Question

The element in the third row and the second column in the inverse matrix of a matrix \(\left[ \begin{array}{ccc}1 & 3 & 3 \\ 3 & 1 & 3 \\ 3 & 3 & 4\end{array} \right]\) is

• A. \(\frac{3}{2}\)
• B. \(\frac{2}{3}\)
• C. \(\frac{-3}{2}\)
• D. \(\frac{-2}{3}\)

Hint:

The (3,2) entry of the inverse is the cofactor C23 divided by the determinant.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
\(A^{-1} = \dfrac{\operatorname{adj}A}{|A|}\), and the adjoint is the transpose of the cofactor matrix. So the entry in row 3, column 2 of \(A^{-1}\) is \(C_{23}/|A|\).

Step 2: Determinant
\[ |A| = 1(4 - 9) - 3(12 - 9) + 3(9 - 3) = -5 - 9 + 18 = 4 \]

Step 3: Cofactor
The minor \(M_{23}\) removes row 2 and column 3, leaving \(\begin{vmatrix} 1 & 3 \\ 3 & 3 \end{vmatrix} = 3 - 9 = -6\).
\[ C_{23} = (-1)^{2+3}(-6) = 6 \]
\[ \text{Entry} = \frac{6}{4} = \frac32 \]
Option (B) 2/3 and (D) come from inverting the fraction, and (C) from missing the sign.

Final Answer:
The required element is \(\frac32\), option (A). \[ \boxed{\frac{3}{2}} \]