A matrix ` A = ({:(2,3),(5,-2):}) ` is such that `A^(-1)= lambda A ` then what is the value of `lambda ` .

Options

A. 19
B. 9
C. ` (1)/(19)`
D. `-(1)/(19)`

Answer

To solve the problem, we need to find the value of \( \lambda \) such that \( A^{-1} = \lambda A \) for the given matrix \( A = \begin{pmatrix} 2 & 3 \\ 5 & -2 \end{pmatrix} \).

### Step 1: Find the determinant of matrix \( A \)

The determinant of a \( 2 \times 2 \) matrix \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by:

\[ \text{det}(A) = ad - bc \]

For our matrix \( A \):

\[ \text{det}(A) = (2)(-2) - (3)(5) = -4 - 15 = -19 \]

### Step 2: Find the adjoint of matrix \( A \)

The adjoint of a \( 2 \times 2 \) matrix \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by:

\[ \text{adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]

For our matrix \( A \):

\[ \text{adj}(A) = \begin{pmatrix} -2 & -3 \\ -5 & 2 \end{pmatrix} \]

### Step 3: Calculate the inverse of matrix \( A \)

The inverse of a matrix \( A \) can be calculated using the formula:

\[ A^{-1} = \frac{\text{adj}(A)}{\text{det}(A)} \]

Substituting the values we found:

\[ A^{-1} = \frac{1}{-19} \begin{pmatrix} -2 & -3 \\ -5 & 2 \end{pmatrix} = \begin{pmatrix} \frac{2}{19} & \frac{3}{19} \\ \frac{5}{19} & -\frac{2}{19} \end{pmatrix} \]

### Step 4: Set up the equation \( A^{-1} = \lambda A \)

We know that:

\[ A^{-1} = \lambda A \]

Substituting the values we have:

\[ \begin{pmatrix} \frac{2}{19} & \frac{3}{19} \\ \frac{5}{19} & -\frac{2}{19} \end{pmatrix} = \lambda \begin{pmatrix} 2 & 3 \\ 5 & -2 \end{pmatrix} \]

### Step 5: Equate the corresponding elements

From the equation above, we can equate the corresponding elements:

1. \( \frac{2}{19} = 2\lambda \)
2. \( \frac{3}{19} = 3\lambda \)
3. \( \frac{5}{19} = 5\lambda \)
4. \( -\frac{2}{19} = -2\lambda \)

From the first equation:

\[ \lambda = \frac{2}{19 \cdot 2} = \frac{1}{19} \]

### Conclusion

Thus, the value of \( \lambda \) is:

\[ \lambda = \frac{1}{19} \]