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: Concept
$det(QPQ^{-1}) = det(Q) \cdot det(P) \cdot det(Q^{-1}) = det(Q) \cdot det(P) \cdot \frac{1}{det(Q)} = det(P)$.

Step 2: Analysis
$P$ is an upper triangular matrix. The determinant of a triangular matrix is the product of its diagonal elements.

Step 3: Calculation
$det(P) = 1 \cdot 2 \cdot 3 = 6$. Final Answer: (E)