Three matrices are given as `A = [{:( alpha ^(2) ,4,6),(9,beta ^(4) , 7),( 1,2,2gamma ^(2)):}] B= [{:( 2beta ^(2),-1,0),( 2,gamma^(2)-2gamma ,1),( 1,9,2alpha-1):}] C= [{:( gamma , 2,1),(1,alpha , 1),(2,0,beta ) :}]` If Tr(A)=Tr (B) -1 and `alpha, beta , gamma in R ` then det( c) can be

Options

A. `-1`
B. `2`
C. 5
D. 0

Answer

To solve the problem, we need to find the determinant of the matrix \( C \) given the conditions on the matrices \( A \) and \( B \). Let's proceed step by step.

### Step 1: Calculate the Trace of Matrices A and B

The trace of a matrix is the sum of its diagonal elements.

For matrix \( A \):
\[ A = \begin{pmatrix} \alpha^2 & 4 & 6 \\ 9 & \beta^4 & 7 \\ 1 & 2 & 2\gamma^2 \end{pmatrix} \]
The trace \( \text{Tr}(A) \) is:
\[ \text{Tr}(A) = \alpha^2 + \beta^4 + 2\gamma^2 \]

For matrix \( B \):
\[ B = \begin{pmatrix} 2\beta^2 & -1 & 0 \\ 2 & \gamma^2 - 2\gamma & 1 \\ 1 & 9 & 2\alpha - 1 \end{pmatrix} \]
The trace \( \text{Tr}(B) \) is:
\[ \text{Tr}(B) = 2\beta^2 + (\gamma^2 - 2\gamma) + (2\alpha - 1) = 2\beta^2 + \gamma^2 - 2\gamma + 2\alpha - 1 \]

### Step 2: Set Up the Equation Based on Given Condition

We are given that:
\[ \text{Tr}(A) = \text{Tr}(B) - 1 \]
Substituting the traces we calculated:
\[ \alpha^2 + \beta^4 + 2\gamma^2 = (2\beta^2 + \gamma^2 - 2\gamma + 2\alpha - 1) - 1 \]
This simplifies to:
\[ \alpha^2 + \beta^4 + 2\gamma^2 = 2\beta^2 + \gamma^2 - 2\gamma + 2\alpha - 2 \]

### Step 3: Rearranging the Equation

Rearranging gives us:
\[ \alpha^2 + \beta^4 - 2\beta^2 + \gamma^2 + 2\gamma - 2\alpha = -2 \]
This can be rewritten as:
\[ \alpha^2 - 2\alpha + \beta^4 - 2\beta^2 + \gamma^2 + 2\gamma + 2 = 0 \]

### Step 4: Completing the Square

We can complete the square for each variable:
1. For \( \alpha \):
\[ \alpha^2 - 2\alpha = (\alpha - 1)^2 - 1 \]
2. For \( \beta \):
\[ \beta^4 - 2\beta^2 = (\beta^2 - 1)^2 - 1 \]
3. For \( \gamma \):
\[ \gamma^2 + 2\gamma = (\gamma + 1)^2 - 1 \]

Substituting these into the equation gives:
\[ (\alpha - 1)^2 + (\beta^2 - 1)^2 + (\gamma + 1)^2 - 3 = 0 \]
Thus, we have:
\[ (\alpha - 1)^2 + (\beta^2 - 1)^2 + (\gamma + 1)^2 = 3 \]

### Step 5: Finding Possible Values for \( \alpha, \beta, \gamma \)

The equation indicates that:
- \( \alpha - 1 = 0 \) implies \( \alpha = 1 \)
- \( \beta^2 - 1 = 0 \) implies \( \beta = \pm 1 \)
- \( \gamma + 1 = 0 \) implies \( \gamma = -1 \)

### Step 6: Substitute Values into Matrix C

Now substituting \( \alpha = 1 \), \( \beta = 1 \) or \( -1 \), and \( \gamma = -1 \) into matrix \( C \):
\[ C = \begin{pmatrix} \gamma & 2 & 1 \\ 1 & \alpha & 1 \\ 2 & 0 & \beta \end{pmatrix} \]
Substituting \( \gamma = -1 \), \( \alpha = 1 \), and \( \beta = 1 \):
\[ C = \begin{pmatrix} -1 & 2 & 1 \\ 1 & 1 & 1 \\ 2 & 0 & 1 \end{pmatrix} \]

### Step 7: Calculate the Determinant of Matrix C

Using the determinant formula for a 3x3 matrix:
\[ \text{det}(C) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
Where:
\[ C = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]
Thus:
\[ \text{det}(C) = -1(1 \cdot 1 - 1 \cdot 0) - 2(1 \cdot 1 - 1 \cdot 2) + 1(1 \cdot 0 - 1 \cdot 2) \]
Calculating this gives:
\[ = -1(1) - 2(1 - 2) + 1(0 - 2) \]
\[ = -1 - 2(-1) - 2 = -1 + 2 - 2 = -1 \]

### Final Result

Thus, the determinant of matrix \( C \) can be:
\[ \text{det}(C) = -1 \]