Review : Matrices & Vectors — Question 10

Question 10 – Solution

Strategy. The powers of the strictly upper triangular part terminate. A matrix update is simultaneous unless a different order is explicitly encoded.

Step 1: Exploit the terminating powers. Multiplication gives $N^2=\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}$ and $N^3=0$. The upper triangular matrix $A$ has determinant $1$. Moreover $$\boxed{A^{-1}=I_3-N+N^2}$$ because $(I_3+N)(I_3-N+N^2)=I_3+N^3=I_3$; the reverse product is the same.

Step 2: Derive the finite power formula. The identity matrix commutes with $N$, so the binomial expansion terminates: $$\boxed{A^m=I_3+mN+\frac{m(m-1)}2N^2.}$$ Alternatively, the formula is true at $m=0$. Multiplication by $I_3+N$ increases the coefficient of $N$ to $m+1$ and that of $N^2$ to $m(m-1)/2+m=m(m+1)/2$, establishing the induction step.

Step 3: Find the exact iterates. Applying the power formula to $v_0$ gives $$\boxed{v_m=\bigl(m(m-1)/2,\ m,\ 1\bigr)^T.}$$ The first coordinate grows quadratically, the second linearly and the third stays fixed. Determinant $1$ preserves oriented volumes of three-dimensional sets, not individual lengths or boundedness under repetition. The plot marks integer-time values only; coincident coordinate values share a location.

Step 4: Diagnose the sequential-update error. The program computes $y_{\mathrm{new}}=y+z$ and then $x_{\mathrm{new}}=x+y+z$. It actually applies $$B=\begin{pmatrix}1&1&1\\0&1&1\\0&0&1\end{pmatrix}=I_3+N+N^2.$$ Starting from $v_0$, its third coordinate remains $1$, its second becomes $m$, and its first accumulates $1+2+\cdots+m=m(m+1)/2$. After two steps the correct vector is $(1,2,1)^T$, while the program gives $(3,2,1)^T$. Both matrices are invertible with determinant $1$, so checking only the determinant would not expose this order-of-update mistake.