If `{1/2(A-A'+1)}^-1=2/lambda[(lambda-13,-lambda/3,lambda/3),(-17,10,-1),(7,-11,5)]` for `A=[(-2,3,4),(5,-4,-3),(7,2,9)]`, then `lambda` is

Answer

To solve the equation

\[ \frac{1}{2}(A - A' + I)^{-1} = \frac{2}{\lambda} \begin{pmatrix} \lambda - 13 & -\frac{\lambda}{3} & \frac{\lambda}{3} \\ -17 & 10 & -1 \\ 7 & -11 & 5 \end{pmatrix} \]

for \( A = \begin{pmatrix} -2 & 3 & 4 \\ 5 & -4 & -3 \\ 7 & 2 & 9 \end{pmatrix} \), we will follow these steps:

### Step 1: Calculate \( A' \) (Transpose of \( A \))
The transpose of matrix \( A \) is obtained by swapping rows with columns:

\[ A' = \begin{pmatrix} -2 & 5 & 7 \\ 3 & -4 & 2 \\ 4 & -3 & 9 \end{pmatrix} \]

### Step 2: Calculate \( A - A' \)
Now, we subtract \( A' \) from \( A \):

\[ A - A' = \begin{pmatrix} -2 & 3 & 4 \\ 5 & -4 & -3 \\ 7 & 2 & 9 \end{pmatrix} - \begin{pmatrix} -2 & 5 & 7 \\ 3 & -4 & 2 \\ 4 & -3 & 9 \end{pmatrix} \]

Calculating this gives:

\[ A - A' = \begin{pmatrix} 0 & -2 & -3 \\ 2 & 0 & -5 \\ 3 & 5 & 0 \end{pmatrix} \]

### Step 3: Add the Identity Matrix \( I \)
The identity matrix \( I \) is:

\[ I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

Now, we add \( I \) to \( A - A' \):

\[ A - A' + I = \begin{pmatrix} 0 & -2 & -3 \\ 2 & 0 & -5 \\ 3 & 5 & 0 \end{pmatrix} + \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -2 & -3 \\ 2 & 1 & -5 \\ 3 & 5 & 1 \end{pmatrix} \]

### Step 4: Multiply by \( \frac{1}{2} \)
Now we multiply the result by \( \frac{1}{2} \):

\[ \frac{1}{2}(A - A' + I) = \frac{1}{2} \begin{pmatrix} 1 & -2 & -3 \\ 2 & 1 & -5 \\ 3 & 5 & 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & -1 & -\frac{3}{2} \\ 1 & \frac{1}{2} & -\frac{5}{2} \\ \frac{3}{2} & \frac{5}{2} & \frac{1}{2} \end{pmatrix} \]

### Step 5: Find the Inverse
Let \( B = \frac{1}{2}(A - A' + I) \). We need to find \( B^{-1} \).

To find the inverse, we can use the formula \( B^{-1} = \frac{1}{\text{det}(B)} \text{adj}(B) \).

First, we calculate the determinant of \( B \).

### Step 6: Calculate the Determinant of \( B \)
Using the determinant formula for a \( 3 \times 3 \) matrix:

\[ \text{det}(B) = b_{11}(b_{22}b_{33} - b_{23}b_{32}) - b_{12}(b_{21}b_{33} - b_{23}b_{31}) + b_{13}(b_{21}b_{32} - b_{22}b_{31}) \]

Substituting the values from \( B \):

\[ \text{det}(B) = \frac{1}{2} \left( \frac{1}{2} \left( \frac{1}{2} \cdot \frac{1}{2} - (-\frac{5}{2})(\frac{5}{2}) \right) - (-1) \left( 1 \cdot \frac{1}{2} - (-\frac{5}{2})(\frac{3}{2}) \right) + (-\frac{3}{2}) \left( 1 \cdot \frac{5}{2} - \frac{1}{2} \cdot \frac{3}{2} \right) \right) \]

After calculating, we find:

\[ \text{det}(B) = \frac{39}{8} \]

### Step 7: Calculate the Adjoint of \( B \)
The adjoint of \( B \) can be calculated by finding the cofactor matrix and then transposing it.

### Step 8: Set Up the Equation
Now we set up the equation:

\[ \frac{1}{2}(A - A' + I)^{-1} = \frac{8}{39} \text{adj}(B) \]

Equating this to the right-hand side given in the problem:

\[ \frac{8}{39} \text{adj}(B) = \frac{2}{\lambda} \begin{pmatrix} \lambda - 13 & -\frac{\lambda}{3} & \frac{\lambda}{3} \\ -17 & 10 & -1 \\ 7 & -11 & 5 \end{pmatrix} \]

### Step 9: Solve for \( \lambda \)
By comparing the elements of the matrices, we can derive the value of \( \lambda \).

After equating and solving the equations, we find:

\[ \lambda = 39 \]

### Final Answer
Thus, the value of \( \lambda \) is

\[ \boxed{39} \]