For the matrix `A=({:(-6, 0, 3), (-2, 1, 1), (-4, -5, 2):}),A*adjA=`

Options

A. `A^(T)`
B. `I_(3)`
C. Orthogonal matrix
D. `O_(3)`

Answer

To solve the problem of finding \( A \times \text{adj}(A) \) for the given matrix \( A \), we will follow these steps:

### Step 1: Define the Matrix \( A \)
The matrix \( A \) is given as:
\[ A = \begin{pmatrix} -6 & 0 & 3 \\ -2 & 1 & 1 \\ -4 & -5 & 2 \end{pmatrix} \]

### Step 2: Calculate the Determinant of \( A \)
To find \( \text{adj}(A) \), we first need to calculate the determinant of \( A \). The determinant of a \( 3 \times 3 \) matrix is calculated as follows:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix \( A \):
- \( a = -6, b = 0, c = 3 \)
- \( d = -2, e = 1, f = 1 \)
- \( g = -4, h = -5, i = 2 \)

Calculating the determinant:
\[ \text{det}(A) = -6 \cdot (1 \cdot 2 - 1 \cdot (-5)) + 0 + 3 \cdot (-2 \cdot (-5) - 1 \cdot (-4)) \]
\[ = -6 \cdot (2 + 5) + 3 \cdot (10 + 4) \]
\[ = -6 \cdot 7 + 3 \cdot 14 \]
\[ = -42 + 42 = 0 \]

### Step 3: Find the Adjoint of \( A \)
Since the determinant of \( A \) is 0, we can conclude that the adjoint of \( A \) will also be the zero matrix. However, let's compute the adjoint for completeness.

1. **Calculate the Minors:**
- Minor of \( -6 \):
\[ M_{11} = \begin{vmatrix} 1 & 1 \\ -5 & 2 \end{vmatrix} = (1)(2) - (1)(-5) = 2 + 5 = 7 \]
- Minor of \( 0 \):
\[ M_{12} = \begin{vmatrix} -2 & 1 \\ -4 & 2 \end{vmatrix} = (-2)(2) - (1)(-4) = -4 + 4 = 0 \]
- Minor of \( 3 \):
\[ M_{13} = \begin{vmatrix} -2 & 1 \\ -4 & -5 \end{vmatrix} = (-2)(-5) - (1)(-4) = 10 + 4 = 14 \]
- Minor of \( -2 \):
\[ M_{21} = \begin{vmatrix} 0 & 3 \\ -5 & 2 \end{vmatrix} = (0)(2) - (3)(-5) = 0 + 15 = 15 \]
- Minor of \( 1 \):
\[ M_{22} = \begin{vmatrix} -6 & 3 \\ -4 & 2 \end{vmatrix} = (-6)(2) - (3)(-4) = -12 + 12 = 0 \]
- Minor of \( 1 \):
\[ M_{23} = \begin{vmatrix} -6 & 0 \\ -4 & -5 \end{vmatrix} = (-6)(-5) - (0)(-4) = 30 \]
- Minor of \( -4 \):
\[ M_{31} = \begin{vmatrix} 0 & 3 \\ 1 & 1 \end{vmatrix} = (0)(1) - (3)(1) = -3 \]
- Minor of \( -5 \):
\[ M_{32} = \begin{vmatrix} -6 & 3 \\ -2 & 1 \end{vmatrix} = (-6)(1) - (3)(-2) = -6 + 6 = 0 \]
- Minor of \( 2 \):
\[ M_{33} = \begin{vmatrix} -6 & 0 \\ -2 & 1 \end{vmatrix} = (-6)(1) - (0)(-2) = -6 \]

2. **Form the Matrix of Minors:**
\[ \text{Minors} = \begin{pmatrix} 7 & 0 & 14 \\ 15 & 0 & 30 \\ -3 & 0 & -6 \end{pmatrix} \]

3. **Calculate the Cofactor Matrix:**
- Apply the checkerboard pattern of signs to the minors:
\[ \text{Cofactors} = \begin{pmatrix} 7 & 0 & 14 \\ -15 & 0 & -30 \\ -3 & 0 & -6 \end{pmatrix} \]

4. **Transpose to Get the Adjoint:**
\[ \text{adj}(A) = \begin{pmatrix} 7 & -15 & -3 \\ 0 & 0 & 0 \\ 14 & -30 & -6 \end{pmatrix} \]

### Step 4: Calculate \( A \times \text{adj}(A) \)
Since \( \text{det}(A) = 0 \), we can conclude that:
\[ A \times \text{adj}(A) = 0 \]
Thus, \( A \times \text{adj}(A) \) results in the null matrix.

### Final Answer:
\[ A \times \text{adj}(A) = \text{Null Matrix} \]