Let `A=[{:(2, -1, 3), (4, 0, 2), (-3, 2, 6):}]`, find det (adj A).

Options

A. 2161
B. 2116
C. 2611
D. 1261

Answer

To find the determinant of the adjoint of the matrix \( A \), we can use the formula:

\[ \text{det}(\text{adj } A) = (\text{det } A)^{n-1} \]

where \( n \) is the order of the matrix \( A \). In this case, \( A \) is a \( 3 \times 3 \) matrix, so \( n = 3 \).

### Step 1: Calculate the determinant of matrix \( A \)

The matrix \( A \) is given as:

\[ A = \begin{pmatrix} 2 & -1 & 3 \\ 4 & 0 & 2 \\ -3 & 2 & 6 \end{pmatrix} \]

We can calculate the determinant using the formula for a \( 3 \times 3 \) matrix:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

where the elements are as follows:

\[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} = \begin{pmatrix} 2 & -1 & 3 \\ 4 & 0 & 2 \\ -3 & 2 & 6 \end{pmatrix} \]

Thus, we have:

- \( a = 2, b = -1, c = 3 \)
- \( d = 4, e = 0, f = 2 \)
- \( g = -3, h = 2, i = 6 \)

Now substituting these values into the determinant formula:

\[ \text{det}(A) = 2(0 \cdot 6 - 2 \cdot 2) - (-1)(4 \cdot 6 - 2 \cdot -3) + 3(4 \cdot 2 - 0 \cdot -3) \]

Calculating each term:

1. \( 0 \cdot 6 - 2 \cdot 2 = 0 - 4 = -4 \)
2. \( 4 \cdot 6 - 2 \cdot -3 = 24 + 6 = 30 \)
3. \( 4 \cdot 2 - 0 \cdot -3 = 8 - 0 = 8 \)

Now substituting back:

\[ \text{det}(A) = 2(-4) + 1(30) + 3(8) \]
\[ = -8 + 30 + 24 \]
\[ = 46 \]

### Step 2: Calculate the determinant of the adjoint of \( A \)

Using the formula we mentioned earlier:

\[ \text{det}(\text{adj } A) = (\text{det } A)^{n-1} = (46)^{3-1} = (46)^{2} \]

Calculating \( 46^2 \):

\[ 46^2 = 2116 \]

### Final Answer

Thus, the determinant of the adjoint of matrix \( A \) is:

\[ \text{det}(\text{adj } A) = 2116 \]