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







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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
















0












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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














0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











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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














  • 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










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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.






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    1 Answer
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    active

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    1












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 24 at 4:29









        Afzal AnsariAfzal Ansari

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