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Let A = 1, -1; 2, -1. Find A², then use it to find A²⁰²⁵…

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 & -1 \\ 2 & -1 \end{bmatrix}

$. Find$ A² $, then use it to find$ A^{2025} $. Give the entry in row$ 1 $, column$ 2 $and the entry in row$ 2 $, column$ 1 $of$ A^{2025}$.

Need a reminder? The key ideaShow

$(AB)_{ij}$ is row $i$ of $A$ multiplied by column $j$ of $B$ , and $AB$ exists when the columns of $A$ equal the rows of $B$ . Matrix multiplication is associative and distributive but not commutative. The entry of $N^k$ counts routes with $k$ steps.

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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