Question

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

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

Answer 1

Answer: (A)

$A$

$A=\left[\begin{array}{ccc}2& 2& 2\\ 2& 2& 2\\ 2& 2& 2\end{array}\right]$
$A^2=\left[\begin{array}{ccc}2& 2& 2\\ 2& 2& 2\\ 2& 2& 2\end{array}\right]\left[\begin{array}{ccc}2& 2& 2\\ 2& 2& 2\\ 2& 2& 2\end{array}\right]=\left[\begin{array}{ccc}12& 12& 12\\ 12& 12& 12\\ 12& 12& 12\end{array}\right]$
$A^3=\left[\begin{array}{ccc}12& 12& 12\\ 12& 12& 12\\ 12& 12& 12\end{array}\right]\left[\begin{array}{ccc}2& 2& 2\\ 2& 2& 2\\ 2& 2& 2\end{array}\right]=\left[\begin{array}{ccc}72& 72& 72\\ 72& 72& 72\\ 72& 72& 72\end{array}\right]$
$A^3-35A=\left[\begin{array}{ccc}72& 72& 72\\ 72& 72& 72\\ 72& 72& 72\end{array}\right]-35\left[\begin{array}{ccc}2& 2& 2\\ 2& 2& 2\\ 2& 2& 2\end{array}\right]$
$=\left[\begin{array}{ccc}72& 72& 72\\ 72& 72& 72\\ 72& 72& 72\end{array}\right]-\left[\begin{array}{ccc}70& 70& 70\\ 70& 70& 70\\ 70& 70& 70\end{array}\right]$
$=\left[\begin{array}{ccc}2& 2& 2\\ 2& 2& 2\\ 2& 2& 2\end{array}\right]$