Question

If, A is a square matrix of order \(3 \times 3\) such that \(A^2 = A\) and I is the unit matrix of order \(3 \times 3\), then the value of \((I-A)^3+A^2+I\) is:

• A. 0
• B. I
• C. 2I
• D. 4I

Answer 1

The Correct Option is C

Given that A is a square matrix of order \(3 \times 3\) such that \(A^2 = A\) (meaning A is an idempotent matrix) and I is the identity matrix of the same order, we need to find the value of \((I-A)^3 + A^2 + I\).

First, note the following relations:

• \(A^2 = A\), so we can substitute \(A\) wherever \(A^2\) appears.
• \((I-A)\) is also a matrix of the same order.

To solve \((I-A)^3 + A^2 + I\), we expand \((I-A)^3\) as follows:

\((I-A)^3 = (I-A)(I-A)(I-A)\).

Expanding further:

• \((I-A)^2 = (I - A)(I - A) = I - A - A + A^2 = I - 2A + A^2\).
• Since \(A^2 = A\), the expression becomes \(I - 2A + A\).
• The simplification gives \(I - A\).

Continuing with \((I-A)^3 = (I-A)(I-A)^2 = (I-A)(I - A)\), we apply the expansion again:

\((I-A)(I - A) = I - A - A + A = I - A\).

Thus, \((I-A)^3 = I - A\).

Substituting back into the main expression:

\((I-A)^3 + A^2 + I = (I-A) + A + I = I - A + A + I = 2I\).

Therefore, the value of the expression is \(2I\).