Question

Let matrix \[ A=\begin{pmatrix} 3 & -4\\ 1 & -1 \end{pmatrix} \] and \(A^{100} = 100B + I\). Find the sum of all the elements in \(B^{100}\).

• A. \(-3\)
• B. \(4\)
• C. \(0\)
• D. \(-2\)

Hint:

If a matrix satisfies a quadratic equation, always try rewriting it in the form \((A-I)^2=0\). Nilpotent matrices make large powers extremely easy!

Answer 1

The Correct Option is C

To solve the given problem, we need to analyze the behavior of matrix \( A \) when raised to the power of 100, i.e., \( A^{100} \), and determine the sum of all elements in \( B^{100} \) where \( A^{100} = 100B + I \).

First, find the eigenvalues of the matrix \( A \).

The matrix \( A \) is:

\( 3 \)	\(-4\)
\( 1 \)	\(-1\)

The characteristic equation of \( A \) is given by:

\(\det(A - \lambda I) = 0\)

Where the identity matrix \( I \) is:

\( 1 \)	\( 0 \)
\( 0 \)	\( 1 \)

The determinant of \( A - \lambda I \) is:

\(\det\begin{pmatrix} 3-\lambda & -4 \\ 1 & -1-\lambda \end{pmatrix} = 0\)

Simplifying the above determinant:

\((3-\lambda)(-1-\lambda) - (-4)(1) = \lambda^2 - 2\lambda + 1 = 0\)

This factors to:

\((\lambda - 1)^2 = 0\)

Therefore, the eigenvalue \( \lambda = 1 \) with algebraic multiplicity 2.

For matrices with eigenvalues all equal to 1 and size 2x2, the powers of the matrices simplify. Specifically, \( A^n \) approaches a scalar multiple of \( I \) plus linear combinations of \( A \) and its infinitesimal changes due to the Jordan form.

Given \( A^{100} = 100B + I \), rearrange as:

\(100B = A^{100} - I\)

Since \( A \) approaches a matrix similar to the identity matrix as \( n \to \infty \) and because eigenvalue is 1:

\(A^{100} \approx I \Rightarrow 100B \approx 0 \Rightarrow B \approx 0\)

Thus, \( B^{100} \approx 0^{100} = 0 \), leading to:

The sum of all elements in \( B^{100} \) is \(0.\)