(34) If $$ \mathrm{A}=\left(\begin{array}{ll}1 & 0 \ 1 & 1\end{array} ight) $$, then the multiplicative inverse of $$ \mathrm{A}^{n} $$ where $$ \mathrm{n} \in \mathbb{Z}^{+} $$is $$ \begin{array}{llll} ext { (a) }\left(\begin{array}{cc}-1 & 0 \ \mathrm{n} & 1\end{array} ight) & ext { (b) }\left(\begin{array}{rr}0 & -1 \ 1 & \mathrm{n}\end{array} ight) & ext { (c) }\left(\begin{array}{ll}1 & 0 \ -\mathrm{n} & 1\end{array} ight) & ext { (d) }\left(\begin{array}{ll}0 & 1 \ 1 & \mathrm{n}\end{array} ight)\end{array} $$

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(34) If $$ \mathrm{A}=\left(\begin{array}{ll}1 & 0 \ 1 & 1\end{array}ight) $$, then the multiplicative inverse of $$ \mathrm{A}^{n} $$ where $$ \mathrm{n} \in \mathbb{Z}^{+} $$is $$ \begin{array}{llll}	ext { (a) }\left(\begin{array}{cc}-1 & 0 \ \mathrm{n} & 1\end{array}ight) & 	ext { (b) }\left(\begin{array}{rr}0 & -1 \ 1 & \mathrm{n}\end{array}ight) & 	ext { (c) }\left(\begin{array}{ll}1 & 0 \ -\mathrm{n} & 1\end{array}ight) & 	ext { (d) }\left(\begin{array}{ll}0 & 1 \ 1 & \mathrm{n}\end{array}ight)\end{array} $$

UpStudy AI · Accepted Answer

We start with the matrix
$$
\mathrm{A}=\begin{pmatrix} 1 & 0 \ 1 & 1 \end{pmatrix}.
$$
It is known that powers of this matrix have the form
$$
\mathrm{A}^n=\begin{pmatrix} 1 & 0 \ n & 1 \end{pmatrix},
$$
since each multiplication adds one more $$1$$ in the lower left entry.

Let
$$
\mathrm{B}=\begin{pmatrix} a & b \ c & d \end{pmatrix}
$$
be the inverse of $$\mathrm{A}^n$$, so that
$$
\mathrm{A}^n\,\mathrm{B}=\begin{pmatrix} 1 & 0 \ n & 1 \end{pmatrix}\begin{pmatrix} a & b \ c & d \end{pmatrix}=\begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}.
$$
Multiplying the matrices gives:
$$
\begin{pmatrix} 1\cdot a+0\cdot c & 1\cdot b+0\cdot d \ n\cdot a+1\cdot c & n\cdot b+1\cdot d \end{pmatrix}=\begin{pmatrix} a & b \ na+c & nb+d \end{pmatrix}.
$$
Setting the product equal to the identity matrix results in the system:
$$
\begin{cases}
a=1,\[1mm]
b=0,\[1mm]
na+c=0,\[1mm]
nb+d=1.
\end{cases}
$$
Substituting $$a=1$$ into the third equation gives:
$$
n\cdot 1+c=0 \quad\Longrightarrow\quad c=-n.
$$
Substituting $$b=0$$ into the fourth equation gives:
$$
n\cdot 0+d=1 \quad\Longrightarrow\quad d=1.
$$
Thus, the inverse of $$\mathrm{A}^n$$ is
$$
\mathrm{B}=\begin{pmatrix} 1 & 0 \ -n & 1 \end{pmatrix}.
$$
This matches option (c).
The inverse of $$ \mathrm{A}^{n} $$ is $$ \begin{pmatrix} 1 & 0 \ -n & 1 \end{pmatrix} $$, which corresponds to option (c).

(34) If , then the multiplicative inverse of where is

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