A^-1 = [1, 2; 0, 1] and B^-1 = [1, 0; 3, 1]. What is (AB)^-1?

Question 102

MediumOriginal question

$A^{-1} = \begin{bmatrix}1 & 2 \\ 0 & 1\end{bmatrix}$ and $B^{-1} = \begin{bmatrix}1 & 0 \\ 3 & 1\end{bmatrix}$ . What is $(AB)^{-1}$ ?

\begin{bmatrix}7 & 2 \\ 3 & 1\end{bmatrix}

Correct answerYour answer

\begin{bmatrix}2 & 2 \\ 3 & 2\end{bmatrix}

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\begin{bmatrix}1 & 2 \\ 3 & 7\end{bmatrix}

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\begin{bmatrix}7 & 16 \\ 3 & 7\end{bmatrix}

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Show the answer and explanation

CorrectNot quiteAnswer C

Correct answer: C

The inverse of a product reverses the order: $(AB)^{-1} = B^{-1}A^{-1}$ .
$B^{-1}A^{-1} = \begin{bmatrix}1 & 0 \\ 3 & 1\end{bmatrix}\begin{bmatrix}1 & 2 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}1 & 2 \\ 3 & 7\end{bmatrix}$ .

Worth noting

$\begin{bmatrix}7 & 2 \\ 3 & 1\end{bmatrix}$ is $A^{-1}B^{-1}$ , in the order of $AB$ and not reversed.

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