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

Step 1: Find a simple property of matrix \(A\)
Compute \(A^2\): \[ A^2 = \begin{pmatrix} 3 & -4\\ 1 & -1 \end{pmatrix} \begin{pmatrix} 3 & -4\\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 5 & -8\\ 2 & -3 \end{pmatrix} \] Now observe: \[ A^2 - A = \begin{pmatrix} 2 & -4\\ 1 & -2 \end{pmatrix} = A - I \] Hence, \[ A^2 - 2A + I = 0 \]
Step 2: Use the identity
\[ A^2 - 2A + I = (A-I)^2 = 0 \] Thus, \(A-I\) is a nilpotent matrix of index 2.
Step 3: Find \(A^{100}\)
Write: \[ A = I + (A-I) \] Then, \[ A^{100} = \left[I + (A-I)\right]^{100} \] Since \((A-I)^2 = 0\), all higher powers vanish: \[ A^{100} = I + 100(A-I) \] \[ A^{100} = 100A - 99I \]
Step 4: Compare with given expression
Given: \[ A^{100} = 100B + I \] Comparing, \[ 100B + I = 100A - 99I \] \[ 100B = 100A - 100I \Rightarrow B = A - I \]
Step 5: Find \(B^{100}\)
Since \(B = A-I\) and \((A-I)^2 = 0\), \[ B^2 = 0 \Rightarrow B^{100} = 0 \]
Step 6: Sum of all elements of \(B^{100}\)
All entries are zero. \[ \boxed{0} \]