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Given $$\mathbf{A}^2=2\mathbf{A}-\mathbf{I}$$ where $\mathbf{A}$ is a $4\times4$ matrix and $\mathbf{I}$ is the $4\times 4$ identity matrix. Express $\mathbf{A}^3$ and $\mathbf{A}^4$ in the form $$k\mathbf{A}+l\mathbf{I}$$ where $k$ and $l$ are scalars.

My attempt at answering this is for $\mathbf{A}^3$$$\begin{align}\mathbf{A}^3=\mathbf{A}^2\mathbf{A} &=(2\mathbf{A}-\mathbf{I})\mathbf{A}\\ &=2\mathbf{A}^2-\mathbf{I}\mathbf{A}\\ &=2(2\mathbf{A}-\mathbf{I})-\mathbf{I}\mathbf{A}\\ &=4\mathbf{A}-2\mathbf{I}-\mathbf{I}\mathbf{A} \end{align}$$

Which is where I get stuck. My question is how do I find $l$?

Jacob
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2 Answers2

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You are almost there! Recall that $AI = IA=A$ for any matrix $A\in \Bbb{R}^{n\times n}$ where $I\in \Bbb{R}^{n\times n}$ is the identity matrix of $\Bbb{R}^{n\times n}$. In this case $n=4$ and therefore

$$4A-2I-IA=4A-2I-A=3A-2I$$

Joaquin Liniado
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$k=3$ and $l=-2$ because you have $3A-2I$.