The Matrix $\left[ \begin{matrix} \lambda & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2 \\ \end{matrix} \right]$ is invertible, if

• A. $\lambda \ne -15$

B. $\lambda \ne -17$

C. $\lambda \ne -16$

D. $\lambda \ne -18$

Hint:
Square matrices with an inverse are known as invertible matrices. Only when the determinant of a matrix is not equal to zero, we say that a square matrix is invertible.

Complete step-by-step answer:
We have given that the matrix is invertible which means its determinant is not equal to zero.
Let $A=\left[ \begin{matrix} \lambda & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2 \\ \end{matrix} \right]$
Therefore,
$|A| \neq 0$
$\begin{vmatrix} \lambda & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2 \\ \end{vmatrix} \neq 0

\Rightarrow [\lambda(0(2)-1(1))-(-1)(-3(2)-1(-1))+4(-3(1)-0(-1))\neq0\\
\Rightarrow -\lambda+1(-6+1)+4(-3-0)\neq0\\
\Rightarrow -\lambda-5-12 \neq 0\\
\Rightarrow -\lambda -17\neq0 \\
\Rightarrow \lambda =-17$

Hence, the option B is correct.

Note:
Only if its determinant is nonzero, or $|A|\neq 0$, is a square matrix $A$ invertible. The $n \times n$ square matrix satisfying the necessary condition for a matrix's inverse to exist is known as an invertible matrix in linear algebra. It is also known as a non-singular or non-degenerate matrix.

Additional Information:
Properties of Invertible Matrix:
If A is non-singular matrix, then
$(A ^{-1})^{-1} = A$
$(A^T)^{-1} = (A^{-1})^T$
$det A^{-1}$ = $det A$
$AB$ is nonsingular and $(AB)^{-1} = B^{-1} A^{-1}$ if $A$ and $B$ are nonsingular matrices.
The scalar multiple $kA$ is invertible and $(kA)^{-1} = A^{-1} /k$ if $k$ is any non-zero scalar.