrix that fails to be invertible using variablesFind the invertible matrixFind values of x so that the matrix is not invertibleInvertible matrix and spanFinding an invertible matrix P and some matrix C
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Find values of $x$ so that the matrix is invertible
$$A=beginpmatrix
x & 0 & x \
x & 2 & 1 \
2x & 0 & 2x \
endpmatrix$$
I know that a matrix is invertible if determinand is not $0$, but I don't know how to find the $x$ values. I feel is a tricky question and this matrix will not be invertible no matter which value $x$ takes, but I don't know how to prove that either.
linear-algebra matrices inverse
edited 8 hours ago
José Carlos Santos
199k25 gold badges158 silver badges276 bronze badges
asked 8 hours ago
Leo PelozoLeo Pelozo
1183 bronze badges
New contributor
Leo Pelozo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Find values of $x$ so that the matrix is invertible
$$A=beginpmatrix
x & 0 & x \
x & 2 & 1 \
2x & 0 & 2x \
endpmatrix$$
I know that a matrix is invertible if determinand is not $0$, but I don't know how to find the $x$ values. I feel is a tricky question and this matrix will not be invertible no matter which value $x$ takes, but I don't know how to prove that either.
linear-algebra matrices inverse
edited 8 hours ago
José Carlos Santos
199k25 gold badges158 silver badges276 bronze badges
asked 8 hours ago
Leo PelozoLeo Pelozo
1183 bronze badges
New contributor
Leo Pelozo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
8
$begingroup$
Your feeling is correct (compare the top and bottom rows)
$endgroup$
– Henry
8 hours ago
add a comment |
$begingroup$
Find values of $x$ so that the matrix is invertible
$$A=beginpmatrix
x & 0 & x \
x & 2 & 1 \
2x & 0 & 2x \
endpmatrix$$
I know that a matrix is invertible if determinand is not $0$, but I don't know how to find the $x$ values. I feel is a tricky question and this matrix will not be invertible no matter which value $x$ takes, but I don't know how to prove that either.
linear-algebra matrices inverse
edited 8 hours ago
José Carlos Santos
199k25 gold badges158 silver badges276 bronze badges
asked 8 hours ago
Leo PelozoLeo Pelozo
1183 bronze badges
New contributor
Leo Pelozo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Find values of $x$ so that the matrix is invertible
$$A=beginpmatrix
x & 0 & x \
x & 2 & 1 \
2x & 0 & 2x \
endpmatrix$$
I know that a matrix is invertible if determinand is not $0$, but I don't know how to find the $x$ values. I feel is a tricky question and this matrix will not be invertible no matter which value $x$ takes, but I don't know how to prove that either.
linear-algebra matrices inverse
linear-algebra matrices inverse
edited 8 hours ago
José Carlos Santos
199k25 gold badges158 silver badges276 bronze badges
asked 8 hours ago
Leo PelozoLeo Pelozo
1183 bronze badges
New contributor
Leo Pelozo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 8 hours ago
José Carlos Santos
199k25 gold badges158 silver badges276 bronze badges
asked 8 hours ago
Leo PelozoLeo Pelozo
1183 bronze badges
New contributor
Leo Pelozo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 8 hours ago
José Carlos Santos
199k25 gold badges158 silver badges276 bronze badges
edited 8 hours ago
José Carlos Santos
199k25 gold badges158 silver badges276 bronze badges
edited 8 hours ago
José Carlos Santos
199k25 gold badges158 silver badges276 bronze badges
199k25 gold badges158 silver badges276 bronze badges
asked 8 hours ago
Leo PelozoLeo Pelozo
1183 bronze badges
New contributor
Leo Pelozo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
Leo PelozoLeo Pelozo
1183 bronze badges
asked 8 hours ago
Leo PelozoLeo Pelozo
1183 bronze badges
1183 bronze badges
New contributor
Leo Pelozo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Leo Pelozo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
8
$begingroup$
Your feeling is correct (compare the top and bottom rows)
$endgroup$
– Henry
8 hours ago
add a comment |
8
$begingroup$
Your feeling is correct (compare the top and bottom rows)
$endgroup$
– Henry
8 hours ago
8
8
$begingroup$
Your feeling is correct (compare the top and bottom rows)
$endgroup$
– Henry
8 hours ago
$begingroup$
Your feeling is correct (compare the top and bottom rows)
$endgroup$
– Henry
8 hours ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
You just have to calculate it determinant:
$$ det(A) = 4x^2 -4x^2 $$
Since it is always $0$ it is never invertibile.
answered 8 hours ago
AquaAqua
56k14 gold badges68 silver badges136 bronze badges
$endgroup$
add a comment |
$begingroup$
If $C_1$, $C_2$, and $C_3$ are the three columns of $A$, then $C_1-frac x2C_2=C_3-frac12C_2$. Therefore, the columns are not linearly independent and so the matrix is not invertible (whatever $x$ is).
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
199k25 gold badges158 silver badges276 bronze badges
$endgroup$
add a comment |
$begingroup$
Note that $forall x, R_3=2R_1implies Rank(A)<3implies det(A)=0$.
answered 7 hours ago
Nitin UniyalNitin Uniyal
4,02512 silver badges23 bronze badges
$endgroup$
add a comment |
$begingroup$
$det(A)=beginvmatrix
x & 0 & x \
x & 2 & 1 \
2x & 0 & 2x \
endvmatrix$
$=beginvmatrix
x & 0 & x \
x & 2 & 1 \
0 & 0 & 0 \
endvmatrix=0$
The first step equals second step by row operations.
answered 8 hours ago
Culver KwanCulver Kwan
12810 bronze badges
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You just have to calculate it determinant:
$$ det(A) = 4x^2 -4x^2 $$
Since it is always $0$ it is never invertibile.
answered 8 hours ago
AquaAqua
56k14 gold badges68 silver badges136 bronze badges
$endgroup$
add a comment |
$begingroup$
You just have to calculate it determinant:
$$ det(A) = 4x^2 -4x^2 $$
Since it is always $0$ it is never invertibile.
answered 8 hours ago
AquaAqua
56k14 gold badges68 silver badges136 bronze badges
$endgroup$
add a comment |
$begingroup$
You just have to calculate it determinant:
$$ det(A) = 4x^2 -4x^2 $$
Since it is always $0$ it is never invertibile.
answered 8 hours ago
AquaAqua
56k14 gold badges68 silver badges136 bronze badges
$endgroup$
You just have to calculate it determinant:
$$ det(A) = 4x^2 -4x^2 $$
Since it is always $0$ it is never invertibile.
answered 8 hours ago
AquaAqua
56k14 gold badges68 silver badges136 bronze badges
answered 8 hours ago
AquaAqua
56k14 gold badges68 silver badges136 bronze badges
answered 8 hours ago
AquaAqua
56k14 gold badges68 silver badges136 bronze badges
answered 8 hours ago
AquaAqua
56k14 gold badges68 silver badges136 bronze badges
56k14 gold badges68 silver badges136 bronze badges
add a comment |
add a comment |
$begingroup$
If $C_1$, $C_2$, and $C_3$ are the three columns of $A$, then $C_1-frac x2C_2=C_3-frac12C_2$. Therefore, the columns are not linearly independent and so the matrix is not invertible (whatever $x$ is).
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
199k25 gold badges158 silver badges276 bronze badges
$endgroup$
add a comment |
$begingroup$
If $C_1$, $C_2$, and $C_3$ are the three columns of $A$, then $C_1-frac x2C_2=C_3-frac12C_2$. Therefore, the columns are not linearly independent and so the matrix is not invertible (whatever $x$ is).
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
199k25 gold badges158 silver badges276 bronze badges
$endgroup$
add a comment |
$begingroup$
If $C_1$, $C_2$, and $C_3$ are the three columns of $A$, then $C_1-frac x2C_2=C_3-frac12C_2$. Therefore, the columns are not linearly independent and so the matrix is not invertible (whatever $x$ is).
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
199k25 gold badges158 silver badges276 bronze badges
$endgroup$
If $C_1$, $C_2$, and $C_3$ are the three columns of $A$, then $C_1-frac x2C_2=C_3-frac12C_2$. Therefore, the columns are not linearly independent and so the matrix is not invertible (whatever $x$ is).
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
199k25 gold badges158 silver badges276 bronze badges
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
199k25 gold badges158 silver badges276 bronze badges
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
199k25 gold badges158 silver badges276 bronze badges
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
199k25 gold badges158 silver badges276 bronze badges
199k25 gold badges158 silver badges276 bronze badges
add a comment |
add a comment |
$begingroup$
Note that $forall x, R_3=2R_1implies Rank(A)<3implies det(A)=0$.
answered 7 hours ago
Nitin UniyalNitin Uniyal
4,02512 silver badges23 bronze badges
$endgroup$
add a comment |
$begingroup$
Note that $forall x, R_3=2R_1implies Rank(A)<3implies det(A)=0$.
answered 7 hours ago
Nitin UniyalNitin Uniyal
4,02512 silver badges23 bronze badges
$endgroup$
add a comment |
$begingroup$
Note that $forall x, R_3=2R_1implies Rank(A)<3implies det(A)=0$.
answered 7 hours ago
Nitin UniyalNitin Uniyal
4,02512 silver badges23 bronze badges
$endgroup$
Note that $forall x, R_3=2R_1implies Rank(A)<3implies det(A)=0$.
answered 7 hours ago
Nitin UniyalNitin Uniyal
4,02512 silver badges23 bronze badges
answered 7 hours ago
Nitin UniyalNitin Uniyal
4,02512 silver badges23 bronze badges
answered 7 hours ago
Nitin UniyalNitin Uniyal
4,02512 silver badges23 bronze badges
answered 7 hours ago
Nitin UniyalNitin Uniyal
4,02512 silver badges23 bronze badges
4,02512 silver badges23 bronze badges
add a comment |
add a comment |
$begingroup$
$det(A)=beginvmatrix
x & 0 & x \
x & 2 & 1 \
2x & 0 & 2x \
endvmatrix$
$=beginvmatrix
x & 0 & x \
x & 2 & 1 \
0 & 0 & 0 \
endvmatrix=0$
The first step equals second step by row operations.
answered 8 hours ago
Culver KwanCulver Kwan
12810 bronze badges
$endgroup$
add a comment |
$begingroup$
$det(A)=beginvmatrix
x & 0 & x \
x & 2 & 1 \
2x & 0 & 2x \
endvmatrix$
$=beginvmatrix
x & 0 & x \
x & 2 & 1 \
0 & 0 & 0 \
endvmatrix=0$
The first step equals second step by row operations.
answered 8 hours ago
Culver KwanCulver Kwan
12810 bronze badges
$endgroup$
add a comment |
$begingroup$
$det(A)=beginvmatrix
x & 0 & x \
x & 2 & 1 \
2x & 0 & 2x \
endvmatrix$
$=beginvmatrix
x & 0 & x \
x & 2 & 1 \
0 & 0 & 0 \
endvmatrix=0$
The first step equals second step by row operations.
answered 8 hours ago
Culver KwanCulver Kwan
12810 bronze badges
$endgroup$
$det(A)=beginvmatrix
x & 0 & x \
x & 2 & 1 \
2x & 0 & 2x \
endvmatrix$
$=beginvmatrix
x & 0 & x \
x & 2 & 1 \
0 & 0 & 0 \
endvmatrix=0$
The first step equals second step by row operations.
answered 8 hours ago
Culver KwanCulver Kwan
12810 bronze badges
answered 8 hours ago
Culver KwanCulver Kwan
12810 bronze badges
answered 8 hours ago
Culver KwanCulver Kwan
12810 bronze badges
answered 8 hours ago
Culver KwanCulver Kwan
12810 bronze badges
12810 bronze badges
add a comment |
add a comment |
Leo Pelozo is a new contributor. Be nice, and check out our Code of Conduct.
Leo Pelozo is a new contributor. Be nice, and check out our Code of Conduct.
Leo Pelozo is a new contributor. Be nice, and check out our Code of Conduct.
Leo Pelozo is a new contributor. Be nice, and check out our Code of Conduct.
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8
$begingroup$
Your feeling is correct (compare the top and bottom rows)
$endgroup$
– Henry
8 hours ago