Matrices & Determinants Question - None

Question

If $A = \begin{bmatrix} 1 & 2 \ 2 & 1 \end{bmatrix}$ and $f(x) = \dfrac{1+x}{1-x}$, then $f(A)$ is

Accepted Answer

Step 1: Find $I - A$

$I - A = \begin{bmatrix} 1-1 & 0-2 \ 0-2 & 1-1 \end{bmatrix} = \begin{bmatrix} 0 & -2 \ -2 & 0 \end{bmatrix}$

Step 2: Find $(I - A)^{-1}$

$\det(I - A) = 0(0) - (-2)(-2) = -4$

$(I - A)^{-1} = -\dfrac{1}{4} \begin{bmatrix} 0 & 2 \ 2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -\dfrac{1}{2} \ -\dfrac{1}{2} & 0 \end{bmatrix}$

Step 3: Find $I + A$

$I + A = \begin{bmatrix} 2 & 2 \ 2 & 2 \end{bmatrix}$

Step 4: Compute $f(A) = (I + A)(I - A)^{-1}$

$f(A) = \begin{bmatrix} 2 & 2 \ 2 & 2 \end{bmatrix} \begin{bmatrix} 0 & -\dfrac{1}{2} \ -\dfrac{1}{2} & 0 \end{bmatrix}$

$= \begin{bmatrix} 2(0) + 2(-\dfrac{1}{2}) & 2(-\dfrac{1}{2}) + 2(0) \ 2(0) + 2(-\dfrac{1}{2}) & 2(-\dfrac{1}{2}) + 2(0) \end{bmatrix}$

$= \begin{bmatrix} -1 & -1 \ -1 & -1 \end{bmatrix}$

$ herefore$ Answer: C

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