If \(A = \begin{bmatrix}0&2&0\\0&0&3\\0&0&0\end{bmatrix}\) and I is the identity matrix of order 3, then \((I - A)^{-1}\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\begin{bmatrix}1&2&6\\0&1&3\\0&0&1\end{bmatrix}\)
The powers of A are
\( \displaystyle A^2 = \begin{bmatrix}0&0&6\\0&0&0\\0&0&0\end{bmatrix} \)
\(A^3 = A^2 A = O\)
Since I commutes with A, the product
\((I - A)(I + A + A^2) = I + A + A^2 - A - A^2 - A^3\)
\((I - A)(I + A + A^2) = I - A^3 = I\)
In the same way \((I + A + A^2)(I - A) = I\). So by the uniqueness of the inverse,
\( \displaystyle (I - A)^{-1} = I + A + A^2 = \begin{bmatrix}1&2&6\\0&1&3\\0&0&1\end{bmatrix} \)
Why the other options are incorrect
Option (1) \(\begin{bmatrix}1&2&0\\0&1&3\\0&0&1\end{bmatrix}\) — Incorrect. This is \(I + A\); the term \(A^2\) is left out, and \((I - A)(I + A) = I - A^2 \ne I\).
Option (2) \(\begin{bmatrix}1&-2&6\\0&1&-3\\0&0&1\end{bmatrix}\) — Incorrect. This is \(I - A + A^2\), the inverse of \(I + A\), not of \(I - A\).
Option (3) \(\begin{bmatrix}1&-2&-6\\0&1&-3\\0&0&1\end{bmatrix}\) — Incorrect. This is \(I - A - A^2\); its product with \(I - A\) has first row \(\begin{bmatrix}1&-4&0\end{bmatrix}\).