If A is an invertible square matrix of order 3 such that \(A^2 = 2A\), then A is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2I
Since A is invertible, \(A^{-1}\) exists. Multiplying both sides of \(A^2 = 2A\) on the left by \(A^{-1}\):
\(A^{-1}(AA) = A^{-1}(2A)\)
\((A^{-1}A)A = 2(A^{-1}A)\)
\(IA = 2I\)
\(A = 2I\)
Check: \((2I)^2 = 4I = 2(2I)\), and 2I is invertible with inverse \(\displaystyle\frac{1}{2}I\).
Why the other options are incorrect
Option (1) O, the zero matrix of order 3 — Incorrect. O satisfies \(A^2 = 2A\), but it is not invertible, since \(OX = O \ne I\) for every X.
Option (2) I, the identity matrix of order 3 — Incorrect. \(I^2 = I\), which is not 2I.
Option (4) \(\displaystyle\frac{1}{2}I\) — Incorrect. Its square is \(\displaystyle\frac{1}{4}I\), while \(2\left(\displaystyle\frac{1}{2}I\right) = I\).