Question

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

• A. $\lambda \ne -15$

B. $\lambda \ne -17$

C. $\lambda \ne -16$

D. $\lambda \ne -18$

Answer 1

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 $A=\left[\begin{array}{ccc}\lambda & -1& 4\\ -3& 0& 1\\ -1& 1& 2\end{array}\right]$
Therefore,
$|A|\ne 0$
$$\begin{gathered} \left|\begin{array}{ccc}\lambda & -1& 4\\ -3& 0& 1\\ -1& 1& 2\end{array}\right|\ne 0\Rightarrow [\lambda (0(2)-1(1))-(-1)(-3(2)-1(-1))+4(-3(1)-0(-1))\ne 0 \\ \Rightarrow -\lambda +1(-6+1)+4(-3-0)\ne 0 \\ \Rightarrow -\lambda -5-12\ne 0 \\ \Rightarrow -\lambda -17\ne 0 \\ \Rightarrow \lambda =-17 \end{gathered}$$

Hence, the option B is correct.

Note:
Only if its determinant is nonzero, or $|A|\ne 0$, 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
$(A^{-1}{)}^{-1}=A$
$(A^T{)}^{-1}=(A^{-1}{)}^T$
$det{A}^{-1}$ = $detA$
$AB$ 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.