Question

The element in 1st row and 2nd column of the inverse of the matrix \(\left[ \begin{array}{ccc}1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0\end{array} \right]\) is ...

• A. \(-\frac{2}{5}\)
• B. \(-\frac{10}{3}\)
• C. \(-\frac{2}{25}\)
• D. \(\frac{1}{25}\)

Hint:

The (1,2) entry of the inverse is the cofactor C21 divided by the determinant. Find both.

Answer 1

The Correct Option is A

Step 1: Formula
\(A^{-1}=\frac{1}{|A|}\,\text{adj}A\), and \((\text{adj}A)_{12}=C_{21}\), the cofactor of the element in row 2, column 1.

Step 2: Find the determinant
\[ |A|=1(0\cdot0-(-5)(5))-3((-3)(0)-(-5)(2))+(-2)((-3)(5)-0\cdot2) \] \[ =1(25)-3(10)-2(-15)=25-30+30=25 \]

Step 3: Find C21
Remove row 2 and column 1: the minor is \(\begin{vmatrix}3&-2\\5&0\end{vmatrix}=0-(-10)=10\). So \(C_{21}=(-1)^{2+1}\cdot10=-10\).

Step 4: Required element
\[ (A^{-1})_{12}=\frac{C_{21}}{|A|}=\frac{-10}{25}=-\frac{2}{5} \]

Step 5: Check the options
Options (B) and (C) come from using a wrong determinant or cofactor. Option (D) has the wrong sign and size. Using \(C_{12}\) instead of \(C_{21}\) is a common slip, because the adjoint is the transpose of the cofactor matrix.

Final Answer:
The element is \(-\frac{2}{5}\), option (A). \[ \boxed{-\dfrac{2}{5}} \]