Question

Let \[ A = \begin{pmatrix} 2 & -1 & 1 \\ -1 & 0 & 2 \\ 1 & -2 & -1 \end{pmatrix} \] and let \( B = \frac{1}{|A|} A \). Then the value of \( |B| \) is equal to:

• A. \( \frac{1}{9} \)
• B. \( \frac{1}{11} \)
• C. \( \frac{1}{81} \)
• D. \( \frac{1}{121} \)
• E. \( 1 \)

Hint:

When a matrix is scaled by a scalar \( k \), the determinant of the matrix is scaled by \( k^n \), where \( n \) is the order of the matrix. In this case, the matrix is scaled by \( \frac{1}{|A|} \), and the determinant of matrix \( B \) is \( \frac{1}{|A|^2} \).

Answer 1

The Correct Option is C

Given matrix \( A \):

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

We are asked to find the determinant of matrix \( B \), which is given as:

\[ B = \frac{1}{|A|} A \]

The determinant of a scalar multiple of a matrix is the scalar raised to the power of the size of the matrix times the determinant of the original matrix. Therefore,

\[ |B| = \left|\frac{1}{|A|} A\right| = \frac{1}{|A|^3} |A| \]

Thus, \( |B| = \frac{1}{|A|^2} \).

Now, let's calculate the determinant of matrix \( A \):

\[ |A| = \begin{vmatrix} 2 & -1 & 1 \\ -1 & 0 & 2 \\ 1 & -2 & -1 \end{vmatrix} \]

We use the cofactor expansion along the first row:

\[ |A| = 2 \begin{vmatrix} 0 & 2 \\ -2 & -1 \end{vmatrix} - (-1) \begin{vmatrix} -1 & 2 \\ 1 & -1 \end{vmatrix} + 1 \begin{vmatrix} -1 & 0 \\ 1 & -2 \end{vmatrix} \]

Calculating the 2x2 determinants:

\[ \begin{vmatrix} 0 & 2 \\ -2 & -1 \end{vmatrix} = (0)(-1) - (2)(-2) = 4 \]

\[ \begin{vmatrix} -1 & 2 \\ 1 & -1 \end{vmatrix} = (-1)(-1) - (2)(1) = 1 - 2 = -1 \]

\[ \begin{vmatrix} -1 & 0 \\ 1 & -2 \end{vmatrix} = (-1)(-2) - (0)(1) = 2 \]

Substitute these into the cofactor expansion:

\[ |A| = 2(4) + 1(-1) + 1(2) = 8 - 1 + 2 = 9 \]

Thus, \( |A| = 9 \).

Now, using the formula for \( |B| \):

\[ |B| = \frac{1}{|A|^2} = \frac{1}{9^2} = \frac{1}{81} \]

Thus, the value of \( |B| \) is \( \frac{1}{81} \).

Thus, the correct answer is option (C), \( \frac{1}{81} \).