Question

For some \(\alpha, \beta \in R\), let \(A = \begin{pmatrix} \alpha & 2 \\1 & 2 \end{pmatrix}\) and \(B = \begin{pmatrix} 1 & 1 \\ \beta & 1 \end{pmatrix}\) be such that \(A^2-4A+2I-B^2-3B+I=O\). Then \((\det(\text{adj}(A^3-B^3)))^2\) is equal to:

Hint:

When faced with a question that seems to have a typo or contradiction, it's important to show why it's flawed.
In an exam where a specific numerical answer is expected, try to deduce the simplest possible intended question that would lead to such an answer.
Working backwards from the answer is a valid strategy in these rare cases.

Answer 1

Correct Answer: 50625

Step 1: Understanding the Question:
We are given a matrix equation that relates two 2x2 matrices, A and B, which depend on parameters \(\alpha\) and \(\beta\).
Our goal is to first find \(\alpha\) and \(\beta\), then compute the matrix \(M = A^3 - B^3\), and finally calculate the value of \((\det(\text{adj}(M)))^2\).
Step 2: Key Formula or Approach:
1. Explicitly compute the matrices in the given equation \(A^2 - 4A + 3I = B^2 + 3B\).
2. Equate the elements of the resulting matrices to find \(\alpha\) and \(\beta\).
3. Use the property \(\det(\text{adj}(M)) = (\det(M))^{n-1}\), where n is the order of the square matrix M. For n=2, this simplifies to \(\det(\text{adj}(M)) = \det(M)\).
4. Therefore, the value to be found is \((\det(A^3-B^3))^2\).
Note: As will be shown, the question as stated in the OCR leads to a mathematical contradiction. This is a common issue in exam papers. The solution will proceed by assuming a plausible intended problem that leads to the provided answer.
Step 3: Detailed Explanation:
Analysis of the given equation
The equation is \(A^2-4A+3I = B^2+3B\).
LHS: \(A^2 = \begin{pmatrix} \alpha^2+2 & 2\alpha+4 \\ \alpha+2 & 6 \end{pmatrix}\).
LHS = \(A^2-4A+3I = \begin{pmatrix} \alpha^2+2-4\alpha+3 & 2\alpha+4-8 \\ \alpha+2-4 & 6-8+3 \end{pmatrix} = \begin{pmatrix} \alpha^2-4\alpha+5 & 2\alpha-4 \\ \alpha-2 & 1 \end{pmatrix}\).
RHS: \(B^2 = \begin{pmatrix} 1+\beta & 2 \\ 2\beta & \beta+1 \end{pmatrix}\).
RHS = \(B^2+3B = \begin{pmatrix} 1+\beta+3 & 2+3 \\ 2\beta+3\beta & \beta+1+3 \end{pmatrix} = \begin{pmatrix} \beta+4 & 5 \\ 5\beta & \beta+4 \end{pmatrix}\).
Equating LHS and RHS:
From element (2,2): \(1 = \beta+4 \implies \beta = -3\).
From element (1,2): \(2\alpha-4 = 5 \implies 2\alpha=9 \implies \alpha=9/2\).
Check for consistency with element (2,1):
LHS: \(\alpha-2 = 9/2 - 2 = 5/2\).
RHS: \(5\beta = 5(-3) = -15\).
Since \(5/2 \neq -15\), the problem statement is contradictory.
Assuming an Intended Problem
Given that the answer is 50625, let's work backward. We need \((\det(A^3-B^3))^2 = 50625\), which implies \(\det(A^3-B^3) = \pm 225\).
A simple way this could happen is if the matrix \(A^3 - B^3\) was a scalar multiple of the identity matrix, say \(A^3 - B^3 = kI\).
Then \(\det(A^3-B^3) = \det(kI) = k^2\).
We would need \(k^2 = \pm 225\). For a real \(k\), we must have \(k^2 = 225\), so \(k=\pm 15\).
Let's assume the intended (but mistyped) problem led to the result \(A^3 - B^3 = 15I\).
Under this assumption, we calculate the required value.
Let \(M = A^3 - B^3 = 15I = \begin{pmatrix} 15 & 0 \\ 0 & 15 \end{pmatrix}\).
The determinant is \(\det(M) = 15 \times 15 = 225\).
We need to find \((\det(\text{adj}(M)))^2\).
Using the property \(\det(\text{adj}(M)) = \det(M)\) for a 2x2 matrix:
\[ \det(\text{adj}(M)) = 225.
\] The required value is:
\[ (225)^2 = 50625.
\] Step 4: Final Answer:
The question statement is flawed as it leads to a contradiction. However, by assuming the problem was intended to result in \(A^3-B^3 = 15I\), we arrive at the given answer of 50625.