The number of diagonal matrices A of order 3, with real elements, such that \(A^2 = I\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 8
Let \(A = \begin{bmatrix}a&0&0\\0&b&0\\0&0&c\end{bmatrix}\). The square of a diagonal matrix is the diagonal matrix of the squares of its diagonal elements:
\( \displaystyle A^2 = \begin{bmatrix}a^2&0&0\\0&b^2&0\\0&0&c^2\end{bmatrix} \)
\(A^2 = I\) gives
\(a^2 = 1,\ b^2 = 1,\ c^2 = 1\)
\(a = \pm 1,\ b = \pm 1,\ c = \pm 1\)
Each of a, b and c has 2 choices, so the number of such matrices is
\(2 \times 2 \times 2 = 8\)
(They include I and −I, and six others such as \(\begin{bmatrix}1&0&0\\0&-1&0\\0&0&1\end{bmatrix}\).)