Let $A$ be a matrix such that $A^2 = I$, where $I$ is an identity matrix. Then $(I + A)^4 - 8A$ is equal to:

Question

For the matrices $ A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} $ and $ B = \begin{bmatrix} -29 & 49 \\ -13 & 18 \end{bmatrix} $, if $ (A^{15} + B) \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix} $, then among the following which one is true?}

Answer 1

Given the matrix equation $A^2 = I$, which defines $A$ as an involutory matrix, we need to calculate $(I + A)^4 - 8A$. The calculation proceeds as follows:

First, simplify $(I + A)^2$: $(I + A)^2 = (I + A)(I + A) = I^2 + IA + AI + A^2 = I + A + A + I = 2I + 2A$, using $A^2 = I$.

Next, compute $(I + A)^4$ using the previous result: $(I + A)^4 = ((I + A)^2)^2 = (2I + 2A)^2$.

Expand $(2I + 2A)^2$: $(2I + 2A)^2 = (2(I + A))(2(I + A)) = 4(I + A)^2 = 4(2I + 2A) = 8I + 8A$.

Finally, substitute this into the expression $(I + A)^4 - 8A$: $(I + A)^4 - 8A = (8I + 8A) - 8A = 8I$.

Therefore, the expression $(I + A)^4 - 8A$ simplifies to $8I$.

The result is $8I$.

Let $A$ be a matrix such that $A^2 = I$, where $I$ is an identity matrix. Then $(I + A)^4 - 8A$ is equal to:

The Correct Option is A

Solution and Explanation

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