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Question: Let P be a square matrix satisfying \({{P}^{2}}=I-P\) where I is the identity matrix of the same dim...

Let P be a square matrix satisfying P2=I−P{{P}^{2}}=I-P where I is the identity matrix of the same dimensions as P. If Pn=5I−8P{{P}^{n}}=5I-8P, then n equals,
[a] 4
[b] 5
[c] 6
[d] 7

Explanation

Solution

- Hint: Multiply the equation P2=I−P{{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 P2{{P}^{2}} by I−PI-P. Hence P3{{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 P2=I−P (i){{P}^{2}}=I-P\text{ (i)}
Multiplying both sides by P, we get
P3=(I−P)P{{P}^{3}}=\left( I-P \right)P.
i.e. P3=IP−P2{{P}^{3}}=IP-{{P}^{2}}
We know that IP = P
Using the above formula, we get
P3=P−P2{{P}^{3}}=P-{{P}^{2}}
Substituting the value of P2{{P}^{2}} from equation (i), we get
P3=P−(I−P) ⇒P3=2P−I \begin{aligned} & {{P}^{3}}=P-\left( I-P \right) \\\ & \Rightarrow {{P}^{3}}=2P-I \\\ \end{aligned}
Multiplying both sides by P, we get
P4=P(2P−I) ⇒P4=2P2−IP \begin{aligned} & {{P}^{4}}=P\left( 2P-I \right) \\\ & \Rightarrow {{P}^{4}}=2{{P}^{2}}-IP \\\ \end{aligned}
We know that IP = P
Using the above formula, we get
P4=2P2−P{{P}^{4}}=2{{P}^{2}}-P
Substituting the value of P2{{P}^{2}} from equation (i), we get
P4=2(I−P)−P ⇒P4=2I−2P−P ⇒P4=2I−3P \begin{aligned} & {{P}^{4}}=2\left( I-P \right)-P \\\ & \Rightarrow {{P}^{4}}=2I-2P-P \\\ & \Rightarrow {{P}^{4}}=2I-3P \\\ \end{aligned}
Multiplying both sides by P, we get
P5=P(2I−3P) ⇒P5=2IP−3P2 \begin{aligned} & {{P}^{5}}=P\left( 2I-3P \right) \\\ & \Rightarrow {{P}^{5}}=2IP-3{{P}^{2}} \\\ \end{aligned}
We know that IP = P
Using the above formula, we get
P5=2P−3P2{{P}^{5}}=2P-3{{P}^{2}}
Substituting the value of P2{{P}^{2}} from equation (i), we get
P5=2P−3(I−P) =2P−3I+3P =5P−3I ⇒P5=5P−3I \begin{aligned} & {{P}^{5}}=2P-3\left( I-P \right) \\\ & =2P-3I+3P \\\ & =5P-3I \\\ & \Rightarrow {{P}^{5}}=5P-3I \\\ \end{aligned}
Multiplying both sides by P, we get
P6=P(5P−3I) =5P2−3IP \begin{aligned} & {{P}^{6}}=P\left( 5P-3I \right) \\\ & =5{{P}^{2}}-3IP \\\ \end{aligned}
We know that IP = P
Using the above formula, we get
P6=5P2−3P{{P}^{6}}=5{{P}^{2}}-3P
Substituting the value of P2{{P}^{2}} from equation (i), we get

& {{P}^{6}}=5\left( I-P \right)-3P \\\ & =5I-5P-3P \\\ & =5I-8P \\\ \end{aligned}$$ 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.