$$ \begin{array}{llll} ext { (a) }-A & ext { (c) }-A^{\prime} \ ext { (33) If } A ext { and } B ext { are two matrices of order } 2 imes 2 ext { and }|A B|=9,\left|A B^{-1} ight|=4\end{array} $$ then $$ |A|+|B| $$ could be equal to $$ \begin{array}{llll} ext { (a) } 5 & ext { (b) } 6 & ext { (c) } \frac{13}{2} & ext { (b) } \frac{15}{2}\end{array} $$ 4) If $$ A=\left(\begin{array}{ll}1 & 0 \ 1 & 1\end{array} ight) $$, then the multiplicative inverse of $$ A^{n} $$ where $$ n \in \mathbb{Z}^{+} $$is $$ \ldots \ldots . $$. (a) $$ \left(\begin{array}{cc}-1 & 0 \ n & 1\end{array} ight) $$ If $$ A $$ is a square matrix,$$ |A|
eq 0 $$ and $$ A^{2}=I $$, then $$ A^{-1}=\ldots \ldots \ldots $$.

Question

$$ \begin{array}{llll}	ext { (a) }-A & 	ext { (c) }-A^{\prime} \ 	ext { (33) If } A 	ext { and } B 	ext { are two matrices of order } 2 	imes 2 	ext { and }|A B|=9,\left|A B^{-1}ight|=4\end{array} $$ then $$ |A|+|B| $$ could be equal to $$ \begin{array}{llll}	ext { (a) } 5 & 	ext { (b) } 6 & 	ext { (c) } \frac{13}{2} & 	ext { (b) } \frac{15}{2}\end{array} $$ 4) If $$ A=\left(\begin{array}{ll}1 & 0 \ 1 & 1\end{array}ight) $$, then the multiplicative inverse of $$ A^{n} $$ where $$ n \in \mathbb{Z}^{+} $$is $$ \ldots \ldots . $$. (a) $$ \left(\begin{array}{cc}-1 & 0 \ n & 1\end{array}ight) $$ If $$ A $$ is a square matrix,$$ |A| 
eq 0 $$ and $$ A^{2}=I $$, then $$ A^{-1}=\ldots \ldots \ldots $$.

UpStudy AI · Accepted Answer

We are given several parts to solve.

1) We are told that for two $$2 \times 2$$ matrices $$A$$ and $$B$$,
$$
|AB|=9 \quad \text{and} \quad \left|AB^{-1}\right| = 4.
$$
Recall that for determinants the following properties hold:
$$
|AB| = |A|\,|B| \quad \text{and} \quad \left|AB^{-1}\right| = |A|\,|B^{-1}| = \frac{|A|}{|B|}.
$$
Let $$x = |A|$$ and $$y = |B|$$. Then we have:
$$
xy = 9 \quad \text{and} \quad \frac{x}{y} = 4.
$$
From the second equation, we have $$x = 4y$$. Substitute this into the first equation:
$$
(4y)y = 9 \implies 4y^2 = 9 \implies y^2 = \frac{9}{4} \implies y = \frac{3}{2}.
$$
Then
$$
x=4y = 4\cdot\frac{3}{2} = 6.
$$
So,
$$
|A| + |B| = 6 + \frac{3}{2} = \frac{12}{2}+\frac{3}{2} = \frac{15}{2}.
$$

2) We are given
$$
A=\begin{pmatrix}1 & 0 \ 1 & 1\end{pmatrix}.
$$
Since $$A$$ has the form $$I+N$$ with the nilpotent matrix
$$
N=\begin{pmatrix}0 & 0 \ 1 & 0\end{pmatrix} \quad \text{and} \quad N^2=0,
$$
we have
$$
A^n = I + nN = \begin{pmatrix}1 & 0 \ n & 1\end{pmatrix}.
$$
To find the inverse of $$A^n$$, we look for a matrix of the form
$$
(A^n)^{-1} = \begin{pmatrix}1 & 0 \ -n & 1\end{pmatrix}.
$$
It is easy to verify by direct multiplication:
$$
\begin{pmatrix}1 & 0 \ n & 1\end{pmatrix}\begin{pmatrix}1 & 0 \ -n & 1\end{pmatrix} 
=\begin{pmatrix}1 & 0 \ n - n & 1\end{pmatrix} 
=\begin{pmatrix}1 & 0 \ 0 & 1\end{pmatrix}.
$$

3) Finally, if $$A$$ is a square matrix with $$|A|\neq 0$$ and
$$
A^2 = I,
$$
then multiplying both sides of $$A^2 = I$$ on the right by $$A^{-1}$$ (which exists since $$|A|\neq 0$$) gives:
$$
A = A^{-1}.
$$

Thus, the answers are:

1) $$|A|+|B| = \frac{15}{2}$$.

2) $$(A^n)^{-1} = \begin{pmatrix}1 & 0 \ -n & 1\end{pmatrix}$$.

3) $$A^{-1} = A$$.
1. $$ |A| + |B| = \frac{15}{2} $$. 2. The inverse of $$ A^n $$ is $$ \begin{pmatrix}1 & 0 \ -n & 1\end{pmatrix} $$. 3. The inverse of $$ A $$ is $$ A $$ itself.

then could be equal to

4) If , then the multiplicative inverse of where is .
(a)
If is a square matrix, and , then .

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then could be equal to 4) If , then the multiplicative inverse of where is . (a) If is a square matrix, and , then .