If \( A=\begin{bmatrix}4&-1&\lambda\\2&0&6\\1&-2&7\end{bmatrix} \), then \( \displaystyle\frac{1}{2}A^{-1} \) exist if
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \lambda\ne 14 \)
\( \frac{1}{2}A^{-1} \) exists exactly when \( A^{-1} \) exists, that is, when \( |A|\ne0 \). Expanding along the second row:
\( \displaystyle |A|=-2\begin{vmatrix}-1&\lambda\\-2&7\end{vmatrix}-6\begin{vmatrix}4&-1\\1&-2\end{vmatrix} \)
\( =-2(-7+2\lambda)-6(-8+1) \)
\( =56-4\lambda \)
So
\( 56-4\lambda\ne0 \)
\( \lambda\ne14 \)