1. Key Concepts and Formulas

• Matrix Multiplication: The fundamental operation used to calculate powers of a matrix. For two matrices $A = [a_{ij}]$ and $B = [b_{jk}]$, their product $C = AB = [c_{ik}]$ is defined as $c_{ik} = \sum_j a_{ij}b_{jk}$.
• Pattern Recognition for Matrix Powers: For finding high powers of a matrix ($P^n$ where $n$ is large), the most efficient method is often to compute the first few powers ($P^2, P^3, P^4, \dots$) and identify a pattern in the elements of the resulting matrices. This pattern can then be generalized to find $P^n$.
• Binomial Theorem for Matrices (Special Case): If a matrix $P$ can be expressed as $P = I + A$, where $I$ is the identity matrix and $A$ is a nilpotent matrix (i.e., $A^k = O$ for some positive integer $k$, where $O$ is the null matrix), and $I$ and $A$ commute ($IA = AI$), then $P^n = (I+A)^n$ can be expanded using the binomial theorem: $(I+A)^n = \binom{n}{0}I + \binom{n}{1}A + \binom{n}{2}A^2 + \dots + \binom{n}{n}A^n$. If $A^k = O$, the series terminates after the $(k-1)$th term.

2. Step-by-Step Solution

Step 1: Understand the Problem and Strategy We are asked to find $P^{50}$ for a given $2 \times 2$ matrix $P$. Directly multiplying the matrix 50 times is impractical. The most effective strategy is to calculate the first few powers of $P$ to identify a recurring pattern.

Given matrix: P = \left[ {\matrix{ 1 & 0 \cr {{1 \over 2}} & 1 \cr } } \right]

Step 2: Calculate Initial Powers of P

Step 2a: Calculate $P^2$ What we are doing: We multiply $P$ by itself to find $P^2$. This is the first step in observing how the matrix elements change. Why we are doing it: We need to see if any elements remain constant or start to follow a sequence.

P^2 = P \cdot P = \left[ {\matrix{ 1 & 0 \cr {{1 \over 2}} & 1 \cr } } \right] \left[ {\matrix{ 1 & 0 \cr {{1 \over 2}} & 1 \cr } } \right] Applying matrix multiplication rules: P^2 = \left[ {\matrix{ (1)(1) + (0)\left({{1 \over 2}}\right) & (1)(0) + (0)(1) \cr \left({{1 \over 2}}\right)(1) + (1)\left({{1 \over 2}}\right) & \left({{1 \over 2}}\right)(0) + (1)(1) \cr } } \right] P^2 = \left[ {\matrix{ 1 + 0 & 0 + 0 \cr {{1 \over 2}} + {{1 \over 2}} & 0 + 1 \cr } } \right] P^2 = \left[ {\matrix{ 1 & 0 \cr 1 & 1 \cr } } \right] Reasoning: We observe that the elements in positions $(1,1)$, $(1,2)$, and $(2,2)$ are $1$, $0$, and $1$ respectively, which are the same as in $P$. Only the element in position $(2,1)$ has changed from $\frac{1}{2}$ to $1$.

**Step 2b: Calculate $P^3$} What we are doing: We multiply $P^2$ by $P$ to find $P^3$. Why we are doing it: Calculating $P^3$ is crucial to confirm if the pattern observed in $P^2$ is consistent and to see how the $(2,1)$ element continues to evolve.

P^3 = P^2 \cdot P = \left[ {\matrix{ 1 & 0 \cr 1 & 1 \cr } } \right] \left[ {\matrix{ 1 & 0 \cr {{1 \over 2}} & 1 \cr } } \right] Applying matrix multiplication rules: P^3 = \left[ {\matrix{ (1)(1) + (0)\left({{1 \over 2}}\right) & (1)(0) + (0)(1) \cr (1)(1) + (1)\left({{1 \over 2}}\right) & (1)(0) + (1)(1) \cr } } \right] P^3 = \left[ {\matrix{ 1 + 0 & 0 + 0 \cr 1 + {{1 \over 2}} & 0 + 1 \cr } } \right] P^3 = \left[ {\matrix{ 1 & 0 \cr {{3 \over 2}} & 1 \cr } } \right] Reasoning: The elements $(1,1)$, $(1,2)$, and $(2,2)$ remain $1$, $0$, and $1$. The $(2,1)$ element has become $\frac{3}{2}$. This confirms a consistent change from $P^1$ to $P^2$ and $P^2$ to $P^3$.

**Step 2c: Calculate $P^4$} What we are doing: We multiply $P^3$ by $P$ to find $P^4$. Why we are doing it: A third or fourth term provides stronger evidence for a pattern, especially for linear or simple arithmetic progressions, reducing the chance of misinterpreting a short sequence.

P^4 = P^3 \cdot P = \left[ {\matrix{ 1 & 0 \cr {{3 \over 2}} & 1 \cr } } \right] \left[ {\matrix{ 1 & 0 \cr {{1 \over 2}} & 1 \cr } } \right] Applying matrix multiplication rules: P^4 = \left[ {\matrix{ (1)(1) + (0)\left({{1 \over 2}}\right) & (1)(0) + (0)(1) \cr \left({{3 \over 2}}\right)(1) + (1)\left({{1 \over 2}}\right) & \left({{3 \over 2}}\right)(0) + (1)(1) \cr } } \right] P^4 = \left[ {\matrix{ 1 + 0 & 0 + 0 \cr {{3 \over 2}} + {{1 \over 2}} & 0 + 1 \cr } } \right] P^4 = \left[ {\matrix{ 1 & 0 \cr {4 \over 2} & 1 \cr } } \right] = \left[ {\matrix{ 1 & 0 \cr 2 & 1 \cr } } \right] Reasoning: The pattern for the $(2,1)$ element is now very clear, while the other elements remain constant.

**Step 3: Identify the Pattern and Generalize $P^n$} What we are doing: We are listing the powers of $P$ and explicitly observing the pattern for each element. Why we are doing it: To derive a general formula for $P^n$.

Let's summarize the powers calculated:

• P^1 = \left[ {\matrix{ 1 & 0 \cr {{1 \over 2}} & 1 \cr } } \right] (The $(2,1)$ element is $1 \times \frac{1}{2}$)
• P^2 = \left[ {\matrix{ 1 & 0 \cr 1 & 1 \cr } } \right] (The $(2,1)$ element is $2 \times \frac{1}{2}$)
• P^3 = \left[ {\matrix{ 1 & 0 \cr {{3 \over 2}} & 1 \cr } } \right] (The $(2,1)$ element is $3 \times \frac{1}{2}$)
• P^4 = \left[ {\matrix{ 1 & 0 \cr 2 & 1 \cr } } \right] (The $(2,1)$ element is $4 \times \frac{1}{2}$)

Reasoning:

• The elements $P_{11}$, $P_{12}$, and $P_{22}$ consistently remain $1$, $0$, and $1$ respectively for all powers.
• The element $P_{21}$ follows an arithmetic progression. For $P^n$, the $(2,1)$ element is $n$ times the original $(2,1)$ element of $P$, which is $n \times \frac{1}{2}$.

Therefore, we can generalize the form of $P^n$ as: P^n = \left[ {\matrix{ 1 & 0 \cr {n \cdot {{1 \over 2}}} & 1 \cr } } \right]

**Step 4: Apply the Pattern to Find $P^{50}$} What we are doing: We are substituting $n=50$ into our generalized formula for $P^n$. Why we are doing it: This directly gives us the required $P^{50}$.

Using the general formula for $P^n$ with $n=50$: P^{50} = \left[ {\matrix{ 1 & 0 \cr {50 \cdot {{1 \over 2}}} & 1 \cr } } \right] P^{50} = \left[ {\matrix{ 1 & 0 \cr {25} & 1 \cr } } \right]

Alternative Method: Using Binomial Expansion for Matrices

What we are doing: We are decomposing $P$ into an identity matrix $I$ and a nilpotent matrix $A$, then applying the binomial theorem. Why we are doing it: This method provides a more rigorous derivation of the general formula and confirms the pattern observed.

Let $P = I + A$, where $I$ is the identity matrix. P = \left[ {\matrix{ 1 & 0 \cr {{1 \over 2}} & 1 \cr } } \right] = \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right] + \left[ {\matrix{ 0 & 0 \cr {{1 \over 2}} & 0 \cr } } \right] So, I = \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right] and A = \left[ {\matrix{ 0 & 0 \cr {{1 \over 2}} & 0 \cr } } \right].

Now, let's find powers of $A$: A^2 = A \cdot A = \left[ {\matrix{ 0 & 0 \cr {{1 \over 2}} & 0 \cr } } \right] \left[ {\matrix{ 0 & 0 \cr {{1 \over 2}} & 0 \cr } } \right] = \left[ {\matrix{ (0)(0) + (0)\left({{1 \over 2}}\right) & (0)(0) + (0)(0) \cr \left({{1 \over 2}}\right)(0) + (0)\left({{1 \over 2}}\right) & \left({{1 \over 2}}\right)(0) + (0)(0) \cr } } \right] = \left[ {\matrix{ 0 & 0 \cr 0 & 0 \cr } } \right] = O Since $A^2$ is the null matrix ($O$), $A$ is a nilpotent matrix of index 2. This means $A^k = O$ for all $k \ge 2$.

Since $I$ commutes with any matrix $A$ ($IA = AI = A$), we can apply the binomial theorem for matrices to $(I+A)^n$: $(I+A)^n = \binom{n}{0}I^n A^0 + \binom{n}{1}I^{n-1} A^1 + \binom{n}{2}I^{n-2} A^2 + \binom{n}{3}I^{n-3} A^3 + \dots$ Because $A^2 = O$, all terms with $A^k$ where $k \ge 2$ will be $O$. So the expansion simplifies significantly: $(I+A)^n = \binom{n}{0}I + \binom{n}{1}A + O + O + \dots$ $(I+A)^n = 1 \cdot I + n \cdot A$ Substituting the matrices $I$ and $A$: P^n = \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right] + n \left[ {\matrix{ 0 & 0 \cr {{1 \over 2}} & 0 \cr } } \right] P^n = \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right] + \left[ {\matrix{ 0 & 0 \cr {{n \cdot {{1 \over 2}}}} & 0 \cr } } \right] P^n = \left[ {\matrix{ 1 & 0 \cr {n \cdot {{1 \over 2}}} & 1 \cr } } \right] This general formula is identical to the one derived through pattern recognition. For $n=50$: P^{50} = \left[ {\matrix{ 1 & 0 \cr {50 \cdot {{1 \over 2}}} & 1 \cr } } \right] = \left[ {\matrix{ 1 & 0 \cr {25} & 1 \cr } } \right]

3. Common Mistakes & Tips

• Matrix Multiplication Errors: The most common mistake is arithmetic errors during matrix multiplication. Always double-check each element calculation.
• Insufficient Pattern Observation: Relying on only two powers ($P^1, P^2$) to identify a pattern can be misleading. Calculate $P^3$ and $P^4$ to ensure the pattern holds consistently.
• Ignoring All Elements: Ensure you observe the pattern for every element in the matrix. Some might remain constant, while others follow specific sequences.
• Overlooking Special Matrix Properties: Always check if the matrix is a sum of an identity and a nilpotent matrix, or if it's diagonal, idempotent ($A^2=A$), or involutory ($A^2=I$). These properties can offer shortcuts.

4. Summary

To find high powers of a matrix, pattern recognition is a highly effective strategy. By calculating the first few powers ($P^2, P^3, P^4$), we observed that the elements at positions $(1,1)$, $(1,2)$, and $(2,2)$ remained constant as $1$, $0$, and $1$ respectively. The element at position $(2,1)$ followed a linear pattern, becoming $n \times \frac{1}{2}$ for $P^n$. This led to the generalized formula P^n = \left[ {\matrix{ 1 & 0 \cr {n \cdot {{1 \over 2}}} & 1 \cr } } \right]. Substituting $n=50$, we found P^{50} = \left[ {\matrix{ 1 & 0 \cr {25} & 1 \cr } } \right]. An alternative method using the binomial expansion for matrices, by decomposing $P$ into $I+A$ where $A$ is nilpotent, confirmed this result.

The final answer is \boxed{\text{\left[ {\matrix{ 1 & 0 \cr {25} & 1 \cr } } \right]}}, which corresponds to option (A).