Find the value of $\lambda $ so that the matrix $\left[ \begin{matrix}
   5-\lambda & \lambda +1 \\
   2 & 4 \\
\end{matrix} \right]$ may be singular.

Hint: We know that the determinant of singular matrix is 0 so find the determinant of the given matrix which will be calculated by multiplying $5-\lambda $ with 4 and then find the multiplication of $\lambda +1$ with 2 and subtract the result of later multiplication with the former one and equate it to 0. Solve this equation and get the value of $\lambda $.

Complete step-by-step answer:
The matrix given in the above problem is:
$\left[ \begin{matrix}
   5-\lambda & \lambda +1 \\
   2 & 4 \\
\end{matrix} \right]$
Let us assume this matrix as “A” so equating the above matrix to “A” we get,
$A=\left[ \begin{matrix}
   5-\lambda & \lambda +1 \\
   2 & 4 \\
\end{matrix} \right]$
It is also mentioned in the above problem that the matrix is a singular matrix and we know that the determinant of a singular matrix is 0.
$\left| A \right|=0$
Finding the determinant of matrix A we get,
$\left| A \right|=\left( 5-\lambda \right)\left( 4 \right)-2\left( \lambda +1 \right)$
Multiplying 4 with $\left( 5-\lambda \right)$ and 2 with $\left( \lambda +1 \right)$ we get,
$\begin{align}
  & \left| A \right|=\left( 20-4\lambda \right)-\left( 2\lambda +2 \right) \\
 & \Rightarrow \left| A \right|=20-4\lambda -2\lambda -2 \\
 & \Rightarrow \left| A \right|=18-6\lambda \\
\end{align}$
Now, equating the above determinant equal to 0 we get,
$\begin{align}
  & \left| A \right|=18-6\lambda =0 \\
 & \Rightarrow 18-6\lambda =0 \\
\end{align}$
Adding $6\lambda $ on both the sides of the above equation we get,
$18=6\lambda $
Dividing 6 on both the sides of the above equation we get,
$\begin{align}
  & \dfrac{18}{6}=\lambda \\
 & \Rightarrow \lambda =3 \\
\end{align}$
From the above solution, we have solved the value of $\lambda $ as 3.
Hence, the value of $\lambda =3$.

Note: You can verify the value of $\lambda $ that you have got above by substituting the value of $\lambda $ in the matrix given and see whether the determinant matrix is coming 0 or not.
The given matrix is:
$\left[ \begin{matrix}
   5-\lambda & \lambda +1 \\
   2 & 4 \\
\end{matrix} \right]$
Substituting the value of $\lambda $ as 3 in the above matrix we get,
$\begin{align}
  & \left[ \begin{matrix}
   5-3 & 3+1 \\
   2 & 4 \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   2 & 4 \\
   2 & 4 \\
\end{matrix} \right] \\
\end{align}$
Now, evaluating the determinant of the above matrix we get,
$\begin{align}
  & 2\left( 4 \right)-4\left( 2 \right) \\
 & =8-8 \\
 & =0 \\
\end{align}$
Hence, on substituting the value of $\lambda =3$ in the matrix we have got the determinant value 0.