To find the inverse of a matrix [tex]\( A \)[/tex], we must follow a series of steps. Here, matrix [tex]\( A \)[/tex] is given by:
[tex]\[ A = \begin{pmatrix} 2 & 0 \\ 9 & 1 \end{pmatrix} \][/tex]
### 1. Determining the Determinant
First, we calculate the determinant of matrix [tex]\( A \)[/tex]. For a 2x2 matrix [tex]\( A \)[/tex] represented as:
[tex]\[ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \][/tex]
The determinant, [tex]\( \text{det}(A) \)[/tex], is calculated as:
[tex]\[ \text{det}(A) = ad - bc \][/tex]
Substituting the values from our matrix [tex]\( A \)[/tex]:
[tex]\[ a = 2, \quad b = 0, \quad c = 9, \quad d = 1 \][/tex]
So,
[tex]\[ \text{det}(A) = (2 \cdot 1) - (0 \cdot 9) = 2 \][/tex]
### 2. Checking Invertibility
Since the determinant is non-zero ([tex]\( \text{det}(A) = 2 \)[/tex]), matrix [tex]\( A \)[/tex] is invertible.
### 3. Finding the Inverse
The formula for the inverse of a 2x2 matrix [tex]\( A \)[/tex] is:
[tex]\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \][/tex]
Using our values:
[tex]\[ A^{-1} = \frac{1}{2} \begin{pmatrix} 1 & 0 \\ -9 & 2 \end{pmatrix} \][/tex]
Now, multiplying by [tex]\( \frac{1}{2} \)[/tex]:
[tex]\[ A^{-1} = \begin{pmatrix} \frac{1}{2} & 0 \\ \frac{-9}{2} & 1 \end{pmatrix} \][/tex]
This simplifies to:
[tex]\[ A^{-1} = \begin{pmatrix} 0.5 & 0 \\ -4.5 & 1 \end{pmatrix} \][/tex]
### 4. Rounding (if necessary)
In this case, all values are already in their simplest form, so rounding is not necessary beyond what has been done.
### Final Answer
The inverse of the matrix [tex]\( A \)[/tex] is:
[tex]\[
\boxed{\begin{pmatrix}
0.5 & 0 \\
-4.5 & 1
\end{pmatrix}}
\][/tex]
This is the final result, and all values are correctly represented to the nearest hundredth if necessary.
