Question

If $P=\left[\begin{array}{cc}\lambda & 0\\ 1& 1\end{array}\right]$ and $Q=\left[\begin{array}{cc}4& 0\\ 3& 1\end{array}\right]$ such that $P^2=Q,$ then $P^3$ is equal to

• A. $\left[\begin{array}{cc}8& 0\\ 7& 1\end{array}\right]$
• B. $\left[\begin{array}{cc}10& 1\\ 8& 0\end{array}\right]$
• C. $\left[\begin{array}{cc}7& 0\\ 8& 1\end{array}\right]$
• D. $\left[\begin{array}{cc}6& 0\\ 4& 1\end{array}\right]$

Answer 1

Answer: (A)

$P^2=\left[\begin{array}{cc}\lambda & 0\\ 1& 1\end{array}\right]\left[\begin{array}{cc}\lambda & 0\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}{\lambda }^2& 0\\ \lambda +1& 1\end{array}\right]=\left[\begin{array}{cc}4& 0\\ 3& 1\end{array}\right]$
So, $\lambda =2$
Hence, $P^3=\left[\begin{array}{cc}2& 0\\ 1& 1\end{array}\right]\left[\begin{array}{cc}4& 0\\ 3& 1\end{array}\right]=\left[\begin{array}{cc}8& 0\\ 7& 1\end{array}\right]$