Examine the product of the two matrices to determine if each is the inverse of the other.

left bracket Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column negative 1 3rd Column 1 2nd Row 1st Column 0 2nd Column 1 3rd Column negative 1 3rd Row 1st Column 4 2nd Column 1 3rd Column 0 EndMatrix right bracket

1

−1

1

0

1

−1

4

1

0

left bracket Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 1 3rd Column 0 2nd Row 1st Column negative 4 2nd Column negative 5 3rd Column 1 3rd Row 1st Column negative 4 2nd Column negative 5 3rd Column 0 EndMatrix right bracket

1

1

0

−4

−5

1

−4

−5

0

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Part 1

Are the matrices inverses of each​ other?

Yes

YesYes

No

No

To determine whether two matrices are inverses of each other, we need to check if their product results in the identity matrix. Specifically, if matrices \( A \) and \( B \) are inverses of each other, then \( A \times B = I \) and \( B \times A = I \), where \( I \) is the identity matrix.

Given matrices:

\[
A =
\begin{pmatrix}
1 & -1 & 1 \\
0 & 1 & -1 \\
4 & 1 & 0
\end{pmatrix}
\quad \text{and} \quad
B =
\begin{pmatrix}
1 & 1 & 0 \\
-4 & -5 & 1 \\
-4 & -5 & 0
\end{pmatrix}
\]

Let's compute the product \( A \times B \):

\[
A \times B =
\begin{pmatrix}
1 & -1 & 1 \\
0 & 1 & -1 \\
4 & 1 & 0
\end{pmatrix}
\cdot
\begin{pmatrix}
1 & 1 & 0 \\
-4 & -5 & 1 \\
-4 & -5 & 0
\end{pmatrix}
\]

### Step-by-Step Multiplication

1. **First row, first column:**
   \[
   (1 \times 1) + (-1 \times -4) + (1 \times -4) = 1 + 4 - 4 = 1
   \]
   
2. **First row, second column:**
   \[
   (1 \times 1) + (-1 \times -5) + (1 \times -5) = 1 + 5 - 5 = 1
   \]

3. **First row, third column:**
   \[
   (1 \times 0) + (-1 \times 1) + (1 \times 0) = 0 - 1 + 0 = -1
   \]

4. **Second row, first column:**
   \[
   (0 \times 1) + (1 \times -4) + (-1 \times -4) = 0 - 4 + 4 = 0
   \]

5. **Second row, second column:**
   \[
   (0 \times 1) + (1 \times -5) + (-1 \times -5) = 0 - 5 + 5 = 0
   \]

6. **Second row, third column:**
   \[
   (0 \times 0) + (1 \times 1) + (-1 \times 0) = 0 + 1 + 0 = 1
   \]

7. **Third row, first column:**
   \[
   (4 \times 1) + (1 \times -4) + (0 \times -4) = 4 - 4 + 0 = 0
   \]

8. **Third row, second column:**
   \[
   (4 \times 1) + (1 \times -5) + (0 \times -5) = 4 - 5 + 0 = -1
   \]

9. **Third row, third column:**
   \[
   (4 \times 0) + (1 \times 1) + (0 \times 0) = 0 + 1 + 0 = 1
   \]

So, the product \( A \times B \) is:

\[
A \times B = 
\begin{pmatrix}
1 & 1 & -1 \\
0 & 0 & 1 \\
0 & -1 & 1
\end{pmatrix}
\]

This result is not the identity matrix. Since \( A \times B \) does not equal the identity matrix, matrices \( A \) and \( B \) are **not** inverses of each other.

### Final Answer:
No, the matrices are not inverses of each other.

Do you have any questions or need further details?

Here are five related questions you might find interesting:
1. What properties must a matrix satisfy to have an inverse?
2. How can we determine if a matrix is singular or non-singular?
3. Can non-square matrices have inverses?
4. What is the process for finding the inverse of a 3x3 matrix?
5. How does the determinant of a matrix relate to its invertibility?

**Tip:** Always check if the product of two matrices results in the identity matrix when verifying if they are inverses.

Explore how to determine if two matrices are inverses of each other through detailed calculations. This problem involves matrix operations and the concept of identity matrices. Suitable for advanced high school students and anyone studying linear algebra.

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