i.e \(\displaystyle{\left|{A}\right|}={0}\)

Here, the given matrix is \[M=\begin{bmatrix}-4&0&-3\\-4&-4&1\\-14+k&-8&-1\end{bmatrix}\]

So we must have

\(\displaystyle{\left|{M}\right|}={0}\)

\[\begin{bmatrix}-4&0&-3\\-4&-4&1\\-14+k&-8&-1\end{bmatrix}=0\]

Now, expanding the determinant by its 1st row

\(\displaystyle-{4}{\left({4}+{8}\right)}-{0}-{3}{\left({32}+{4}{\left(-{14}+{k}\right)}\right)}={0}\)

\(\displaystyle-{48}-{3}{\left({32}-{96}+{4}{k}\right)}={0}\)

\(\displaystyle-{48}-{3}{\left(-{64}+{4}{k}\right)}={0}\)

\(\displaystyle-{48}+{192}-{12}{k}={0}\)

\(\displaystyle{144}-{12}{k}={0}\)

\(\displaystyle{12}{k}={144}\)

\(\displaystyle{k}={12}\)

So ,the required value of k=12 for which the matric is singular.