Question

Let $A = \begin{pmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{pmatrix}$ is the adjoint of a $3 \times 3$ matrix $A$ and $|A| = 4$, then $\alpha$ is equal to:

• A. 4
• B. 5
• C. 11
• D. 0

Answer 1

The Correct Option is C

The adjoint (or adjugate) of a matrix $A$ is related to the determinant of $A$ by the property: \[ \text{adj}(A) \cdot A = |A| \cdot I, \] where $|A|$ is the determinant of $A$, and $I$ is the identity matrix. Here, $P$ is the adjoint of $A$, and $|A| = 4$. Thus: \[ |P| = |A|^{n-1} = 4^{3-1} = 4^2 = 16. \] Compute $|P|$ directly using the determinant formula: \[ |P| = \begin{vmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{vmatrix}. \] Expanding along the first row: \[ |P| = 1 \cdot \begin{vmatrix} 3 & 3 \\ 4 & 4 \end{vmatrix} - \alpha \cdot \begin{vmatrix} 1 & 3 \\ 2 & 4 \end{vmatrix} + 3 \cdot \begin{vmatrix} 1 & 3 \\ 2 & 4 \end{vmatrix}. \] Simplify each minor: \[ \begin{vmatrix} 3 & 3 \\ 4 & 4 \end{vmatrix} = (3)(4) - (3)(4) = 0, \] \[ \begin{vmatrix} 1 & 3 \\ 2 & 4 \end{vmatrix} = (1)(4) - (3)(2) = 4 - 6 = -2. \] Substitute back: \[ |P| = 1 \cdot 0 - \alpha \cdot (-2) + 3 \cdot (-2) = 0 + 2\alpha - 6 = 2\alpha - 6. \] Equating $|P|$ to 16: \[ 2\alpha - 6 = 16 \implies 2\alpha = 22 \implies \alpha = 11. \] Hence, $\alpha = 11$.