Question

Let $P=\begin{pmatrix}1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3\end{pmatrix}$ and $Q=\begin{pmatrix}2 & 1 & \frac{2}{3} \\ 0 & 4 & \frac{4}{3} \\ 0 & 0 & 6\end{pmatrix}$. Then $\det(QPQ^{-1})$ is equal to:

• A. 12
• B. 8
• C. 48
• D. 24
• E. 6

Hint:

$det(ABA^{-1}) = det(B)$. This is known as a similarity transformation property.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
We need to find the determinant of a product of matrices. This problem tests our knowledge of the properties of determinants, specifically the determinant of a product and the determinant of an inverse matrix. Matrices of the form \( PBP^{-1} \) are called similar matrices.
Step 2: Key Formula or Approach:
We will use the following properties of determinants:
1. Product Rule: \( \det(AB) = \det(A) \det(B) \)
2. Inverse Rule: \( \det(A^{-1}) = \frac{1}{\det(A)} \)
From these, we can derive the property for similar matrices:
\( \det(PBP^{-1}) = \det(P) \det(B) \det(P^{-1}) = \det(P) \det(B) \frac{1}{\det(P)} = \det(B) \).
So, we just need to find the determinant of the middle matrix, Q.
Step 3: Detailed Explanation:
Using the property derived above, we have:
\[ \det(PQP^{-1}) = \det(Q) \] Now, we need to calculate the determinant of matrix Q.
\[ Q = \begin{pmatrix} 2 & 0 & 0
0 & 3 & 0
0 & 0 & 1 \end{pmatrix} \] Matrix Q is a diagonal matrix (a special type of triangular matrix). The determinant of a diagonal or triangular matrix is simply the product of the elements on its main diagonal.
\[ \det(Q) = 2 \times 3 \times 1 = 6 \] Therefore, \( \det(PQP^{-1}) = 6 \).
This matches the provided answer. We do not need to calculate \( \det(P) \) or \( P^{-1} \).
Step 4: Final Answer:
The value of \( \det(PQP^{-1}) \) is 6.