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

1. Understand the problem:

We are given that matrix \( P \) is the adjoint of a \( 3 \times 3 \) matrix \( A \), and the determinant of \( A \) is 4. We need to find the value of \( \alpha \) in matrix \( P \).

2. Recall the relationship between a matrix and its adjoint:

For any square matrix \( A \), the following relationship holds:

\[ A \cdot \text{adj}(A) = \text{adj}(A) \cdot A = |A| \cdot I \]

where \( \text{adj}(A) \) is the adjoint of \( A \), \( |A| \) is its determinant, and \( I \) is the identity matrix.

3. Apply the property to the given matrices:

Since \( P = \text{adj}(A) \) and \( |A| = 4 \), we have:

\[ A \cdot P = 4I \]

Taking determinants on both sides:

\[ |A \cdot P| = |4I| \]

\[ |A| \cdot |P| = 4^3 \cdot |I| = 64 \]

Since \( |A| = 4 \), we get:

\[ 4 \cdot |P| = 64 \implies |P| = 16 \]

4. Compute the determinant of \( P \):

Given \( P = \begin{bmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{bmatrix} \), its determinant is:

\[ |P| = 1 \cdot (3 \cdot 4 - 3 \cdot 4) - \alpha \cdot (1 \cdot 4 - 3 \cdot 2) + 3 \cdot (1 \cdot 4 - 3 \cdot 2) \]

Simplify each term:

\[ |P| = 1 \cdot (12 - 12) - \alpha \cdot (4 - 6) + 3 \cdot (4 - 6) \]

\[ |P| = 0 + 2\alpha - 6 = 2\alpha - 6 \]

5. Solve for \( \alpha \):

From step 3, we know \( |P| = 16 \):

\[ 2\alpha - 6 = 16 \implies 2\alpha = 22 \implies \alpha = 11 \]

Correct Answer: (C) 11

Answer 2

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, $\alpha = 11$.