0
$\begingroup$

Please consider the 2x2 matrix below:

$\left[\begin{array}{ccc} 1 & 2 \\ 3 & 4 \end{array}\right]$

According to the definition given here and here, the cofactor matrix becomes:

$\left[\begin{array}{ccc} a_{22} & -a_{12} \\ -a_{21} & a_{11} \end{array}\right] = \left[\begin{array}{ccc} 4 & -2 \\ -3 & 1\end{array}\right]$

However, when I follow the practice given here and here, I do obtain the following cofactor matrix, which is the transpose of the above:

$\left[\begin{array}{ccc} a_{22} & -a_{21} \\ -a_{12} & a_{11} \end{array}\right] = \left[\begin{array}{ccc} 4 & -3 \\ -2 & 1\end{array}\right]$

The difference arises from the off-diagonal locations of $a_{12}$ and $a_{21}$.

Are these two cofactors equivalent to each other in some way?

$\endgroup$
3
$\begingroup$

The first "here" link is wrong. The second one, contrary to your thinking, gives the correct interpretation.

The rule is simple: to obtain the minor/cofactor of any element, strike out the whole row and column that contain it. Hence it cannot contain the element self.

$\endgroup$
2
  • 1
    $\begingroup$ Exactly. What the first link calls the cofactor matrix is actually the adjugate matrix. $\endgroup$ – darij grinberg Feb 14 '20 at 10:39
  • 1
    $\begingroup$ @darijgrinberg: I guess that there are other anomalies on that page. $\endgroup$ – Yves Daoust Feb 14 '20 at 10:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.