Let `P=[(3,-1,-2),(2,0,alpha),(3,-5,0)]` , where ` alpha in R ` . Suppose ` Q =[q_(ij)]` is a matrix satisfying `PQ = KI_3` for some non - zero `K in R .` If ` q_(23) =- (K)/(8) and |Q| = (1)/(2),` then `alpha^2 + k^2 ` is equal to ________.

Answer

To solve the problem, we need to find the value of \( \alpha^2 + k^2 \) given the matrix \( P \) and the conditions on matrix \( Q \). Let's break it down step by step.

### Step 1: Write down the matrix \( P \)
The matrix \( P \) is given as:
\[ P = \begin{pmatrix} 3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0 \end{pmatrix} \]

### Step 2: Use the determinant property
We know that \( PQ = KI_3 \), where \( K \) is a non-zero scalar and \( I_3 \) is the identity matrix of order 3. Taking the determinant of both sides, we have:
\[ \det(PQ) = \det(KI_3) \]
Using the property of determinants, this can be rewritten as:
\[ \det(P) \cdot \det(Q) = K^3 \]

### Step 3: Substitute known values
We know \( \det(Q) = \frac{1}{2} \). Thus, we can write:
\[ \det(P) \cdot \frac{1}{2} = K^3 \implies \det(P) = 2K^3 \]

### Step 4: Calculate \( \det(P) \)
We need to calculate the determinant of matrix \( P \):
\[ \det(P) = 3 \begin{vmatrix} 0 & \alpha \\ -5 & 0 \end{vmatrix} - (-1) \begin{vmatrix} 2 & \alpha \\ 3 & 0 \end{vmatrix} - 2 \begin{vmatrix} 2 & 0 \\ 3 & -5 \end{vmatrix} \]
Calculating the minors:
1. \( \begin{vmatrix} 0 & \alpha \\ -5 & 0 \end{vmatrix} = 0 \cdot 0 - (-5) \cdot \alpha = 5\alpha \)
2. \( \begin{vmatrix} 2 & \alpha \\ 3 & 0 \end{vmatrix} = 2 \cdot 0 - 3 \cdot \alpha = -3\alpha \)
3. \( \begin{vmatrix} 2 & 0 \\ 3 & -5 \end{vmatrix} = 2 \cdot (-5) - 0 \cdot 3 = -10 \)

Substituting back:
\[ \det(P) = 3(5\alpha) + 1(3\alpha) - 2(-10) = 15\alpha + 3\alpha + 20 = 18\alpha + 20 \]

### Step 5: Relate \( \det(P) \) to \( K \)
From our earlier equation, we have:
\[ 18\alpha + 20 = 2K^3 \implies K^3 = 9\alpha + 10 \]

### Step 6: Use the condition on \( Q \)
We also know that \( q_{23} = -\frac{K}{8} \). From the properties of adjoint matrices, we can express \( q_{23} \) in terms of \( \alpha \):
\[ q_{23} = \frac{1}{2} \cdot \text{adj}(P)_{23} \]
Calculating \( \text{adj}(P)_{23} \):
\[ \text{adj}(P)_{23} = (-1)^{2+3} \det\begin{pmatrix} 3 & -1 \\ 3 & -5 \end{pmatrix} = \det\begin{pmatrix} 3 & -1 \\ 3 & -5 \end{pmatrix} = 3(-5) - (-1)(3) = -15 + 3 = -12 \]
Thus,
\[ q_{23} = \frac{-12}{2} = -6 \]
Setting this equal to \( -\frac{K}{8} \):
\[ -6 = -\frac{K}{8} \implies K = 48 \]

### Step 7: Substitute \( K \) back to find \( \alpha \)
Substituting \( K = 48 \) into \( K^3 = 9\alpha + 10 \):
\[ 48^3 = 9\alpha + 10 \]
Calculating \( 48^3 = 110592 \):
\[ 110592 = 9\alpha + 10 \implies 9\alpha = 110582 \implies \alpha = \frac{110582}{9} \approx 12286.89 \]

### Step 8: Calculate \( \alpha^2 + k^2 \)
Now we can compute \( \alpha^2 + K^2 \):
\[ K^2 = 48^2 = 2304 \]
Calculating \( \alpha^2 \):
\[ \alpha^2 \approx 12286.89^2 \approx 150000000 \]
Thus,
\[ \alpha^2 + K^2 \approx 150000000 + 2304 \approx 150000000 \]

### Final Answer
The value of \( \alpha^2 + k^2 \) is approximately \( 150000000 \).