Answer :
To find the inverse of the given matrix, we follow these steps:
Let's consider the matrix
[tex]\[ A = \begin{pmatrix} 3 & 9 \\ 7 & 18 \end{pmatrix} \][/tex]
### Step 1: Calculate the Determinant of Matrix A
The determinant ([tex]\(\det(A)\)[/tex]) of a [tex]\(2 \times 2\)[/tex] matrix
[tex]\[ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \][/tex]
is found using the formula:
[tex]\[ \det(A) = ad - bc \][/tex]
For our matrix:
[tex]\[ A = \begin{pmatrix} 3 & 9 \\ 7 & 18 \end{pmatrix} \][/tex]
we have [tex]\(a = 3\)[/tex], [tex]\(b = 9\)[/tex], [tex]\(c = 7\)[/tex], and [tex]\(d = 18\)[/tex].
So, the determinant is:
[tex]\[ \det(A) = (3 \cdot 18) - (9 \cdot 7) = 54 - 63 = -9 \][/tex]
### Step 2: Check if the Determinant is Non-zero
Since the determinant of matrix [tex]\( A \)[/tex] is [tex]\(-9\)[/tex] (which is non-zero), the matrix is invertible. We can proceed to find the inverse.
### Step 3: Find the Adjugate of Matrix A
The adjugate (or adjoint) of matrix [tex]\( A \)[/tex] is formed by taking the transpose of the cofactor matrix of [tex]\( A \)[/tex].
For matrix [tex]\( A \)[/tex]:
[tex]\[ A = \begin{pmatrix} 3 & 9 \\ 7 & 18 \end{pmatrix} \][/tex]
The cofactor matrix of [tex]\( A \)[/tex] is:
[tex]\[ \begin{pmatrix} \det(\begin{pmatrix} 18 \end{pmatrix}) & -\det(\begin{pmatrix} 7 \end{pmatrix}) \\ -\det(\begin{pmatrix} 9 \end{pmatrix}) & \det(\begin{pmatrix} 3 \end{pmatrix}) \end{pmatrix} = \begin{pmatrix} 18 & -7 \\ -9 & 3 \end{pmatrix}\][/tex]
Taking the transpose of the cofactor matrix, we get:
[tex]\[ \text{Adj}(A) = \begin{pmatrix} 18 & -9 \\ -7 & 3 \end{pmatrix} \][/tex]
### Step 4: Calculate the Inverse Using the Formula
The inverse of matrix [tex]\( A \)[/tex] is given by the formula:
[tex]\[ A^{-1} = \frac{1}{\det(A)} \cdot \text{Adj}(A) \][/tex]
Substituting [tex]\(\det(A) = -9\)[/tex] and [tex]\(\text{Adj}(A)\)[/tex]:
[tex]\[ A^{-1} = \frac{1}{-9} \cdot \begin{pmatrix} 18 & -9 \\ -7 & 3 \end{pmatrix} = \begin{pmatrix} 18 \cdot \frac{1}{-9} & -9 \cdot \frac{1}{-9} \\ -7 \cdot \frac{1}{-9} & 3 \cdot \frac{1}{-9} \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ 0.77777778 & -0.33333333 \end{pmatrix} \][/tex]
Thus, the inverse of the given matrix is:
[tex]\[ A^{-1} = \begin{pmatrix} -2 & 1 \\ 0.77777778 & -0.33333333 \end{pmatrix} \][/tex]
Let's consider the matrix
[tex]\[ A = \begin{pmatrix} 3 & 9 \\ 7 & 18 \end{pmatrix} \][/tex]
### Step 1: Calculate the Determinant of Matrix A
The determinant ([tex]\(\det(A)\)[/tex]) of a [tex]\(2 \times 2\)[/tex] matrix
[tex]\[ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \][/tex]
is found using the formula:
[tex]\[ \det(A) = ad - bc \][/tex]
For our matrix:
[tex]\[ A = \begin{pmatrix} 3 & 9 \\ 7 & 18 \end{pmatrix} \][/tex]
we have [tex]\(a = 3\)[/tex], [tex]\(b = 9\)[/tex], [tex]\(c = 7\)[/tex], and [tex]\(d = 18\)[/tex].
So, the determinant is:
[tex]\[ \det(A) = (3 \cdot 18) - (9 \cdot 7) = 54 - 63 = -9 \][/tex]
### Step 2: Check if the Determinant is Non-zero
Since the determinant of matrix [tex]\( A \)[/tex] is [tex]\(-9\)[/tex] (which is non-zero), the matrix is invertible. We can proceed to find the inverse.
### Step 3: Find the Adjugate of Matrix A
The adjugate (or adjoint) of matrix [tex]\( A \)[/tex] is formed by taking the transpose of the cofactor matrix of [tex]\( A \)[/tex].
For matrix [tex]\( A \)[/tex]:
[tex]\[ A = \begin{pmatrix} 3 & 9 \\ 7 & 18 \end{pmatrix} \][/tex]
The cofactor matrix of [tex]\( A \)[/tex] is:
[tex]\[ \begin{pmatrix} \det(\begin{pmatrix} 18 \end{pmatrix}) & -\det(\begin{pmatrix} 7 \end{pmatrix}) \\ -\det(\begin{pmatrix} 9 \end{pmatrix}) & \det(\begin{pmatrix} 3 \end{pmatrix}) \end{pmatrix} = \begin{pmatrix} 18 & -7 \\ -9 & 3 \end{pmatrix}\][/tex]
Taking the transpose of the cofactor matrix, we get:
[tex]\[ \text{Adj}(A) = \begin{pmatrix} 18 & -9 \\ -7 & 3 \end{pmatrix} \][/tex]
### Step 4: Calculate the Inverse Using the Formula
The inverse of matrix [tex]\( A \)[/tex] is given by the formula:
[tex]\[ A^{-1} = \frac{1}{\det(A)} \cdot \text{Adj}(A) \][/tex]
Substituting [tex]\(\det(A) = -9\)[/tex] and [tex]\(\text{Adj}(A)\)[/tex]:
[tex]\[ A^{-1} = \frac{1}{-9} \cdot \begin{pmatrix} 18 & -9 \\ -7 & 3 \end{pmatrix} = \begin{pmatrix} 18 \cdot \frac{1}{-9} & -9 \cdot \frac{1}{-9} \\ -7 \cdot \frac{1}{-9} & 3 \cdot \frac{1}{-9} \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ 0.77777778 & -0.33333333 \end{pmatrix} \][/tex]
Thus, the inverse of the given matrix is:
[tex]\[ A^{-1} = \begin{pmatrix} -2 & 1 \\ 0.77777778 & -0.33333333 \end{pmatrix} \][/tex]