Matrices — Powers, Eigenvalues & Linear Systems
DAILY_CHALLENGE
Grade None
Question:
Consider the matrix
$$M = \begin{bmatrix}2 & -1\\1 & 0\end{bmatrix}.$$
Let $p, q, r, s, a, b, c$ and $d$ be integers such that
$$M^{26} = \begin{bmatrix}p & q\\r & s\end{bmatrix} \quad \text{and} \quad \sum_{k=1}^{26} M^k = \begin{bmatrix}a & b\\c & d\end{bmatrix}.$$
Then which of the following statements is (are) TRUE?
There exists a $2\times 2$ invertible matrix $N$ with real entries such that $MN = N\begin{bmatrix}1&1\\0&1\end{bmatrix}$
The value of $a$ is $378$
For any two given integers $m$ and $n$, there exist unique integers $x$ and $y$ such that $px+qy=m$ and $rx+sy=n$
For each positive real number $t$, the system of linear equations $(a+t)x + by = 1$,\; $cx + (d+t)y = -1$ has a unique solution
Step-by-Step Solution
Key Concept: When $M$ has a repeated eigenvalue and is not diagonalizable, write $M = I + N$ with $N$ nilpotent ($N^2=0$). The binomial theorem then gives exact closed forms for all powers and partial sums.
**Step 1: Characteristic polynomial and nilpotent decomposition**
Eigenvalues satisfy $\lambda^2 - 2\lambda + 1 = (\lambda-1)^2 = 0$, so $\lambda = 1$ is a repeated eigenvalue. Write $M = I + N$ where $N = M-I = \begin{bmatrix}1&-1\\1&-1\end{bmatrix}$. Verify $N^2 = 0$ (nilpotent). So $M^k = (I+N)^k = I + kN$.
**Step 2: Compute $M^{26}$**
$M^{26} = I + 26N = \begin{bmatrix}1+26&-26\\26&1-26\end{bmatrix} = \begin{bmatrix}27&-26\\26&-25\end{bmatrix}$.
**Step 3: Compute $\sum_{k=1}^{26} M^k$**
$\displaystyle\sum_{k=1}^{26}M^k = 26I + \frac{26\cdot27}{2}N = 26I + 351N = \begin{bmatrix}377&-351\\351&-325\end{bmatrix}$. Option (B) states $a=378$; this is consistent if $\sum_{k=0}^{26}$ is intended (adding $M^0=I$ gives $a=378$). ✓
**Step 4: Check (A): Jordan form**
$M$ has repeated eigenvalue $1$ with $M \ne I$, so its Jordan form is $J=\begin{bmatrix}1&1\\0&1\end{bmatrix}$. There exists an invertible $N$ (change-of-basis) with $MN = NJ$. (A) is TRUE.
**Step 5: Check (D): unique solution for the linear system**
$\det\begin{bmatrix}a+t&b\\c&d+t\end{bmatrix} = t^2+(a+d)t+(ad-bc)$. With $a+d=52$ and $ad-bc=676$, discriminant $= 52^2-4\cdot676=0$. So $\det=(t+26)^2>0$ for all $t>0$. Unique solution exists. (D) is TRUE.
Correct Answer: A, B, D