Question

If \( P = \begin{bmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{bmatrix} \) is the adjoint of a matrix \( A \) and \( \det(A) = 4 \), then the value of \( \alpha \) is:

• A. $3$
• B. $22$
• C. $11$
• D. $4$

Hint:

Use the identity $ \det(adj}(A)) = (\det(A))^{n-1} $. When computing determinants with unknowns, expand carefully and solve algebraically.

Answer 1

The Correct Option is C

Step 1: Recall the relationship between adjoint and determinant.
We know that: $$ A \cdot adj(A) = \det(A) \cdot I \quad \text{and} \quad \det(adj(A)) = (\det(A))^{n-1}, $$ where \( n \) is the size of the matrix. Given: - \( P = adj(A) = \begin{bmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{bmatrix} \), - \( \det(A) = 4 \), - \( n = 3 \). So: $$ \det(P) = (\det(A))^2 = 4^2 = 16. $$ Step 2: Compute \( \det(P) \) explicitly.
Compute: $$ \det(P) = \begin{vmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{vmatrix}. $$ Expand along the first row: $$ \det(P) = 1 \cdot \begin{vmatrix} 3 & 3 \\ 4 & 4 \end{vmatrix} - \alpha \cdot \begin{vmatrix} 1 & 3 \\ 2 & 4 \end{vmatrix} + 3 \cdot \begin{vmatrix} 1 & 3 \\ 2 & 4 \end{vmatrix}. $$ Now compute the minors:
\( \begin{vmatrix} 3 & 3 \\ 4 & 4 \end{vmatrix} = 12 - 12 = 0 \),
\( \begin{vmatrix} 1 & 3 \\ 2 & 4 \end{vmatrix} = 4 - 6 = -2 \).
Substitute back: $$ \det(P) = 1 \cdot 0 - \alpha \cdot (-2) + 3 \cdot (-2) = 0 + 2\alpha - 6 = 2\alpha - 6. $$ Set equal to 16: $$ 2\alpha - 6 = 16 \Rightarrow 2\alpha = 22 \Rightarrow \alpha = 11. $$ Step 3: Final Answer.
$$ \boxed{11} $$