If $A^2=A$, then simplify $(I+A)^2 - 7A$, where $A$ is a square matrix.

Question

Let the foot of the perpendicular from the point $ (1, 2, 4) $ on the line $ \frac{x+2}{4} = \frac{y-1}{2} = \frac{z+1}{3} $ be $ P $. Then the distance of $ P $ from the plane $ 3x + 4y + 12z + 23 = 0 $ is:

Answer 1

Step 1: Write out the expansion term by term without skipping any cross term:
$(I+A)(I+A) = I\cdot I + I\cdot A + A\cdot I + A\cdot A$.

Step 2: Simplify each of the four terms individually:
$I\cdot I = I$; $I\cdot A = A$; $A\cdot I = A$; $A\cdot A = A^2 = A$ (given). Adding all four: $I+A+A+A = I+3A$.

Step 3: Now subtract $7A$ from this result:
$(I+A)^2-7A = (I+3A)-7A = I-4A$.

Final Answer:
The simplified form is $I-4A$. \[ \boxed{I - 4A} \]

If \(A^2=A\), then simplify \((I+A)^2 - 7A\), where \(A\) is a square matrix.

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Solution and Explanation

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