Question

If A = 1 2 1 1 and B= 0 -1 1 2 then show that (AB)^-1 = B^-1 A^-1 help meeeee plessss ​

Answer 1

$A = \begin{bmatrix}1&2\\1&1\end{bmatrix} \implies A^{-1} = \dfrac1{\det(A)}\begin{bmatrix}1&-1\\-2&1\end{bmatrix} = \begin{bmatrix}-1&1\\2&-1\end{bmatrix}$

where det(A) = 1×1 - 2×1 = -1.

$B = \begin{bmatrix}0&-1\\1&2\end{bmatrix} \implies B^{-1} = \dfrac1{\det(B)}\begin{bmatrix}2&1\\-1&0\end{bmatrix} = \begin{bmatrix}2&1\\-1&0\end{bmatrix}$

where det(B) = 0×2 - (-1)×1 = 1. Then

$B^{-1}A^{-1} = \begin{bmatrix}2&1\\-1&0\end{bmatrix} \begin{bmatrix}-1&1\\2&-1\end{bmatrix} = \begin{bmatrix}-1&3\\1&-2\end{bmatrix}$

On the other side, we have

$AB = \begin{bmatrix}1&2\\1&1\end{bmatrix} \begin{bmatrix}0&-1\\1&2\end{bmatrix} = \begin{bmatrix}2&3\\1&1\end{bmatrix}$

and det(AB) = det(A) det(B) = (-1)×1 = -1. So

$(AB)^{-1} = \dfrac1{\det(AB)}\begin{bmatrix}1&-3\\-1&2\end{bmatrix} = \begin{bmatrix}-1&3\\1&-2\end{bmatrix}$

and both matrices are clearly the same.

More generally, we have by definition of inverse,

$(AB)(AB)^{-1} = I$

where $I$ is the identity matrix. Multiply on the left by A ⁻¹ to get

$A^{-1}(AB)(AB)^{-1} = A^{-1}I = A^{-1}$

Multiplication of matrices is associative, so we can regroup terms as

$(A^{-1}A)B(AB)^{-1} = A^{-1} \\\\ B(AB)^{-1} = A^{-1}$

Now multiply again on the left by B ⁻¹ and do the same thing:

$B^{-1}\left(B(AB)^{-1}\right) = (B^{-1}B)(AB)^{-1} = B^{-1}A^{-1} \\\\ (AB)^{-1} = B^{-1}A^{-1}$