\(\rm A=\begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix}\) where a ∈ ℕ, then is A 100 - A 50 - 2A 25 equal to? where I is the identity matrix.
-2I
Understanding the Matrix Problem
The question asks us to evaluate a specific expression involving powers of a given matrix \(A = \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix}\), where \(a\) is a natural number. The expression is \(A^{100} - A^{50} - 2A^{25}\), and we need to find which of the given options it equals, in terms of the identity matrix \(I\).
To solve this, we first need to understand how to calculate powers of the matrix \(A\).
Calculating Powers of Matrix A
Let's calculate the first few powers of the matrix \(A\):
For n=1:
\[ A^1 = \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} \]For n=2:
\[ A^2 = A \times A = \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} \]Performing the matrix multiplication:
\[ A^2 = \begin{bmatrix} (1)(1) + (a)(0) & (1)(a) + (a)(1) \\ (0)(1) + (1)(0) & (0)(a) + (1)(1) \end{bmatrix} = \begin{bmatrix} 1+0 & a+a \\ 0+0 & 0+1 \end{bmatrix} = \begin{bmatrix} 1 & 2a \\ 0 & 1 \end{bmatrix} \]For n=3:
\[ A^3 = A^2 \times A = \begin{bmatrix} 1 & 2a \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} \]Performing the matrix multiplication:
\[ A^3 = \begin{bmatrix} (1)(1) + (2a)(0) & (1)(a) + (2a)(1) \\ (0)(1) + (1)(0) & (0)(a) + (1)(1) \end{bmatrix} = \begin{bmatrix} 1+0 & a+2a \\ 0+0 & 0+1 \end{bmatrix} = \begin{bmatrix} 1 & 3a \\ 0 & 1 \end{bmatrix} \]Observing the pattern, it appears that for any positive integer \(n\), the matrix \(A^n\) is given by:
\[ A^n = \begin{bmatrix} 1 & na \\ 0 & 1 \end{bmatrix} \]This type of matrix is a special case often related to shear transformations or related to the Jordan form of a matrix. The formula \(A^n = \begin{bmatrix} 1 & na \\ 0 & 1 \end{bmatrix}\) holds true and can be proven by induction.
Evaluating the Matrix Expression
Now we use the general formula for \(A^n\) to find the specific powers required in the expression \(A^{100} - A^{50} - 2A^{25}\).
- For \(n=100\): \(A^{100} = \begin{bmatrix} 1 & 100a \\ 0 & 1 \end{bmatrix}\)
- For \(n=50\): \(A^{50} = \begin{bmatrix} 1 & 50a \\ 0 & 1 \end{bmatrix}\)
- For \(n=25\): \(A^{25} = \begin{bmatrix} 1 & 25a \\ 0 & 1 \end{bmatrix}\)
Now substitute these into the expression:
\[ A^{100} - A^{50} - 2A^{25} = \begin{bmatrix} 1 & 100a \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 50a \\ 0 & 1 \end{bmatrix} - 2 \begin{bmatrix} 1 & 25a \\ 0 & 1 \end{bmatrix} \]First, multiply the matrix \(A^{25}\) by 2:
\[ 2 \begin{bmatrix} 1 & 25a \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 \times 1 & 2 \times 25a \\ 2 \times 0 & 2 \times 1 \end{bmatrix} = \begin{bmatrix} 2 & 50a \\ 0 & 2 \end{bmatrix} \]Now perform the matrix subtraction and addition:
\[ A^{100} - A^{50} - 2A^{25} = \begin{bmatrix} 1 & 100a \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 50a \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 50a \\ 0 & 2 \end{bmatrix} \]We perform the operations element by element:
- Top-left element: \(1 - 1 - 2 = -2\)
- Top-right element: \(100a - 50a - 50a = 100a - 100a = 0\)
- Bottom-left element: \(0 - 0 - 0 = 0\)
- Bottom-right element: \(1 - 1 - 2 = -2\)
The resulting matrix is:
\[ \begin{bmatrix} -2 & 0 \\ 0 & -2 \end{bmatrix} \]We can factor out -2 from this matrix:
\[ \begin{bmatrix} -2 & 0 \\ 0 & -2 \end{bmatrix} = -2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]The matrix \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) is the 2x2 identity matrix, denoted by \(I\).
Therefore, the expression \(A^{100} - A^{50} - 2A^{25}\) is equal to \(-2I\).
Comparing with Options
Let's compare our result \(-2I\) with the given options:
- Option 1: \(-2I\)
- Option 2: \(-I\)
- Option 3: \(2I\)
- Option 4: \(I\)
Our result matches Option 1.
| Term | Matrix Form |
|---|---|
| \(A^{100}\) | \(\begin{bmatrix} 1 & 100a \\ 0 & 1 \end{bmatrix}\) |
| \(A^{50}\) | \(\begin{bmatrix} 1 & 50a \\ 0 & 1 \end{bmatrix}\) |
| \(A^{25}\) | \(\begin{bmatrix} 1 & 25a \\ 0 & 1 \end{bmatrix}\) |
| \(2A^{25}\) | \(\begin{bmatrix} 2 & 50a \\ 0 & 2 \end{bmatrix}\) |
| \(A^{100} - A^{50} - 2A^{25}\) | \(\begin{bmatrix} 1-1-2 & 100a-50a-50a \\ 0-0-0 & 1-1-2 \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 0 & -2 \end{bmatrix}\) |
| Simplified Result | \(-2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = -2I\) |
Revision Table: Key Matrix Concepts
| Concept | Description | Example (for 2x2) |
|---|---|---|
| Matrix Addition/Subtraction | Add/subtract corresponding elements. Matrices must have the same dimensions. | \(\begin{bmatrix} a & b \\ c & d \end{bmatrix} + \begin{bmatrix} e & f \\ g & h \end{bmatrix} = \begin{bmatrix} a+e & b+f \\ c+g & d+h \end{bmatrix}\) |
| Scalar Multiplication | Multiply every element in the matrix by the scalar value. | \(k \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} ka & kb \\ kc & kd \end{bmatrix}\) |
| Matrix Multiplication | The element in row \(i\) and column \(j\) of the product is the dot product of row \(i\) of the first matrix and column \(j\) of the second matrix. Dimensions must be compatible (m x n and n x p results in m x p). | \(\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} e & f \\ g & h \end{bmatrix} = \begin{bmatrix} ae+bg & af+bh \\ ce+dg & cf+dh \end{bmatrix}\) |
| Identity Matrix (I) | A square matrix with 1s on the main diagonal and 0s elsewhere. Acts like the number 1 in matrix multiplication (\(AI = IA = A\)). | \(I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) (for 2x2) |
Additional Information: Special Matrices and Powers
The matrix \(A = \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix}\) is an example of a unipotent matrix if \(a \ne 0\). These matrices have 1s on the main diagonal and 0s below. The eigenvalues of such matrices are all 1.
For a general 2x2 matrix of the form \(N = \begin{bmatrix} 0 & a \\ 0 & 0 \end{bmatrix}\), we have \(N^2 = \begin{bmatrix} 0 & a \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & a \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\). Such a matrix is called nilpotent. We can write the given matrix \(A\) as \(A = I + N\), where \(I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) and \(N = \begin{bmatrix} 0 & a \\ 0 & 0 \end{bmatrix}\).
Since \(I\) and \(N\) commute (\(IN = NI = N\)), we can use the binomial expansion for \((I+N)^n\):
\[ (I+N)^n = \sum_{k=0}^{n} \binom{n}{k} I^{n-k} N^k \]Since \(I^m = I\) for any m, and \(N^k = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\) for \(k \ge 2\), the sum simplifies:
\[ A^n = (I+N)^n = \binom{n}{0} I^n N^0 + \binom{n}{1} I^{n-1} N^1 + \binom{n}{2} I^{n-2} N^2 + ... \] \[ A^n = 1 \cdot I \cdot I + n \cdot I \cdot N + 0 + ... \] \[ A^n = I + nN \]Substituting the matrices for I and N:
\[ A^n = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + n \begin{bmatrix} 0 & a \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & na \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & na \\ 0 & 1 \end{bmatrix} \]This confirms the pattern we observed earlier and used in the solution. This method using binomial expansion is a more formal way to derive the formula for \(A^n\) for this specific type of matrix.
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The product of matrices (PQ)–1P is
