Question

Find the inverse of the given matrix.
Aequals
left bracket Start 3 By 3 Matrix 1st Row 1st Column 6 2nd Column 6 3rd Column negative 1 2nd Row 1st Column 0 2nd Column 3 3rd Column negative 1 3rd Row 1st Column negative 1 2nd Column negative 6 3rd Column 1 EndMatrix right bracket

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Part 1
Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.
A.
The inverse of the given matrix is Upper A Superscript negative 1equals
  
enter your response here
​(Type an integer or a simplified fraction for each matrix​ element.)
B.
The given matrix does not have an inverse.

Answer 1

To find the inverse of the matrix \( A \), we use the formula for the inverse of a \( 3 \times 3 \) matrix. The given matrix is:

\[
A = \begin{pmatrix} 6 & 6 & -1 \\ 0 & 3 & -1 \\ -1 & -6 & 1 \end{pmatrix}
\]

To find the inverse, we first need to calculate the determinant of \( A \). The determinant \( \text{det}(A) \) can be calculated using the formula for the determinant of a \( 3 \times 3 \) matrix:

\[
\text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg)
\]
where the matrix is:
\[
\begin{pmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{pmatrix}
\]
For matrix \( A \):
- \( a = 6 \), \( b = 6 \), \( c = -1 \)
- \( d = 0 \), \( e = 3 \), \( f = -1 \)
- \( g = -1 \), \( h = -6 \), \( i = 1 \)

Calculating the determinant:

\[
\text{det}(A) = 6(3 \cdot 1 - (-1)(-6)) - 6(0 \cdot 1 - (-1)(-1)) + (-1)(0 \cdot (-6) - 3 \cdot -1)
\]
\[
= 6(3 - 6) - 6(0 - 1) + (-1)(0 + 3)
\]
\[
= 6(-3) - 6(-1) - 3
\]
\[
= -18 + 6 - 3
\]
\[
= -15
\]

Since the determinant is not zero (\( \text{det}(A) \neq 0 \)), the inverse exists.

Next, we find the inverse using the adjugate method. The formula for the inverse of a matrix is given by:

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

We first calculate the adjugate of \( A \). The adjugate is the transpose of the cofactor matrix of \( A \).

Calculating the cofactor matrix:

- Cofactor \( C_{11} = \text{det}\begin{pmatrix} 3 & -1 \\ -6 & 1 \end{pmatrix} = 3 \cdot 1 - (-1)(-6) = 3 - 6 = -3 \)
- Cofactor \( C_{12} = -\text{det}\begin{pmatrix} 0 & -1 \\ -1 & 1 \end{pmatrix} = - (0 \cdot 1 - (-1)(-1)) = -(-1) = 1 \)
- Cofactor \( C_{13} = \text{det}\begin{pmatrix} 0 & 3 \\ -1 & -6 \end{pmatrix} = 0 \cdot (-6) - 3 \cdot (-1) = 0 + 3 = 3 \)

- Cofactor \( C_{21} = -\text{det}\begin{pmatrix} 6 & -1 \\ -6 & 1 \end{pmatrix} = - (6 \cdot 1 - (-1)(-6)) = -(6 - 6) = 0 \)
- Cofactor \( C_{22} = \text{det}\begin{pmatrix} 6 & -1 \\ -1 & 1 \end{pmatrix} = 6 \cdot 1 - (-1)(-1) = 6 - 1 = 5 \)
- Cofactor \( C_{23} = -\text{det}\begin{pmatrix} 6 & 6 \\ -1 & -6 \end{pmatrix} = - (6 \cdot -6 - 6 \cdot -1) = -(-36 + 6) = 30 \)

- Cofactor \( C_{31} = \text{det}\begin{pmatrix} 6 & -1 \\ 3 & -1 \end{pmatrix} = 6 \cdot -1 - (-1)(3) = -6 + 3 = -3 \)
- Cofactor \( C_{32} = -\text{det}\begin{pmatrix} 6 & -1 \\ 0 & -1 \end{pmatrix} = - (6 \cdot -1 - (-1)(0)) = -(-6) = 6 \)
- Cofactor \( C_{33} = \text{det}\begin{pmatrix} 6 & 6 \\ 0 & 3 \end{pmatrix} = 6 \cdot 3 - 6 \cdot 0 = 18 \)

Assembling the cofactor matrix:

\[
\text{Cof}(A) = \begin{pmatrix} -3 & 1 & 3 \\ 0 & 5 & 30 \\ -3 & 6 & 18 \end{pmatrix}
\]

Then taking the transpose:

\[
\text{adj}(A) = \begin{pmatrix} -3 & 0 & -3 \\ 1 & 5 & 6 \\ 3 & 30 & 18 \end{pmatrix}
\]

Now we can calculate the inverse:

\[
A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) = \frac{1}{-15} \cdot \begin{pmatrix} -3 & 0 & -3 \\ 1 & 5 & 6 \\ 3 & 30 & 18 \end{pmatrix}
\]

So, the final result is:

\[
A^{-1} = \begin{pmatrix} \frac{3}{15} & 0 & \frac{3}{15} \\ -\frac{1}{15} & -\frac{5}{15} & -\frac{6}{15} \\ -\frac{3}{15} & -\frac{30}{15} & -\frac{18}{15} \end{pmatrix}
\]

Simplifying:

\[
A^{-1} = \begin{pmatrix} \frac{1}{5} & 0 & \frac{1}{5} \\ -\frac{1}{15} & -\frac{1}{3} & -\frac{2}{5} \\ -\frac{1}{5} & -2 & -\frac{6}{5} \end{pmatrix}
\]

Thus, the correct option is A, and the answer is:

\[
A^{-1} = \begin{pmatrix} \frac{1}{5} & 0 & \frac{1}{5} \\ -\frac{1}{15} & -\frac{1}{3} & -\frac{2}{5} \\ -\frac{1}{5} & -2 & -\frac{6}{5} \end{pmatrix}
\]

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