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:
\[ A^{-1} = \begin{pmatrix} \frac{1}{5} & 0 & \frac{1}{5} \ -\frac{1}{15} & -\frac{1}{3} & -\frac{2}{5} \ -\frac{1}{5} & -2 & -\frac{6}{5} \end{pmatrix} \]
Feel free to enter these values into the answer box for your question.