To find the inverse of a 2x2 matrix, we use the following formula for a matrix:
[tex]\[
A = \begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix}
\][/tex]
The inverse [tex]\( A^{-1} \)[/tex] is given by:
[tex]\[
A^{-1} = \frac{1}{ad - bc} \begin{pmatrix}
d & -b \\
-c & a \\
\end{pmatrix}
\][/tex]
where [tex]\( ad - bc \)[/tex] is the determinant of the matrix [tex]\( A \)[/tex].
Now, let's apply this to the given matrix:
[tex]\[
A = \begin{pmatrix}
-4 & 11 \\
2 & -6 \\
\end{pmatrix}
\][/tex]
First, we calculate the determinant of matrix [tex]\( A \)[/tex]:
[tex]\[
\text{det}(A) = (-4) \cdot (-6) - (11) \cdot (2) = 24 - 22 = 2
\][/tex]
Next, we use the formula for the inverse matrix, substituting [tex]\( a = -4 \)[/tex], [tex]\( b = 11 \)[/tex], [tex]\( c = 2 \)[/tex], and [tex]\( d = -6 \)[/tex]:
[tex]\[
A^{-1} = \frac{1}{2} \begin{pmatrix}
-6 & -11 \\
-2 & -4 \\
\end{pmatrix}
\][/tex]
Multiplying each element of the matrix by [tex]\( \frac{1}{2} \)[/tex]:
[tex]\[
A^{-1} = \begin{pmatrix}
-3 & -5.5 \\
-1 & -2 \\
\end{pmatrix}
\][/tex]
Since the problem asks us to round to the nearest hundredth if necessary, we check if any rounding is needed. In this case, the elements are already at their nearest hundredth.
Therefore, the inverse of the given matrix is:
[tex]\[
\begin{pmatrix}
-3.00 & -5.50 \\
-1.00 & -2.00 \\
\end{pmatrix}
\][/tex]
So, the inverse matrix is:
[tex]\[
\begin{pmatrix}
-3 & -5.5 \\
-1 & -2 \\
\end{pmatrix}
\][/tex]
