0
$\begingroup$

Is there another method (involving determinants) for finding the inverse of a matrix than making the matrix equal to what it is supposed to look like in reduced row echelon form? Please explain. Thanks

$\endgroup$
1
$\begingroup$

Inverse of matrix (suppose A)

$$A^{-1} =\frac{ adjoint(A)}{det(A)}$$

*$A^{-1}$ Exist iff $det(A)≠0$

$\endgroup$
1
$\begingroup$

For a $(2\times2)$-matrix you may use the following formula:

$$\frac{1}{ad-bc}\cdot \begin{pmatrix} d & -b \\ -c & a \\ \end{pmatrix}. $$

$\endgroup$

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.