Discussion
Let \(P = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{pmatrix}\) and \(Q = \begin{pmatrix} 2 & 1 & 2/3 \\ 0 & 4 & 4/3 \\ 0 & 0 & 6 \end{pmatrix}\). Then the \(\det(QPQ^{-1})\) is equal to
Answer: E
The determinant is 6.
Determinants multiply, and \(\det(Q^{-1}) = \frac{1}{\det Q}\).
\[ \begin{aligned} \det(QPQ^{-1}) &= \det Q \cdot \det P \cdot \frac{1}{\det Q} \\ &= \det P \end{aligned} \]
P is upper triangular, so its determinant is the product of the diagonal: \(1 \times 2 \times 3 = 6\). Q never needs to be inverted. 48 is det Q, and 24, 12 or 8 come from multiplying in parts of det Q that cancel.
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