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Part 1
Find Upper A Superscript negative 1​, or explain why it does not exist.
Aequalsleft bracket Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 1 3rd Column 1 2nd Row 1st Column 2 2nd Column 1 3rd Column negative 1 3rd Row 1st Column 5 2nd Column 1 3rd Column 0 EndMatrix right bracket
 
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Part 1
Select the correct choice below and fill in any answer boxes present in your choice.
A.
Upper A Superscript negative 1equals
  
enter your response here ​(Simplify your​ answers.)
B.
Upper A Superscript negative 1 does not exist because A in the augmented matrix​ [A|I] cannot be transformed to I using the​ Gauss-Jordan method.

To find \( A^{-1} \), we need to determine if matrix \( A \) is invertible. If it is, we can calculate the inverse; if not, we will explain why the inverse does not exist.

### Matrix \( A \) is:
\[
A = \begin{bmatrix} 
1 & 1 & 1 \\
2 & 1 & -1 \\
5 & 1 & 0 
\end{bmatrix}
\]

To check whether the matrix \( A \) is invertible, we need to calculate its **determinant**. A matrix is invertible if and only if its determinant is non-zero.

### Step 1: Calculate the determinant of \( A \)

The determinant of a \( 3 \times 3 \) matrix is given by:
\[
\text{det}(A) = a(ei − fh) − b(di − fg) + c(dh − eg)
\]
For our matrix \( A \), we assign:
\[
a = 1, b = 1, c = 1, d = 2, e = 1, f = -1, g = 5, h = 1, i = 0
\]
Substitute these values into the formula:

\[
\text{det}(A) = 1[(1 \cdot 0) - (-1 \cdot 1)] - 1[(2 \cdot 0) - (-1 \cdot 5)] + 1[(2 \cdot 1) - (1 \cdot 5)]
\]
\[
\text{det}(A) = 1[0 + 1] - 1[0 + 5] + 1[2 - 5]
\]
\[
\text{det}(A) = 1 - 5 - 3
\]
\[
\text{det}(A) = -7
\]

### Step 2: Conclusion

Since \( \text{det}(A) = -7 \), which is not equal to zero, matrix \( A \) **is invertible**.

### Step 3: Inverse of \( A \)

Now, let's find \( A^{-1} \). The inverse of a matrix \( A \) is given by:

\[
A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A)
\]

Where \( \text{adj}(A) \) is the adjugate (transpose of the cofactor matrix). Let's calculate that.

The determinant of matrix \( A \) is indeed \(-7\), confirming that the matrix is invertible. The inverse of matrix \( A \), \( A^{-1} \), is:

\[
A^{-1} = \frac{1}{-7} \times
\begin{bmatrix}
1 & 5 & -6 \\
1 & 5 & -4 \\
-1 & 2 & -1
\end{bmatrix}
\]

Thus, \( A^{-1} \) is:

\[
A^{-1} = \begin{bmatrix}
-1/7 & -5/7 & 6/7 \\
-1/7 & -5/7 & 4/7 \\
1/7 & -2/7 & 1/7
\end{bmatrix}
\]

Would you like further details or explanations?

### Related Questions:
1. How do you find the determinant of larger matrices?
2. What is the Gauss-Jordan method, and how does it relate to finding matrix inverses?
3. What happens when the determinant of a matrix is zero?
4. How can you verify if a matrix is invertible using row reduction?
5. What are some applications of matrix inverses in real-world problems?

**Tip**: When calculating matrix inverses by hand, always check the determinant first to avoid unnecessary computations if it's zero!

Find A^(-1), or explain why it does not exist.
A = [ [1, 1, 1], [2, 1, -1], [5, 1, 0] ]
Select the correct choice below and fill in any answer boxes present in your choice.
A.
A^(-1) = (Simplify your answers)
B.
A^(-1) does not exist because A in the augmented matrix [A|I] cannot be transformed to I using the Gauss-Jordan method.

Learn how to find the inverse of a 3x3 matrix step-by-step. This guide covers calculating the determinant, checking invertibility, and applying the adjugate method to obtain the inverse. Ideal for high school and college students studying linear algebra.

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