Question

If \(B(\text{adj } B) = \begin{bmatrix \frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{3} \end{bmatrix}\), then the value of \(\det (B^{-1}) =\)}

• A. \(\frac{1}{3}\)
• B. \(\frac{1}{9}\)
• C. \(3\)
• D. \(9\)

Hint:

Whenever a diagonal matrix with identical diagonal elements \(k\) is given as \(B(\text{adj } B)\), simply identify \(|B| = k\). The determinant of the inverse is then instantly found as \(\frac{1}{k}\).

Answer 1

The Correct Option is C

Concept:
• For any non-singular square matrix \(B\) of order \(n\), the fundamental relation between the matrix and its adjoint is given by \(B(\text{adj } B) = |B| I_n\), where \(I_n\) is the identity matrix of order \(n\).
• The determinant of the inverse of a matrix is the reciprocal of the determinant of the matrix: \(\det(B^{-1}) = |B^{-1}| = \frac{1}{|B|}\).

Step 1: Rewrite the given matrix in terms of the identity matrix
The given equation is: \[ B(\text{adj } B) = \begin{bmatrix} \frac{1}{3} & 0 & 0 0 & \frac{1}{3} & 0 0 & 0 & \frac{1}{3} \end{bmatrix} \] Factoring out the scalar \(\frac{1}{3}\) from the matrix: \[ B(\text{adj } B) = \frac{1}{3} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \frac{1}{3} I_3 \]

Step 2: Find the determinant of \(B\)
Comparing this with the property \(B(\text{adj } B) = |B| I_3\): \[ |B| I_3 = \frac{1}{3} I_3 \] Therefore: \[ |B| = \frac{1}{3} \]

Step 3: Calculate \(\det(B^{-1})\)
Using the property of determinants for inverse matrices: \[ \det(B^{-1}) = \frac{1}{|B|} = \frac{1}{\frac{1}{3}} = 3 \]