Question

If $A=\left[\begin{array}{ccc}2& 2& 2\\ 2& 2& 2\\ 2& 2& 2\end{array}\right]$, then $A^3-35A=$

• A. $A$
• B. $2A$
• C. $3A$
• D. $4A$

Answer 1

The correct option is A $A$

$A^2=A\cdot A=\left[\begin{array}{ccc}2& 2& 2\\ 2& 2& 2\\ 2& 2& 2\end{array}\right]\left[\begin{array}{ccc}2& 2& 2\\ 2& 2& 2\\ 2& 2& 2\end{array}\right]=\left[\begin{array}{ccc}12& 12& 12\\ 12& 12& 12\\ 12& 12& 12\end{array}\right]A^3=A^2\cdot A=\left[\begin{array}{ccc}12& 12& 12\\ 12& 12& 12\\ 12& 12& 12\end{array}\right]\left[\begin{array}{ccc}2& 2& 2\\ 2& 2& 2\\ 2& 2& 2\end{array}\right]=\left[\begin{array}{ccc}72& 72& 72\\ 72& 72& 72\\ 72& 72& 72\end{array}\right]\Rightarrow A^3-35A=\left[\begin{array}{ccc}72& 72& 72\\ 72& 72& 72\\ 72& 72& 72\end{array}\right]-35\left[\begin{array}{ccc}2& 2& 2\\ 2& 2& 2\\ 2& 2& 2\end{array}\right]=\left[\begin{array}{ccc}2& 2& 2\\ 2& 2& 2\\ 2& 2& 2\end{array}\right]=A$