select the correct answer.
$a = \begin{bmatrix}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1\end{bmatrix}$ and $ab = \begin{bmatrix}-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{bmatrix}$.
which matrix is matrix $b$?
$\bigcirc$ a. $\begin{bmatrix}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1\end{bmatrix}$
$\bigcirc$ b. $\begin{bmatrix}-1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{bmatrix}$
$\bigcirc$ c. $\begin{bmatrix}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{bmatrix}$
$\bigcirc$ d. $\begin{bmatrix}-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1\end{bmatrix}$

Question

Accepted Answer

# Explanation:
## Step1: Recall Matrix Multiplication Inverse
To find $ B $ when $ A $ and $ AB $ are known, we use the property that if $ A $ is invertible (determinant non - zero), then $ B=A^{-1}(AB) $. First, find the inverse of $ A $. For a diagonal matrix $ A=\begin{bmatrix}a&0&0\0&b&0\0&0&c\end{bmatrix} $, the inverse is $ A^{-1}=\begin{bmatrix}\frac{1}{a}&0&0\0&\frac{1}{b}&0\0&0&\frac{1}{c}\end{bmatrix} $ (since $ \det(A) = a\times b\times c
eq0 $ here as $ a = 1,b=- 1,c = 1 $, so $ \det(A)=-1
eq0 $). So $ A^{-1}=\begin{bmatrix}1&0&0\0&-1&0\0&0&1\end{bmatrix} $ (because $ \frac{1}{1}=1,\frac{1}{-1}=-1,\frac{1}{1}=1 $).

## Step2: Multiply $ A^{-1} $ with $ AB $
We know that $ AB=\begin{bmatrix}-1&0&0\0&-1&0\0&0&-1\end{bmatrix} $. Now, multiply $ A^{-1} $ and $ AB $:

$$
\begin{align*}
A^{-1}(AB)&=\begin{bmatrix}1&0&0\0&-1&0\0&0&1\end{bmatrix}\begin{bmatrix}-1&0&0\0&-1&0\0&0&-1\end{bmatrix}\
&=\begin{bmatrix}1\times(-1)+0\times0 + 0\times0&1\times0+0\times(-1)+0\times0&1\times0+0\times0 + 0\times(-1)\
0\times(-1)+(-1)\times0+0\times0&0\times0+(-1)\times(-1)+0\times0&0\times0+(-1)\times0 + 0\times(-1)\
0\times(-1)+0\times0+1\times0&0\times0+0\times(-1)+1\times0&0\times0+0\times0+1\times(-1)
\end{bmatrix}\
&=\begin{bmatrix}-1&0&0\0&1&0\0&0&-1\end{bmatrix}
\end{align*}
$$

We can also check by multiplying $ A $ with each option:

- Option A: $ A\times A=\begin{bmatrix}1&0&0\0&-1&0\0&0&1\end{bmatrix}\begin{bmatrix}1&0&0\0&-1&0\0&0&1\end{bmatrix}=\begin{bmatrix}1&0&0\0&1&0\0&0&1\end{bmatrix}
eq AB $
- Option B: $ A\times B=\begin{bmatrix}1&0&0\0&-1&0\0&0&1\end{bmatrix}\begin{bmatrix}-1&0&0\0&1&0\0&0&-1\end{bmatrix}=\begin{bmatrix}-1&0&0\0&-1&0\0&0&-1\end{bmatrix}=AB $
- Option C: $ A\times C=\begin{bmatrix}1&0&0\0&-1&0\0&0&1\end{bmatrix}\begin{bmatrix}1&0&0\0&-1&0\0&0&-1\end{bmatrix}=\begin{bmatrix}1&0&0\0&1&0\0&0&-1\end{bmatrix}
eq AB $
- Option D: $ A\times D=\begin{bmatrix}1&0&0\0&-1&0\0&0&1\end{bmatrix}\begin{bmatrix}-1&0&0\0&-1&0\0&0&1\end{bmatrix}=\begin{bmatrix}-1&0&0\0&1&0\0&0&1\end{bmatrix}
eq AB $

# Answer: B. $\begin{bmatrix}-1&0&0\0&1&0\0&0&-1\end{bmatrix}$

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QUESTION IMAGE

Question

Explanation:

Step1: Recall Matrix Multiplication Inverse

Step2: Multiply \( A^{-1} \) with \( AB \)

Answer: