Key Concept: Idempotent Matrices

A square matrix $M$ is called idempotent if $M^2 = M$. A fundamental property of idempotent matrices is that for any positive integer $k$, $M^k = M$. This property significantly simplifies calculations involving higher powers of such matrices.

Step-by-step Derivations

1. Determine the Nature of Matrices $A$ and $B$

First, we need to calculate the squares of matrices $A$ and $B$ to identify any patterns, particularly if they are idempotent. This step is crucial because it simplifies the terms $A^n$ and $B^m$ in the given equation.

For matrix $A$: $A = \begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix}$ Calculate $A^2$: $A^2 = A \cdot A = \begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix}$ $A^2 = \begin{pmatrix} (2)(2) + (-2)(1) & (2)(-2) + (-2)(-1) \\ (1)(2) + (-1)(1) & (1)(-2) + (-1)(-1) \end{pmatrix}$ $A^2 = \begin{pmatrix} 4 - 2 & -4 + 2 \\ 2 - 1 & -2 + 1 \end{pmatrix} = \begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix}$ Since $A^2 = A$, matrix $A$ is an idempotent matrix. Therefore, for any positive integer $n \ge 1$, we have $A^n = A$.

For matrix $B$: $B = \begin{pmatrix} -1 & 2 \\ -1 & 2 \end{pmatrix}$ Calculate $B^2$: $B^2 = B \cdot B = \begin{pmatrix} -1 & 2 \\ -1 & 2 \end{pmatrix} \begin{pmatrix} -1 & 2 \\ -1 & 2 \end{pmatrix}$ $B^2 = \begin{pmatrix} (-1)(-1) + (2)(-1) & (-1)(2) + (2)(2) \\ (-1)(-1) + (2)(-1) & (-1)(2) + (2)(2) \end{pmatrix}$ $B^2 = \begin{pmatrix} 1 - 2 & -2 + 4 \\ 1 - 2 & -2 + 4 \end{pmatrix} = \begin{pmatrix} -1 & 2 \\ -1 & 2 \end{pmatrix}$ Since $B^2 = B$, matrix $B$ is also an idempotent matrix. Therefore, for any positive integer $m \ge 1$, we have $B^m = B$.

Tip: Always check for idempotent, nilpotent ($M^k = 0$), or periodic ($M^k = I$) properties when dealing with higher powers of matrices. This can drastically simplify calculations.

2. Simplify the Matrix Equation

The given equation is $nA^n + mB^m = I$, where $I$ is the $2 \times 2$ identity matrix, $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. Using the idempotent property derived in Step 1, we can substitute $A^n = A$ and $B^m = B$ into the equation: $nA + mB = I$ Now, substitute the actual matrices $A$ and $B$: $n \begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix} + m \begin{pmatrix} -1 & 2 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

3. Perform Scalar Multiplication and Matrix Addition

Multiply each matrix by its scalar coefficient ($n$ and $m$ respectively): $\begin{pmatrix} 2n & -2n \\ n & -n \end{pmatrix} + \begin{pmatrix} -m & 2m \\ -m & 2m \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ Now, add the two matrices on the left side: $\begin{pmatrix} 2n - m & -2n + 2m \\ n - m & -n + 2m \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

4. Form and Solve a System of Linear Equations

For two matrices to be equal, their corresponding elements must be equal. This gives us a system of four linear equations:

1. $2n - m = 1$ (from element at row 1, col 1)
2. $-2n + 2m = 0$ (from element at row 1, col 2)
3. $n - m = 0$ (from element at row 2, col 1)
4. $-n + 2m = 1$ (from element at row 2, col 2)

Let's solve this system: From equation (2), $-2n + 2m = 0 \implies 2m = 2n \implies m = n$. From equation (3), $n - m = 0 \implies n = m$. Both equations (2) and (3) consistently tell us that $n$ must be equal to $m$.

Now, substitute $m = n$ into equation (1): $2n - n = 1$ $n = 1$

Since $m=n$, we also get $m=1$. So, the unique solution to the system of equations is $(n, m) = (1, 1)$.

Let's verify this solution with the remaining equation (4): $-n + 2m = -1 + 2(1) = -1 + 2 = 1$. This is consistent with equation (4). Therefore, $(n, m) = (1, 1)$ is the unique solution to the matrix equation.

Common Mistake: Sometimes students only equate a subset of the elements, which might lead to incorrect or multiple solutions if the system is not fully constrained. Always check consistency across all elements.

5. Check the Domain and Count the Elements

The problem states that $n, m \in \{1, 2, \ldots, 10\}$. Our unique solution is $(n, m) = (1, 1)$. Both $n=1$ and $m=1$ fall within the specified set $\{1, 2, \ldots, 10\}$. Since there is only one pair $(1, 1)$ that satisfies all the conditions, the number of elements in the set is 1.

Alternative Approach (Using $A+B=I$): Observe that $A+B = \begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix} + \begin{pmatrix} -1 & 2 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$. So, we have $nA + mB = I$ and $A+B=I$. Subtracting the second equation from the first: $(n-1)A + (m-1)B = 0$ Let $X = n-1$ and $Y = m-1$. $XA + YB = 0$ $X \begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix} + Y \begin{pmatrix} -1 & 2 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$ This gives the system: $2X - Y = 0$ $X - Y = 0$ The determinant of the coefficients for $X,Y$ in these two equations is $2(-1) - (-1)(1) = -2+1 = -1 \ne 0$. This implies a unique solution. From $X-Y=0$, we get $X=Y$. Substitute into $2X-Y=0$: $2X-X=0 \implies X=0$. Since $X=Y$, then $Y=0$. So $X=0$ and $Y=0$. This means $n-1=0 \implies n=1$ and $m-1=0 \implies m=1$. Again, this leads to the unique solution $(n,m) = (1,1)$.

Summary/Key Takeaway

This problem emphasizes the importance of recognizing special matrix properties, particularly idempotency. By identifying that $A^n=A$ and $B^m=B$, the complex matrix power equation simplifies to a straightforward system of linear equations. Solving this system and verifying the solution against the given domain provides the number of valid pairs. In this case, there is only one such pair, $(1,1)$.

The number of elements in the set $\{(n, m) : n, m \in \{1, 2, .........., 10\} \text{ and } nA^n + mB^m = I\}$ is 1.