1
$\begingroup$

I have a question about this matrix: $\begin{bmatrix}a & b\\c & d\end{bmatrix}$

Show that if $D = ad − cb$ does not equal $0$, then $A^{-1}$ = $D^{-1}*\begin{bmatrix}d & -b\\-c & a\end{bmatrix}$

How do I show this? Can someone help?

$\endgroup$
1
  • 4
    $\begingroup$ Hint: multiply the first matrix by the second. $\endgroup$ Dec 14, 2017 at 13:40

1 Answer 1

2
$\begingroup$

By definition, the inverse matrix $A^{-1}$ satisfies $$ A A^{-1} = A^{-1} A = I $$ where $I$ is the (in this case, $2\times 2$) identity matrix. To show the matrix $B = \frac{1}{D}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ is equal to $A^{-1}$, it suffices to show $AB = I$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .