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:
- 0
- I
- 2I
- 4I
The Correct Option is C
Solution and Explanation
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 \(2I\).
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