Question

If $\mathrm{A}$ is a square matrix such that ${\mathrm{A}}^2=\mathrm{I}$, then ${(\mathrm{A}-\mathrm{I})}^3+{(\mathrm{A}+\mathrm{I})}^3-7\mathrm{A}$ is equal to ________?

Answer 1

Evaluate the given expression:

The given expression is ${(\mathrm{A}-\mathrm{I})}^3+{(\mathrm{A}+\mathrm{I})}^3-7\mathrm{A}$ .

${(\mathrm{A}-\mathrm{I})}^3+{(\mathrm{A}+\mathrm{I})}^3-7\mathrm{A}={\mathrm{A}}^3-{\mathrm{I}}^3-3\mathrm{AI}(\mathrm{A}-\mathrm{I})+{\mathrm{A}}^3+{\mathrm{I}}^3+3\mathrm{AI}(\mathrm{A}+\mathrm{I})-7\mathrm{A}$ $$\begin{gathered} \left[{(\mathrm{a}–\mathrm{b})}^3={\mathrm{a}}^3–{\mathrm{b}}^3–3\mathrm{ab}(\mathrm{a}–\mathrm{b})\right] \\ \left[{(\mathrm{a}+\mathrm{b})}^3={\mathrm{a}}^3+{\mathrm{b}}^3+3\mathrm{ab}(\mathrm{a}+\mathrm{b})\right] \end{gathered}$$

$$\begin{gathered} =2{\mathrm{A}}^3+6{\mathrm{AI}}^2-7\mathrm{A} \\ =2\left({\mathrm{A}}^2\right)\mathrm{A}+6\mathrm{A}-7\mathrm{A}\left[\because {\mathrm{I}}^2=\mathrm{I}\mathrm{and}\mathrm{AI}=\mathrm{IA}=\mathrm{A}\right] \\ =2\left(\mathrm{I}\right)\mathrm{A}+6\mathrm{A}-7\mathrm{A}\left[\because {\mathrm{A}}^2=\mathrm{I}\right] \\ =2\mathrm{A}+6\mathrm{A}-7\mathrm{A}\left[\because \mathrm{AI}=\mathrm{IA}=\mathrm{A}\right] \\ =8\mathrm{A}-7\mathrm{A} \\ =\mathrm{A} \end{gathered}$$

Hence, if ${\mathrm{A}}^2=\mathrm{I}$, then ${(\mathrm{A}-\mathrm{I})}^3+{(\mathrm{A}+\mathrm{I})}^3-7\mathrm{A}$ is equal to $A$.