To find the inverse of the matrix [tex]\( A \)[/tex], we need to follow a few key steps. The matrix [tex]\( A \)[/tex] is given as:
[tex]\[
A = \begin{pmatrix}
6 & 1 \\
11 & 2
\end{pmatrix}
\][/tex]
First, we calculate the determinant of [tex]\( A \)[/tex]. For a [tex]\( 2 \times 2 \)[/tex] matrix [tex]\( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex], the determinant is given by:
[tex]\[
\text{det}(A) = ad - bc
\][/tex]
Plugging in the values from matrix [tex]\( A \)[/tex]:
[tex]\[
\text{det}(A) = (6 \times 2) - (1 \times 11) = 12 - 11 = 1
\][/tex]
Since the determinant is 1, which is non-zero, the matrix [tex]\( A \)[/tex] is invertible.
Next, for a [tex]\( 2 \times 2 \)[/tex] matrix [tex]\( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex], the inverse [tex]\( A^{-1} \)[/tex] is given by:
[tex]\[
A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}
\][/tex]
Applying this to matrix [tex]\( A \)[/tex]:
[tex]\[
A^{-1} = \frac{1}{1} \begin{pmatrix} 2 & -1 \\ -11 & 6 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -11 & 6 \end{pmatrix}
\][/tex]
So, the inverse of matrix [tex]\( A \)[/tex] should be:
[tex]\[
A^{-1} = \begin{pmatrix} 2 & -1 \\ -11 & 6 \end{pmatrix}
\][/tex]
However, the numerical values may contain minor discrepancies due to floating-point arithmetic precision inherent in calculations. Specifically, the result we found is:
[tex]\[
A^{-1} = \begin{pmatrix} 2.0000000000000018 & -1.0000000000000009 \\ -11.00000000000001 & 6.000000000000005 \end{pmatrix}
\][/tex]
Thus, the inverse of the matrix [tex]\( A \)[/tex] is precisely:
[tex]\[
A^{-1} = \begin{pmatrix}
2.0000000000000018 & -1.0000000000000009 \\
-11.00000000000001 & 6.000000000000005
\end{pmatrix}
\][/tex]
