Question

If $P=\begin{bmatrix} \lambda & 0 \\ 1 & 1 \end{bmatrix}$ and $Q=\begin{bmatrix} 4 & 0 \\ 3 & 1 \end{bmatrix}$ such that $P^{2}=Q,$ then $P^{3}$ is equal to

• A. $\begin{bmatrix} 8 & 0 \\ 7 & 1 \end{bmatrix}$
• B. $\begin{bmatrix} 10 & 1 \\ 8 & 0 \end{bmatrix}$
• C. $\begin{bmatrix} 7 & 0 \\ 8 & 1 \end{bmatrix}$
• D. $\begin{bmatrix} 6 & 0 \\ 4 & 1 \end{bmatrix}$

Answer 1

Answer: (A)

$P^{2}=\begin{bmatrix} \lambda & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} \lambda & 0 \\ 1 & 1 \end{bmatrix}=\begin{bmatrix} \lambda ^{2} & 0 \\ \lambda +1 & 1 \end{bmatrix}=\begin{bmatrix} 4 & 0 \\ 3 & 1 \end{bmatrix}$
So, $\lambda =2$
Hence, $P^{3}=\begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 4 & 0 \\ 3 & 1 \end{bmatrix}=\begin{bmatrix} 8 & 0 \\ 7 & 1 \end{bmatrix}$