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Let A = 1, 2; 0, 1 and B = 2, 1; 1, 1. Find (AB)⁻¹ as B⁻¹A⁻¹…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Let $A =

\displaystyle \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}

$and$ B =

\displaystyle \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}

$. Find$ (AB)^{-1} $as$ B^{-1}A^{-1} $. Give its entries in row$ 1 $, column$ 2 $and in row$ 2 $, column$ 2$.

Need a reminder? The key ideaShow

$A^{-1} = \dfrac{\text{adj}\,A}{|A|}$ exists exactly when $|A| \ne 0$ , and $A\,\text{adj}\,A = |A|I$ . For $2 \times 2$ , swap the diagonal and negate the other two entries. Also $(AB)^{-1} = B^{-1}A^{-1}$ , $|A^{-1}| = \dfrac{1}{|A|}$ and $|\text{adj}\,A| = |A|^{n-1}$ .

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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