If `B = [(2,3),(5,-2)]`, then find the inverse of `B`

Options

A. B
B. `(1)/(19) B`
C. `(1)/(5) B`
D. `4B`

Answer

To find the inverse of the matrix \( B = \begin{pmatrix} 2 & 3 \\ 5 & -2 \end{pmatrix} \), we will follow these steps:

### Step 1: Calculate the Determinant of Matrix B
The determinant of a 2x2 matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is calculated using the formula:
\[ \text{det}(B) = ad - bc \]
For our matrix \( B \):
- \( a = 2 \)
- \( b = 3 \)
- \( c = 5 \)
- \( d = -2 \)

Now substituting these values into the determinant formula:
\[ \text{det}(B) = (2)(-2) - (3)(5) = -4 - 15 = -19 \]

### Step 2: Calculate the Adjoint of Matrix B
The adjoint of a matrix is the transpose of the cofactor matrix. We will find the cofactors for each element of matrix \( B \).

1. **Cofactor \( C_{11} \)**:
\[ C_{11} = (-1)^{1+1} \cdot M_{11} = 1 \cdot (-2) = -2 \]
(Minor \( M_{11} \) is the determinant of the matrix formed by removing the first row and first column, which is just \(-2\)).

2. **Cofactor \( C_{12} \)**:
\[ C_{12} = (-1)^{1+2} \cdot M_{12} = -1 \cdot 5 = -5 \]
(Minor \( M_{12} \) is the determinant of the matrix formed by removing the first row and second column, which is \( 5 \)).

3. **Cofactor \( C_{21} \)**:
\[ C_{21} = (-1)^{2+1} \cdot M_{21} = -1 \cdot 3 = -3 \]
(Minor \( M_{21} \) is the determinant of the matrix formed by removing the second row and first column, which is \( 3 \)).

4. **Cofactor \( C_{22} \)**:
\[ C_{22} = (-1)^{2+2} \cdot M_{22} = 1 \cdot 2 = 2 \]
(Minor \( M_{22} \) is the determinant of the matrix formed by removing the second row and second column, which is \( 2 \)).

Now, we can write the cofactor matrix:
\[ \text{Cofactor Matrix} = \begin{pmatrix} -2 & -5 \\ -3 & 2 \end{pmatrix} \]

Taking the transpose of the cofactor matrix gives us the adjoint:
\[ \text{adj}(B) = \begin{pmatrix} -2 & -3 \\ -5 & 2 \end{pmatrix} \]

### Step 3: Calculate the Inverse of Matrix B
The inverse of matrix \( B \) is given by the formula:
\[ B^{-1} = \frac{1}{\text{det}(B)} \cdot \text{adj}(B) \]
Substituting the values we found:
\[ B^{-1} = \frac{1}{-19} \cdot \begin{pmatrix} -2 & -3 \\ -5 & 2 \end{pmatrix} \]
This simplifies to:
\[ B^{-1} = \begin{pmatrix} \frac{2}{19} & \frac{3}{19} \\ \frac{5}{19} & -\frac{2}{19} \end{pmatrix} \]

### Final Answer
Thus, the inverse of matrix \( B \) is:
\[ B^{-1} = \begin{pmatrix} \frac{2}{19} & \frac{3}{19} \\ \frac{5}{19} & -\frac{2}{19} \end{pmatrix} \]