If A = \(\begin{bmatrix} 2 & 3 \\ 3 & 5 \end{bmatrix}\) then value of det(\(A^{2025} - 3A^{2024} + A^{2023}\)) :
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The Cayley-Hamilton theorem (\(A^2 - \text{tr}(A)A + \det(A)I = 0\) for a 2x2 matrix) is extremely useful for simplifying matrix polynomials and finding powers of matrices.
Step 1: Understanding the Question:
We are asked to find the determinant of a matrix polynomial. A powerful tool for this kind of problem is the Cayley-Hamilton theorem, which relates a matrix to its characteristic equation.
Step 2: Simplifying the Expression:
First, we can factor out the lowest power of A from the expression inside the determinant.
\[ \det(A^{2025} - 3A^{2024} + A^{2023}) = \det(A^{2023}(A^2 - 3A + I)) \]
Using the property \(\det(XY) = \det(X)\det(Y)\), this becomes:
\[ \det(A^{2023}) \cdot \det(A^2 - 3A + I) = (\det(A))^{2023} \cdot \det(A^2 - 3A + I) \]
Step 3: Applying the Cayley-Hamilton Theorem:
The characteristic equation of a 2x2 matrix A = \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\) is \(\lambda^2 - \text{tr}(A)\lambda + \det(A) = 0\).
For our matrix A, \(\text{tr}(A) = 2+5 = 7\) and \(\det(A) = (2)(5) - (3)(3) = 10 - 9 = 1\).
So the characteristic equation is \(\lambda^2 - 7\lambda + 1 = 0\).
By the Cayley-Hamilton theorem, the matrix A satisfies its own characteristic equation:
\[ A^2 - 7A + I = \mathbf{0} \]
where \(\mathbf{0}\) is the 2x2 zero matrix.
Step 4: Final Calculation:
Now we can simplify the term \(\det(A^2 - 3A + I)\). From the Cayley-Hamilton result, we can write \(A^2 = 7A - I\). However, a more direct approach is:
\[ A^2 - 3A + I = (A^2 - 7A + I) + 4A \]
Since \(A^2 - 7A + I = \mathbf{0}\),
\[ A^2 - 3A + I = \mathbf{0} + 4A = 4A \]
Now, substitute this back into our expression from Step 2:
\[ (\det(A))^{2023} \cdot \det(4A) \]
We know \(\det(A) = 1\). For a 2x2 matrix, \(\det(kA) = k^2\det(A)\).
So, \(\det(4A) = 4^2\det(A) = 16 \cdot 1 = 16\).
The final value is:
\[ (1)^{2023} \cdot 16 = 1 \cdot 16 = 16 \]
