Key Concept: The Cayley-Hamilton Theorem

The Cayley-Hamilton Theorem states that every square matrix satisfies its own characteristic equation. For a square matrix $A$ of order $n \times n$, its characteristic equation is given by $\det(A - \lambda I) = 0$, where $I$ is the identity matrix of the same order and $\lambda$ is a scalar variable. If the characteristic equation is $\lambda^n + c_{n-1}\lambda^{n-1} + \dots + c_1\lambda + c_0 = 0$, then according to the theorem, the matrix $A$ satisfies $A^n + c_{n-1}A^{n-1} + \dots + c_1A + c_0I = O$, where $O$ is the null matrix of order $n \times n$. This theorem is incredibly useful for expressing higher powers of a matrix in terms of lower powers and the identity matrix, significantly simplifying calculations.

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Step-by-Step Solution

Step 1: Determine the Characteristic Equation of Matrix P

First, we need to find the characteristic equation of the given matrix $P$. The matrix is $P = \left[ {\begin{matrix} 2 & { - 1} \\ 5 & { - 3} \\ \end{matrix} } \right]$. The characteristic equation is given by $\det(P - \lambda I) = 0$, where $I$ is the $2 \times 2$ identity matrix $\left[ {\begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} } \right]$.

$P - \lambda I = \left[ {\begin{matrix} 2 & { - 1} \\ 5 & { - 3} \\ \end{matrix} } \right] - \lambda \left[ {\begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} } \right] = \left[ {\begin{matrix} 2 - \lambda & { - 1} \\ 5 & { - 3 - \lambda} \\ \end{matrix} } \right]$

Now, we calculate the determinant: $\det(P - \lambda I) = (2 - \lambda)(-3 - \lambda) - (-1)(5) = 0$ Expand the terms: $(2)(-3) + (2)(-\lambda) + (-\lambda)(-3) + (-\lambda)(-\lambda) + 5 = 0$ $-6 - 2\lambda + 3\lambda + \lambda^2 + 5 = 0$ Combine like terms: $\lambda^2 + \lambda - 1 = 0$ This is the characteristic equation of matrix $P$.

Step 2: Apply the Cayley-Hamilton Theorem to Express $P^2$

According to the Cayley-Hamilton Theorem, the matrix $P$ must satisfy its own characteristic equation. This means we can substitute $P$ for $\lambda$ and $I$ for the constant term (multiplied by the identity matrix, as $c_0$ in the general form $c_0 I$) in the characteristic equation.

So, from $\lambda^2 + \lambda - 1 = 0$, we get: $P^2 + P - I = O$ Here, $O$ represents the $2 \times 2$ null matrix. This equation is fundamental as it allows us to express $P^2$ in terms of $P$ and $I$. Rearranging the equation to isolate $P^2$: $P^2 = I - P$ This relation is crucial for simplifying higher powers of $P$.

Step 3: Calculate Higher Powers of P in terms of P and I

Our goal is to find $n$ such that $P^n = 5I - 8P$. We will systematically calculate powers of $P$ using the relation $P^2 = I - P$.

• Calculate $P^3$ (optional, but good for understanding the pattern): $P^3 = P \cdot P^2 = P(I - P)$ $P^3 = PI - P^2$ Since $PI = P$ and substituting $P^2 = I - P$: $P^3 = P - (I - P)$ $P^3 = P - I + P$ $P^3 = 2P - I$

• Calculate $P^4$: We can calculate $P^4$ by squaring $P^2$: $P^4 = (P^2)^2 = (I - P)^2$ Expand the square (remembering that for matrices, $(A-B)^2 = A^2 - AB - BA + B^2$. Since $I$ commutes with any matrix $P$, $IP = PI = P$, so we can use the familiar $(A-B)^2 = A^2 - 2AB + B^2$ pattern): $P^4 = I^2 - 2IP + P^2$ Since $I^2 = I$ and $IP = P$: $P^4 = I - 2P + P^2$ Now, substitute $P^2 = I - P$ into this expression: $P^4 = I - 2P + (I - P)$ $P^4 = I - 2P + I - P$ $P^4 = 2I - 3P$

• Calculate $P^6$: We can calculate $P^6$ by multiplying $P^4$ by $P^2$: $P^6 = P^4 \cdot P^2$ Substitute the expressions we found for $P^4$ and $P^2$: $P^6 = (2I - 3P)(I - P)$ Now, carefully expand this matrix product using the distributive property. Remember that $I \cdot I = I$, $I \cdot P = P$, and $P \cdot I = P$: $P^6 = 2I \cdot I - 2I \cdot P - 3P \cdot I + 3P \cdot P$ $P^6 = 2I - 2P - 3P + 3P^2$ Combine the $P$ terms: $P^6 = 2I - 5P + 3P^2$ Finally, substitute $P^2 = I - P$ one last time into this expression: $P^6 = 2I - 5P + 3(I - P)$ $P^6 = 2I - 5P + 3I - 3P$ Combine the $I$ terms and $P$ terms: $P^6 = (2+3)I + (-5-3)P$ $P^6 = 5I - 8P$

Step 4: Determine the Value of n

We are given the condition $P^n = 5I - 8P$. From our calculations, we found that $P^6 = 5I - 8P$. By comparing these two equations, we can conclude that: $n = 6$

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Tips and Common Mistakes

• Tip 1: Efficiency of Cayley-Hamilton: This theorem is extremely powerful for problems involving higher powers of matrices. Directly calculating $P^2, P^3, P^4, \dots$ by matrix multiplication would be much more tedious and error-prone.
• Tip 2: Identity Matrix Properties: Always remember that $I \cdot A = A \cdot I = A$ for any matrix $A$ of compatible dimensions. Also, $I^k = I$ for any positive integer $k$.
• Common Mistake 1: Forgetting the Identity Matrix: When applying Cayley-Hamilton, the constant term in the characteristic equation must be multiplied by the identity matrix ($c_0$ becomes $c_0 I$). Forgetting this is a common error.
• Common Mistake 2: Algebraic Errors: Be careful with signs and combining terms during expansion and substitution. A small arithmetic error can lead to a completely different result.
• Alternative (but less efficient) Method: One could calculate $P^2$, then $P^3$, $P^4$, etc., directly by matrix multiplication until the desired form is reached. However, this is computationally intensive and not recommended for JEE.

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Summary and Key Takeaway

This problem beautifully demonstrates the utility of the Cayley-Hamilton Theorem. By first finding the characteristic equation of the matrix $P$ and then applying the theorem, we obtained a fundamental relation ($P^2 = I - P$) that allowed us to express all higher powers of $P$ as a linear combination of $P$ and $I$. This systematic approach made it straightforward to determine the value of $n$ for the given equation $P^n = 5I - 8P$. The key takeaway is to always consider using the Cayley-Hamilton Theorem when dealing with higher powers of matrices, as it often provides a much more elegant and efficient solution than direct computation.