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Find adj A and A⁻¹ for A = 2, 3, 1; 1, 2, 1; 1, 1, 1…

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

Find $\text{adj}\,A$ and $A^{-1}$ for $A =

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

$. Give$ |A| $, and the entries of$ \text{adj}\,A $in row$ 1 $, column$ 1 $, in row$ 1 $, column$ 2 $, and in row$ 3 $, column$ 1$.

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

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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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