Question

The sum of three numbers is 6. Twice the third number, when added to the first number gives 7. On adding the sum of the second and third numbers to thrice the first number, we get 12. The above situation can be represented in matrix form as \( AX = B \). Then \( | \text{adj } A | \) is equal to

• A. \( -4 \)
• B. 4
• C. \( -64 \)
• D. 16

Hint:

For an \(n \times n\) matrix, \( |\text{adj } A| = |A|^{n-1} \). For \(3 \times 3\), it becomes \( |A|^2 \).

Answer 1

The Correct Option is D

Step 1: Assume the variables.
Let the three numbers be \( x, y, z \).

Step 2: Form the equations.
From the given conditions:
\[ x + y + z = 6, \] \[ x + 2z = 7, \] \[ 3x + y + z = 12. \]

Step 3: Write in matrix form \( AX = B \).
\[ A = \begin{pmatrix} 1 & 1 & 1 1 & 0 & 2 3 & 1 & 1 \end{pmatrix}, \quad X = \begin{pmatrix} x y z \end{pmatrix}, \quad B = \begin{pmatrix} 6 7 12 \end{pmatrix}. \]

Step 4: Find determinant of matrix \( A \).
\[ |A| = \begin{vmatrix} 1 & 1 & 1 1 & 0 & 2 3 & 1 & 1 \end{vmatrix}. \]
Expanding along first row:
\[ |A| = 1\begin{vmatrix} 0 & 2 1 & 1 \end{vmatrix} -1\begin{vmatrix} 1 & 2 3 & 1 \end{vmatrix} +1\begin{vmatrix} 1 & 0 3 & 1 \end{vmatrix}. \]
\[ = 1(0-2) -1(1-6) +1(1-0). \]
\[ = (-2) -(-5) + 1. \]
\[ = -2 + 5 + 1 = 4. \]

Step 5: Use property of adjoint.
For a \(3 \times 3\) matrix:
\[ |\text{adj } A| = |A|^{n-1} = |A|^2. \]

Step 6: Substitute value.
\[ |\text{adj } A| = 4^2. \]
\[ = 16. \]

Step 7: Final conclusion.
Thus, the required value is 16.
Final Answer:
\[ \boxed{16}. \]