To find the inverse of the given matrix, we start with the following 2x2 matrix:
[tex]\[
A = \begin{pmatrix}
-6 & -12 \\
-9 & -16
\end{pmatrix}
\][/tex]
The formula for the inverse of a 2x2 matrix
[tex]\[
A = \begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
\][/tex]
is:
[tex]\[
A^{-1} = \frac{1}{ad - bc} \begin{pmatrix}
d & -b \\
-c & a
\end{pmatrix}
\][/tex]
In our case:
[tex]\[
a = -6, \quad b = -12, \quad c = -9, \quad d = -16
\][/tex]
First, calculate the determinant of [tex]\(A\)[/tex]:
[tex]\[
\text{det}(A) = (-6)(-16) - (-12)(-9)
\][/tex]
[tex]\[
\text{det}(A) = 96 - 108 = -12
\][/tex]
Since the determinant is non-zero, the inverse of the matrix exists. Next, use the formula to find the inverse:
[tex]\[
A^{-1} = \frac{1}{-12} \begin{pmatrix}
-16 & 12 \\
9 & -6
\end{pmatrix}
\][/tex]
Now, multiply each element of the matrix by [tex]\(\frac{1}{-12}\)[/tex]:
[tex]\[
A^{-1} = \begin{pmatrix}
\frac{-16}{-12} & \frac{12}{-12} \\
\frac{9}{-12} & \frac{-6}{-12}
\end{pmatrix}
\][/tex]
Simplify the fractions:
[tex]\[
A^{-1} = \begin{pmatrix}
\frac{4}{3} & -1 \\
-\frac{3}{4} & \frac{1}{2}
\end{pmatrix}
\][/tex]
Next, convert the fractions to decimals and round to the nearest hundredth if necessary:
[tex]\[
\frac{4}{3} \approx 1.33, \quad -1 \text{ remains } -1, \quad -\frac{3}{4} = -0.75, \quad \frac{1}{2} = 0.5
\][/tex]
Thus, the inverse matrix rounded to the nearest hundredth is:
[tex]\[
A^{-1} = \begin{pmatrix}
1.33 & -1 \\
-0.75 & 0.5
\end{pmatrix}
\][/tex]
Hence, the inverse of the matrix
[tex]\[
\begin{pmatrix}
-6 & -12 \\
-9 & -16
\end{pmatrix}
\][/tex]
is
[tex]\[
\boxed{\begin{pmatrix}
1.33 & -1 \\
-0.75 & 0.5
\end{pmatrix}}
\][/tex]
