Question:
Let $A$ be a matrix such that $A^2 = I$, where $I$ is an identity matrix. Then $(I + A)^4 - 8A$ is equal to:
Let $A$ be a matrix such that $A^2 = I$, where $I$ is an identity matrix. Then $(I + A)^4 - 8A$ is equal to:
Updated On: Mar 27, 2025
$8 I$
$5 I$
- 8(I + A)
- 5(I - A)
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The Correct Option is A
Solution and Explanation
Given \(A^2 = I\), the identity matrix, we analyze \((I + A)^4\):
\[(I + A)^2 = I^2 + 2IA + A^2 = I + 2A + I = 2I + 2A\]
\[(I + A)^4 = ((I + A)^2)^2 = (2I + 2A)^2\]
Expanding \((2I + 2A)^2\):
\[(2I + 2A)^2 = 4I^2 + 8IA + 4A^2\]
Since \(A^2 = I\), substitute \(4A^2 = 4I\):
\[(2I + 2A)^2 = 4I + 8A + 4I = 8I + 8A\]
Now compute \((I + A)^4 - 8A\):
\[(I + A)^4 - 8A = (8I + 8A) - 8A = 8I\]
Thus, the result is: \[8I\]
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