Matrices & Determinants
Matrix powers and determinants
Grade 12
Question:
<p>Let \(P = \begin{bmatrix} 1 & n & 0 \\ 0 & 1 & 0 \\ 0 & n & 1 \end{bmatrix}\) and \(B^n = QP^nQ^T\). Which of the following are correct?</p>
<p>(a) det(<em>A</em>) = 1</p>
<p>(b) det(<em>B</em>) = 1</p>
<p>(c) \(P^n = \begin{bmatrix} 1 & n & 0 \\ 0 & 1 & 0 \\ 0 & n & 1 \end{bmatrix}\)</p>
<p>(d) \(B^n = QP^nQ^T\)</p>
Step-by-Step Solution
Key Concept: Recognize that P is a special matrix where diagonalization or direct computation of P^n yields a pattern; the key is computing P^n directly by observing that P = I + N where N is nilpotent, so P^n follows a binomial expansion pattern.
**Step 1:** Write $P$ in decomposed form:
$$P = I + N \text{ where } N = \begin{bmatrix} 0 & n & 0 \\ 0 & 0 & 0 \\ 0 & n & 0 \end{bmatrix}$$
(the non-identity part)
**Step 2:** Check nilpotency: $N^2 = 0$ (since the product of columns of $N$ results in the zero matrix)
**Step 3:** Apply binomial expansion:
$$P^n = (I + N)^n = I + nN$$
(since $N^2 = 0$, all higher powers vanish)
$$P^n = \begin{bmatrix} 1 & n^2 & 0 \\ 0 & 1 & 0 \\ 0 & n^2 & 1 \end{bmatrix}$$
**Step 4:** Verify determinants:
- $\det(P) = 1$ (upper triangular with diagonal entries all equal to 1)
- $\det(P^n) = 1$ (same reason)
- Since $B^n = QP^nQ^T$, we have $\det(B^n) = \det(Q)\det(P^n)\det(Q^T) = \det(P^n) = 1$
**Step 5:** The relation $B^n = QP^nQ^T$ represents a similarity transformation. Verify each option against the computed form of $P^n$.
**Answer:** Options (a), (b), and (d) are correct.
Correct Answer: a,b,c,d