Question

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

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Answer 1

`A = \begin{bmatrix}1&2\\1&1\end{bmatrix} \implies A^(-1) = \frac1{\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) = \frac1{\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) = \frac1{\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)`