Mixing
Is the formula $A^{-1}=frac{1}{det (A) }text{Adj}(A)$ really true for 2 x 2 matrices?
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Given a 2 x 2 matrix $A=begin{bmatrix} a & b \ c & d end{bmatrix}$, we have the inverse formula $A^{-1}=frac{1}{ad-bc}begin{bmatrix} d & -b \ -c & a end{bmatrix}$.
But isn't $text{Adj}(A)$ equal to $begin{bmatrix} d & -c \ -b & a end{bmatrix}$? So don't we have $frac{1}{det (A)}text{Adj}(A)=frac{1}{ad-bc}begin{bmatrix} d & -c \ -b & a end{bmatrix} ne begin{bmatrix} d & -b \ -c & a end{bmatrix}$?
linear-algebra matrices linear-transformations determinant inverse
asked Jan 24 at 2:11
RasputinRasputin
446211
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add a comment |
$begingroup$
Given a 2 x 2 matrix $A=begin{bmatrix} a & b \ c & d end{bmatrix}$, we have the inverse formula $A^{-1}=frac{1}{ad-bc}begin{bmatrix} d & -b \ -c & a end{bmatrix}$.
But isn't $text{Adj}(A)$ equal to $begin{bmatrix} d & -c \ -b & a end{bmatrix}$? So don't we have $frac{1}{det (A)}text{Adj}(A)=frac{1}{ad-bc}begin{bmatrix} d & -c \ -b & a end{bmatrix} ne begin{bmatrix} d & -b \ -c & a end{bmatrix}$?
linear-algebra matrices linear-transformations determinant inverse
asked Jan 24 at 2:11
RasputinRasputin
446211
$endgroup$
3
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$operatorname{Adj}(A)$ is the transpose of what you wrote.
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– anomaly
Jan 24 at 2:13
2
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Huh? No, that's not the adjugate. The cited formula for the inverse is correct.
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– Randall
Jan 24 at 2:13
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Thanks @anomaly, that explains it
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– Rasputin
Jan 24 at 2:15
add a comment |
$begingroup$
Given a 2 x 2 matrix $A=begin{bmatrix} a & b \ c & d end{bmatrix}$, we have the inverse formula $A^{-1}=frac{1}{ad-bc}begin{bmatrix} d & -b \ -c & a end{bmatrix}$.
But isn't $text{Adj}(A)$ equal to $begin{bmatrix} d & -c \ -b & a end{bmatrix}$? So don't we have $frac{1}{det (A)}text{Adj}(A)=frac{1}{ad-bc}begin{bmatrix} d & -c \ -b & a end{bmatrix} ne begin{bmatrix} d & -b \ -c & a end{bmatrix}$?
linear-algebra matrices linear-transformations determinant inverse
asked Jan 24 at 2:11
RasputinRasputin
446211
$endgroup$
Given a 2 x 2 matrix $A=begin{bmatrix} a & b \ c & d end{bmatrix}$, we have the inverse formula $A^{-1}=frac{1}{ad-bc}begin{bmatrix} d & -b \ -c & a end{bmatrix}$.
But isn't $text{Adj}(A)$ equal to $begin{bmatrix} d & -c \ -b & a end{bmatrix}$? So don't we have $frac{1}{det (A)}text{Adj}(A)=frac{1}{ad-bc}begin{bmatrix} d & -c \ -b & a end{bmatrix} ne begin{bmatrix} d & -b \ -c & a end{bmatrix}$?
linear-algebra matrices linear-transformations determinant inverse
linear-algebra matrices linear-transformations determinant inverse
asked Jan 24 at 2:11
RasputinRasputin
446211
asked Jan 24 at 2:11
RasputinRasputin
446211
asked Jan 24 at 2:11
RasputinRasputin
446211
asked Jan 24 at 2:11
RasputinRasputin
446211
asked Jan 24 at 2:11
RasputinRasputin
446211
446211
3
$begingroup$
$operatorname{Adj}(A)$ is the transpose of what you wrote.
$endgroup$
– anomaly
Jan 24 at 2:13
2
$begingroup$
Huh? No, that's not the adjugate. The cited formula for the inverse is correct.
$endgroup$
– Randall
Jan 24 at 2:13
$begingroup$
Thanks @anomaly, that explains it
$endgroup$
– Rasputin
Jan 24 at 2:15
add a comment |
3
$begingroup$
$operatorname{Adj}(A)$ is the transpose of what you wrote.
$endgroup$
– anomaly
Jan 24 at 2:13
2
$begingroup$
Huh? No, that's not the adjugate. The cited formula for the inverse is correct.
$endgroup$
– Randall
Jan 24 at 2:13
$begingroup$
Thanks @anomaly, that explains it
$endgroup$
– Rasputin
Jan 24 at 2:15
3
3
$begingroup$
$operatorname{Adj}(A)$ is the transpose of what you wrote.
$endgroup$
– anomaly
Jan 24 at 2:13
$begingroup$
$operatorname{Adj}(A)$ is the transpose of what you wrote.
$endgroup$
– anomaly
Jan 24 at 2:13
2
2
$begingroup$
Huh? No, that's not the adjugate. The cited formula for the inverse is correct.
$endgroup$
– Randall
Jan 24 at 2:13
$begingroup$
Huh? No, that's not the adjugate. The cited formula for the inverse is correct.
$endgroup$
– Randall
Jan 24 at 2:13
$begingroup$
Thanks @anomaly, that explains it
$endgroup$
– Rasputin
Jan 24 at 2:15
$begingroup$
Thanks @anomaly, that explains it
$endgroup$
– Rasputin
Jan 24 at 2:15
add a comment |
1 Answer
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The formula is true for all square invertible matrices. For adjoint A you have to take transpose of 'cofactor matrix of A'. You have not taken transpose.
answered Jan 24 at 4:29
Afzal AnsariAfzal Ansari
856
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add a comment |
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1 Answer
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$begingroup$
The formula is true for all square invertible matrices. For adjoint A you have to take transpose of 'cofactor matrix of A'. You have not taken transpose.
answered Jan 24 at 4:29
Afzal AnsariAfzal Ansari
856
$endgroup$
add a comment |
$begingroup$
The formula is true for all square invertible matrices. For adjoint A you have to take transpose of 'cofactor matrix of A'. You have not taken transpose.
answered Jan 24 at 4:29
Afzal AnsariAfzal Ansari
856
$endgroup$
add a comment |
$begingroup$
The formula is true for all square invertible matrices. For adjoint A you have to take transpose of 'cofactor matrix of A'. You have not taken transpose.
answered Jan 24 at 4:29
Afzal AnsariAfzal Ansari
856
$endgroup$
The formula is true for all square invertible matrices. For adjoint A you have to take transpose of 'cofactor matrix of A'. You have not taken transpose.
answered Jan 24 at 4:29
Afzal AnsariAfzal Ansari
856
answered Jan 24 at 4:29
Afzal AnsariAfzal Ansari
856
answered Jan 24 at 4:29
Afzal AnsariAfzal Ansari
856
answered Jan 24 at 4:29
Afzal AnsariAfzal Ansari
856
856
add a comment |
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3
$begingroup$
$operatorname{Adj}(A)$ is the transpose of what you wrote.
$endgroup$
– anomaly
Jan 24 at 2:13
2
$begingroup$
Huh? No, that's not the adjugate. The cited formula for the inverse is correct.
$endgroup$
– Randall
Jan 24 at 2:13
$begingroup$
Thanks @anomaly, that explains it
$endgroup$
– Rasputin
Jan 24 at 2:15