Question

State true/false:

If $A=\left[\begin{array}{ccc}1& -1& 1\\ 2& -1& 0\\ 1& 0& 0\end{array}\right],$ then solve is $A^3=I$?

• A. True
• B. False

Answer 1

The correct option is A True

Given

$A=\left[\begin{array}{ccc}1& -1& 1\\ 2& -1& 0\\ 1& 0& 0\end{array}\right]$

Calculating $A^3$

$A^2=\left[\begin{array}{ccc}1& -1& 1\\ 2& -1& 0\\ 1& 0& 0\end{array}\right]\left[\begin{array}{ccc}1& -1& 1\\ 2& -1& 0\\ 1& 0& 0\end{array}\right]$

$A^2=\left[\begin{array}{ccc}(1\times 1+2\times -1+1\times 1)& (1\times -1+(-1\times -1)+(1\times 0))& 1\times 1+-1\times 0+1\times 0\\ 2\times 1+-1\times 2+0\times 1& 2\times -1-1\times -1+0\times 0& 1\times 1+1\times 0+0\times 0\\ 1\times 1+2\times 0+0\times 1& -1\times 1+-1\times 0+0\times 0& 1\times 1+0\times 0+0\times 0\end{array}\right]$

$A^2=\left[\begin{array}{ccc}1-2+1& -1+1+0& 1+0+0\\ 2-2+0& -2+1+0& 2+0+0\\ 1+0+0& -1+0+0& 1+0+0\end{array}\right]=\left[\begin{array}{ccc}0& 0& 1\\ 0& -1& 2\\ 1& -1& 1\end{array}\right]$

$A^3=A^2\times A=\left[\begin{array}{ccc}0& 0& 1\\ 0& -1& 2\\ 1& -1& 1\end{array}\right]\left[\begin{array}{ccc}1& -1& 1\\ 2& -1& 0\\ 1& 0& 0\end{array}\right]$

$A^3=\left[\begin{array}{ccc}0\times 1+0\times 2+1\times 1& 0\times -1+0\times -1+1\times 0& 0\times 1+0\times 0+1\times 0\\ 0\times 1+-1\times 2+2\times 1& 0\times -1+-1\times -1+2\times 0& 0\times 1+(-1\times 0)+2\times 0\\ 1\times 1+2\times -1+1\times 1& -1\times 1+(-1\times -1)+1\times 0& 1\times 1+(-1\times 0)+(1\times 0)\end{array}\right]$

$A^3=\left[\begin{array}{ccc}0+0+1& 0+0+0& 0+0+0\\ 0-2+2& 0+1+0& 0+0+0\\ 1-2+1& -1+1+0& 1+0+0\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=I$

$\Rightarrow A^3=I$

Hence, given statement is true.