Question

Let \( P = \begin{pmatrix} 1 & 2 \\ -1 & 4 \end{pmatrix} \) and \( Q = P^3 - 2P^2 - 4P + 13I_2 \), where \( I_2 \) denotes the identity matrix of order 2. Then the determinant of \( Q \) is equal to ________ (answer in integer).

Hint:

When dealing with matrix operations, carefully apply the basic operations and simplify the resulting matrices step by step.

Answer 1

Step 1: Characteristic Polynomial of \( P \)
The characteristic polynomial is found by: \[ \det(P - \lambda I) = \begin{vmatrix} 1 - \lambda & 2 \\ -1 & 4 - \lambda \end{vmatrix} = (1 - \lambda)(4 - \lambda) + 2 = \lambda^2 - 5\lambda + 6 \] Thus, the characteristic equation is: \[ P^2 - 5P + 6I = 0 \] Step 2: Express \( P^3 \) and \( P^2 \) in Terms of \( P \) and \( I \)
Using the characteristic equation: \[ P^2 = 5P - 6I \] Multiply by \( P \) to get \( P^3 \): \[ P^3 = 5P^2 - 6P = 5(5P - 6I) - 6P = 19P - 30I \] Step 3: Simplify \( Q \)
Substitute \( P^3 \) and \( P^2 \) into \( Q \): \[ Q = P^3 - 2P^2 - 4P + 13I = (19P - 30I) - 2(5P - 6I) - 4P + 13I \] Simplify: \[ Q = 5P - 5I = 5(P - I) \] Step 4: Compute Determinant of \( Q \)
First, find \( P - I \): \[ P - I = \begin{pmatrix} 0 & 2 \\ -1 & 3 \end{pmatrix} \] Then: \[ \det(P - I) = 0 \cdot 3 - 2 \cdot (-1) = 2 \] Finally: \[ \det(Q) = \det(5(P - I)) = 5^2 \cdot \det(P - I) = 25 \times 2 = 50 \] Verification
Compute \( Q \) directly: \[ Q = \begin{pmatrix} 0 & 10 \\ -5 & 15 \end{pmatrix} \implies \det(Q) = 0 \cdot 15 - 10 \cdot (-5) = 50 \] Final Answer
The determinant of \( Q \) is \(\boxed{50}\).