Let \(A = \begin{bmatrix}1&2\\-1&-2\end{bmatrix}\) and let I be the identity matrix of order 2. Consider the following statements:
(A) \(A^2 = -A\)
(B) A is an invertible matrix.
(C) \(A^{10} = -A\)
(D) \((I + A)^2 = I + A\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (C) and (D) only
Squaring A row by column:
\( \displaystyle A^2 = \begin{bmatrix}1 - 2&2 - 4\\-1 + 2&-2 + 4\end{bmatrix} = \begin{bmatrix}-1&-2\\1&2\end{bmatrix} = -A \)
Each further multiplication by A changes the sign:
\(A^3 = A^2A = -A^2 = A\)
\(A^4 = A^3A = A^2 = -A\)
so every even power of A equals \(-A\). Also
\((I + A)^2 = I + 2A + A^2 = I + 2A - A = I + A\)
For a \(2 \times 2\) matrix to be invertible, \(ad - bc\) must be non-zero; here
\(ad - bc = 1(-2) - 2(-1) = 0\)
So \(A^2 = -A\), \(A^{10} = -A\), \((I + A)^2 = I + A\), and \(ad - bc = 0\), so A has no inverse.
Checking the statements
(A) \(A^2 = -A\) — Correct. \(A^2 = \begin{bmatrix}-1&-2\\1&2\end{bmatrix} = -A\).
(B) A is an invertible matrix. — Incorrect. \(ad - bc = 0\); also, if \(A^{-1}\) existed, \(A^2 = -A\) would give \(A = -I\), which is false.
(C) \(A^{10} = -A\) — Correct. 10 is even, and every even power of A is \(-A\).
(D) \((I + A)^2 = I + A\) — Correct. \(I + 2A + A^2 = I + 2A - A = I + A\).