Let \( A=\begin{bmatrix}1 & -2 & 3\\1 & 2 & 1\\\lambda & 2 & -3\end{bmatrix} \). If \( A^{-1} \) does not exist, then \( \lambda= \)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → −1
\( A^{-1} \) does not exist exactly when \( |A|=0 \). Expanding along the first row:
\( \displaystyle |A|=1\begin{vmatrix}2 & 1\\2 & -3\end{vmatrix}-(-2)\begin{vmatrix}1 & 1\\\lambda & -3\end{vmatrix}+3\begin{vmatrix}1 & 2\\\lambda & 2\end{vmatrix} \)
\( \displaystyle =1(-6-2)+2(-3-\lambda)+3(2-2\lambda) \)
\( \displaystyle =-8-6-2\lambda+6-6\lambda \)
\( \displaystyle =-8-8\lambda \)
Setting \( |A|=0 \):
\( \displaystyle -8-8\lambda=0 \)
\( \displaystyle \lambda=-1 \)