Key Concepts and Formulas

• Definition of Symmetric and Skew-Symmetric Matrices:
  • A square matrix $X$ is symmetric if its transpose equals itself: $X^T = X$.
  • A square matrix $X$ is skew-symmetric if its transpose equals its negative: $X^T = -X$.
• Properties of the Identity Matrix ($I$):
  • For any positive integer $k$, $I^k = I$.
  • A scalar multiple of the identity matrix, $kI$, is always symmetric, as $(kI)^T = kI^T = kI$. It is an identity matrix only if $k=1$. It is skew-symmetric only if $k=0$.
• Sum of a Geometric Series:
  • The sum of the first $n$ terms of a geometric series $a + ar + ar^2 + \dots + ar^{n-1}$ is given by the formula $S_n = a \frac{r^n - 1}{r - 1}$, where $a$ is the first term and $r$ is the common ratio.

Step-by-Step Solution

Step 1: Discovering the Pattern in Powers of Matrix A

The first and most crucial step is to analyze the given matrix $A$ and its powers. This will significantly simplify the calculation of the sums $M$ and $N$.

Given matrix: $A = \left[ {\begin{matrix} 0 & { - 2} \\ 2 & 0 \end{matrix}} \right]$

Let's compute $A^2$: $A^2 = A \cdot A = \left[ {\begin{matrix} 0 & { - 2} \\ 2 & 0 \end{matrix}} \right] \left[ {\begin{matrix} 0 & { - 2} \\ 2 & 0 \end{matrix}} \right]$ $A^2 = \left[ {\begin{matrix} (0)(0) + (-2)(2) & (0)(-2) + (-2)(0) \\ (2)(0) + (0)(2) & (2)(-2) + (0)(0) \end{matrix}} \right]$ $A^2 = \left[ {\begin{matrix} { - 4} & 0 \\ 0 & { - 4} \end{matrix}} \right]$ We can factor out $-4$ from this matrix: $A^2 = -4 \left[ {\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}} \right] = -4I$ where $I$ is the $2 \times 2$ identity matrix.

Why this is important: When $A^2$ is a scalar multiple of the identity matrix, it simplifies all higher powers of $A$ significantly.

• Even powers of A: We can write $A^{2k}$ as $(A^2)^k$. $A^{2k} = (A^2)^k = (-4I)^k = (-4)^k I^k = (-4)^k I$
• Odd powers of A: We can write $A^{2k-1}$ as $A^{2(k-1)} \cdot A$. $A^{2k-1} = (A^2)^{k-1} \cdot A = (-4I)^{k-1} \cdot A = (-4)^{k-1} I \cdot A = (-4)^{k-1} A$ These general forms for $A^{2k}$ and $A^{2k-1}$ will be instrumental in calculating $M$ and $N$.

Step 2: Calculating Matrix M (Sum of Even Powers)

The matrix $M$ is defined as the sum of the first 10 even powers of $A$: $M = \sum_{k=1}^{10} {{A^{2k}}} = A^2 + A^4 + A^6 + \dots + A^{20}$ Using the pattern we found in Step 1, $A^{2k} = (-4)^k I$: $M = (-4)^1 I + (-4)^2 I + (-4)^3 I + \dots + (-4)^{10} I$ We can factor out the identity matrix $I$: $M = I \left( (-4)^1 + (-4)^2 + (-4)^3 + \dots + (-4)^{10} \right)$ The expression in the parenthesis is a geometric series. Why we use a geometric series formula: Summing 10 terms individually would be tedious and error-prone. The geometric series formula provides a quick and accurate way to find the sum. In our case:

• First term $a = -4$
• Common ratio $r = -4$
• Number of terms $n = 10$

Using the formula $S_n = a \frac{r^n - 1}{r - 1}$: $S_{10} = (-4) \frac{(-4)^{10} - 1}{(-4) - 1} = (-4) \frac{4^{10} - 1}{-5} = \frac{4}{5} (4^{10} - 1)$ Therefore, matrix $M$ is: $M = \frac{4}{5} (4^{10} - 1) I$

Step 3: Calculating Matrix N (Sum of Odd Powers)

The matrix $N$ is defined as the sum of the first 10 odd powers of $A$: $N = \sum_{k=1}^{10} {{A^{2k - 1}}} = A^1 + A^3 + A^5 + \dots + A^{19}$ Using the pattern we found in Step 1, $A^{2k-1} = (-4)^{k-1} A$: $N = (-4)^0 A + (-4)^1 A + (-4)^2 A + \dots + (-4)^9 A$ We can factor out matrix $A$: $N = A \left( (-4)^0 + (-4)^1 + (-4)^2 + \dots + (-4)^9 \right)$ Again, the expression in the parenthesis is a geometric series. Why we use a geometric series formula: For the same reasons as with $M$, it simplifies the summation. In this series:

• First term $a = (-4)^0 = 1$
• Common ratio $r = -4$
• Number of terms $n = 10$

Using the formula $S_n = a \frac{r^n - 1}{r - 1}$: $S_{10} = (1) \frac{(-4)^{10} - 1}{(-4) - 1} = \frac{4^{10} - 1}{-5} = -\frac{1}{5} (4^{10} - 1)$ Therefore, matrix $N$ is: $N = -\frac{1}{5} (4^{10} - 1) A$

Step 4: Computing the Product $MN^2$

Now we need to find the product $MN^2$. First, let's calculate $N^2$: $N^2 = \left( -\frac{1}{5} (4^{10} - 1) A \right)^2$ Why we square the entire expression: The scalar part and the matrix part both get squared. $N^2 = \left( -\frac{1}{5} (4^{10} - 1) \right)^2 A^2$ $N^2 = \frac{1}{25} (4^{10} - 1)^2 A^2$ Now, substitute $A^2 = -4I$ (from Step 1): $N^2 = \frac{1}{25} (4^{10} - 1)^2 (-4I)$ $N^2 = -\frac{4}{25} (4^{10} - 1)^2 I$

Finally, calculate $MN^2$: $MN^2 = \left( \frac{4}{5} (4^{10} - 1) I \right) \left( -\frac{4}{25} (4^{10} - 1)^2 I \right)$ Why we multiply the scalar coefficients and the matrices separately: Scalar multiplication commutes with matrix multiplication, and $I \cdot I = I$. $MN^2 = \left( \frac{4}{5} \right) \left( -\frac{4}{25} \right) \left( (4^{10} - 1) (4^{10} - 1)^2 \right) (I \cdot I)$ $MN^2 = -\frac{16}{125} (4^{10} - 1)^3 I$

Let $C = -\frac{16}{125} (4^{10} - 1)^3$. Since $4^{10} - 1$ is a large positive number, $C$ is a non-zero scalar. So, $MN^2 = C I$.

Step 5: Determining the Nature of $MN^2$

We have found that $MN^2 = C I$, where $C = -\frac{16}{125} (4^{10} - 1)^3$. Let's analyze the properties of this matrix:

1. Is it an identity matrix? For $MN^2$ to be an identity matrix, the scalar $C$ must be equal to 1. Since $C = -\frac{16}{125} (4^{10} - 1)^3$ is clearly not equal to 1 (it's a negative number with a large magnitude), $MN^2$ is not an identity matrix.

2. Is it a symmetric matrix? A matrix $X$ is symmetric if $X^T = X$. For $MN^2 = CI$: $(CI)^T = C I^T = C I$ Since $(MN^2)^T = MN^2$, $MN^2$ is a symmetric matrix.

3. Is it a skew-symmetric matrix? A matrix $X$ is skew-symmetric if $X^T = -X$. For $MN^2 = CI$: $(CI)^T = CI$ For it to be skew-symmetric, we would need $CI = -CI$, which implies $2CI = 0$. This can only happen if $C=0$ (or $I$ is a zero matrix, which it isn't). Since $C = -\frac{16}{125} (4^{10} - 1)^3 \ne 0$, $MN^2$ is not a skew-symmetric matrix.

Combining these observations, $MN^2$ is a non-identity symmetric matrix.

Common Mistakes & Tips

• Overlooking Patterns: Always look for patterns in matrix powers, especially when dealing with matrices like $A$ where $A^2$ is a scalar multiple of the identity matrix ($kI$). This is a common simplification strategy in matrix problems.
• Incorrect Geometric Series Application: Ensure you correctly identify the first term ($a$), common ratio ($r$), and number of terms ($n$) for the geometric series formula. A common error is using $n=10$ when the terms are $A^1, A^3, \dots, A^{19}$ and the powers of the common ratio start from $0$.
• Misclassifying $kI$: Remember that a matrix of the form $kI$ is inherently symmetric for any scalar $k$. It is only an identity matrix if $k=1$, and only skew-symmetric if $k=0$. Do not confuse these distinct properties.

Summary

We began by analyzing the powers of matrix $A$, finding that $A^2 = -4I$, which allowed us to express all even powers as $A^{2k} = (-4)^k I$ and all odd powers as $A^{2k-1} = (-4)^{k-1} A$. We then calculated the sums $M$ and $N$ using the formula for a geometric series, resulting in $M = C_M I$ and $N = C_N A$ for certain scalar constants $C_M$ and $C_N$. Finally, we computed the product $MN^2$, which simplified to the form $C I$, where $C = -\frac{16}{125} (4^{10} - 1)^3$. Since $C$ is a non-zero scalar not equal to 1, $MN^2$ is a symmetric matrix but not an identity matrix, nor is it skew-symmetric.

The final answer is $\boxed{A}$ which corresponds to option (A).