Question

Let P be a square matrix satisfying ${{P}^{2}}=I-P$ where I is the identity matrix of the same dimensions as P. If ${{P}^{n}}=5I-8P$, then n equals,
[a] 4
[b] 5
[c] 6
[d] 7

Answer 1

- Hint: Multiply the equation ${{P}^{2}}=I-P$ by P and Express higher powers of P in terms of I and P only. Start by multiplying both sides of the equation by P and replace ${{P}^{2}}$ by $I-P$. Hence ${{P}^{3}}$ will be expressed in terms of I and P only. Again multiply both sides by P and repeat the same process. Stop when RHS becomes 5I -8P.

Complete step-by-step solution -

We know ${{P}^{2}}=I-P\text{ (i)}$
Multiplying both sides by P, we get
${{P}^{3}}=\left( I-P \right)P$.
i.e. ${{P}^{3}}=IP-{{P}^{2}}$
We know that IP = P
Using the above formula, we get
${{P}^{3}}=P-{{P}^{2}}$
Substituting the value of ${{P}^{2}}$ from equation (i), we get
$\begin{align}
  & {{P}^{3}}=P-\left( I-P \right) \\
 & \Rightarrow {{P}^{3}}=2P-I \\
\end{align}$
Multiplying both sides by P, we get
$\begin{align}
  & {{P}^{4}}=P\left( 2P-I \right) \\
 & \Rightarrow {{P}^{4}}=2{{P}^{2}}-IP \\
\end{align}$
We know that IP = P
Using the above formula, we get
${{P}^{4}}=2{{P}^{2}}-P$
Substituting the value of ${{P}^{2}}$ from equation (i), we get
$\begin{align}
  & {{P}^{4}}=2\left( I-P \right)-P \\
 & \Rightarrow {{P}^{4}}=2I-2P-P \\
 & \Rightarrow {{P}^{4}}=2I-3P \\
\end{align}$
Multiplying both sides by P, we get
$\begin{align}
  & {{P}^{5}}=P\left( 2I-3P \right) \\
 & \Rightarrow {{P}^{5}}=2IP-3{{P}^{2}} \\
\end{align}$
We know that IP = P
Using the above formula, we get
${{P}^{5}}=2P-3{{P}^{2}}$
Substituting the value of ${{P}^{2}}$ from equation (i), we get
$\begin{align}
  & {{P}^{5}}=2P-3\left( I-P \right) \\
 & =2P-3I+3P \\
 & =5P-3I \\
 & \Rightarrow {{P}^{5}}=5P-3I \\
\end{align}$
Multiplying both sides by P, we get
$\begin{align}
  & {{P}^{6}}=P\left( 5P-3I \right) \\
 & =5{{P}^{2}}-3IP \\
\end{align}$
We know that IP = P
Using the above formula, we get
${{P}^{6}}=5{{P}^{2}}-3P$
Substituting the value of ${{P}^{2}}$ from equation (i), we get
\[\begin{align}
  & {{P}^{6}}=5\left( I-P \right)-3P \\
 & =5I-5P-3P \\
 & =5I-8P \\
\end{align}\]
Hence we have ${{P}^{6}}=5I-8P$
Hence the value of n = 6.

Note: [1]This solution is a continued application of the recursive relation ${{P}^{n+2}}={{P}^{n}}-{{P}^{n+1}}$. This relation can be obtained by multiplying ${{P}^{n}}$ on both sides of the equation ${{P}^{2}}=I-P$.
[2] These matrix relations can also be obtained by the application of Cayley-Hamilton theorem, which says that every square matrix satisfies its characteristic polynomial.
[3] The roots of the characteristic polynomial are called eigenvalues which play a significant role in matrix algebra.
[4] Matrices also play a fundamental role in studying systems of equations.