Let `A=[(x,2,-3),(-1,3,-2),(2,-1,1)]` be a matrix and `|adj(adjA)|=(12)^(4)`, then the sum of all the values of x is equal to

Options

A. `-24`
B. 24
C. `-18`
D. 1

Answer

To solve the problem, we need to find the values of \( x \) in the matrix \( A \) such that \( |adj(adjA)| = (12)^4 \). Let's go through the solution step by step.

### Step 1: Understanding the relationship between the adjoint and determinant
We know that for any square matrix \( A \) of order \( n \):
\[ |adj(A)| = |A|^{n-1} \]
For our matrix \( A \), which is \( 3 \times 3 \) (so \( n = 3 \)):
\[ |adj(A)| = |A|^{3-1} = |A|^2 \]

### Step 2: Finding \( |adj(adjA)| \)
Using the property of the adjoint again:
\[ |adj(adjA)| = |adj(A)|^{n-1} = |adj(A)|^{2} \]
Substituting the expression we found earlier:
\[ |adj(adjA)| = (|A|^2)^2 = |A|^4 \]

### Step 3: Setting up the equation
We are given that:
\[ |adj(adjA)| = (12)^4 \]
Thus, we can equate:
\[ |A|^4 = (12)^4 \]
Taking the fourth root of both sides gives us:
\[ |A| = 12 \quad \text{or} \quad |A| = -12 \]
Since the determinant can be negative or positive, we can consider both cases.

### Step 4: Finding the determinant of matrix \( A \)
The matrix \( A \) is given as:
\[ A = \begin{pmatrix} x & 2 & -3 \\ -1 & 3 & -2 \\ 2 & -1 & 1 \end{pmatrix} \]
We calculate the determinant \( |A| \) using the formula for the determinant of a \( 3 \times 3 \) matrix:
\[ |A| = x \begin{vmatrix} 3 & -2 \\ -1 & 1 \end{vmatrix} - 2 \begin{vmatrix} -1 & -2 \\ 2 & 1 \end{vmatrix} - 3 \begin{vmatrix} -1 & 3 \\ 2 & -1 \end{vmatrix} \]

Calculating the minors:
1. \( \begin{vmatrix} 3 & -2 \\ -1 & 1 \end{vmatrix} = (3)(1) - (-2)(-1) = 3 - 2 = 1 \)
2. \( \begin{vmatrix} -1 & -2 \\ 2 & 1 \end{vmatrix} = (-1)(1) - (-2)(2) = -1 + 4 = 3 \)
3. \( \begin{vmatrix} -1 & 3 \\ 2 & -1 \end{vmatrix} = (-1)(-1) - (3)(2) = 1 - 6 = -5 \)

Substituting back into the determinant formula:
\[ |A| = x(1) - 2(3) - 3(-5) = x - 6 + 15 = x + 9 \]

### Step 5: Setting up the equations
Now we have:
\[ |A| = x + 9 \]
We set this equal to the two cases we derived:
1. \( x + 9 = 12 \)
2. \( x + 9 = -12 \)

### Step 6: Solving for \( x \)
1. From \( x + 9 = 12 \):
\[ x = 12 - 9 = 3 \]
2. From \( x + 9 = -12 \):
\[ x = -12 - 9 = -21 \]

### Step 7: Finding the sum of all values of \( x \)
Now we sum the values obtained:
\[ 3 + (-21) = -18 \]

### Final Answer
The sum of all the values of \( x \) is:
\[ \boxed{-18} \]