If \(A = \begin{bmatrix}3&9\\-1&-3\end{bmatrix}\), then \(I - A + A^2 - A^3 + \dots\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\begin{bmatrix}-2&-9\\1&4\end{bmatrix}\)
The series stops early if some power of A is the zero matrix. The elements of \(A^2\) are
\(3(3) + 9(-1) = 0,\quad 3(9) + 9(-3) = 0\)
\((-1)(3) + (-3)(-1) = 0,\quad (-1)(9) + (-3)(-3) = 0\)
So \(A^2 = O\), and then \(A^3 = A^2A = O\), and every higher power is O. The series reduces to
\(I - A + A^2 - A^3 + \dots = I - A\)
\( \displaystyle I - A = \begin{bmatrix}1 - 3&0 - 9\\0 + 1&1 + 3\end{bmatrix} = \begin{bmatrix}-2&-9\\1&4\end{bmatrix} \)