If the matrix \(\begin{bmatrix}2&-1&3\\\lambda&0&7\\-1&1&4\end{bmatrix}\) is not invertible, then value of \(\lambda\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1
A square matrix is not invertible exactly when its determinant is zero. Expanding along the first row:
\( \displaystyle \begin{vmatrix}2&-1&3\\\lambda&0&7\\-1&1&4\end{vmatrix} = 2\begin{vmatrix}0&7\\1&4\end{vmatrix} + 1\begin{vmatrix}\lambda&7\\-1&4\end{vmatrix} + 3\begin{vmatrix}\lambda&0\\-1&1\end{vmatrix} \)
\( = 2(0 - 7) + (4\lambda + 7) + 3\lambda \)
\( = 7\lambda - 7 \)
Setting it equal to zero:
\(7\lambda - 7 = 0\)
\(\lambda = 1\)