If the matrix $\begin{bmatrix}2&-1&3\\λ&0&7\\-1&1&4\end{bmatrix}$ is not invertible, then value of $λ$ is

1

The correct answer is Option (3) → 1

Matrix:

$A=\begin{pmatrix} 2 & -1 & 3 \\ \lambda & 0 & 7 \\ -1 & 1 & 4 \end{pmatrix}$

For the matrix to be not invertible:

$\det(A)=0$

Compute determinant:

$\det(A) = 2\begin{vmatrix}0 & 7 \\ 1 & 4\end{vmatrix} - (-1)\begin{vmatrix}\lambda & 7 \\ -1 & 4\end{vmatrix} + 3\begin{vmatrix}\lambda & 0 \\ -1 & 1\end{vmatrix}$

$= 2(0\cdot4 - 7\cdot1) + 1(\lambda\cdot4 - 7(-1)) + 3(\lambda\cdot1 - 0)$

$= 2(-7) + (4\lambda + 7) + 3\lambda$

$= -14 + 4\lambda + 7 + 3\lambda$

$= 7\lambda - 7$

Set determinant = 0:

$7\lambda - 7 = 0$

$7\lambda = 7$

$\lambda = 1$

The matrix is not invertible when $\lambda = 1$.