Question

# For what value of k, the matrix $\left( \begin{matrix} 2-k & 4 \\ -5 & 1 \\\end{matrix} \right)$ is not invertible?

Hint:Find the determinant of the matrix. As itâ€™s not invertible it is equal to zero. Consider the matrix A. Find the determinant of A and equal it to zero. From that, find the value of k.

A square matrix is said to be invertible if its inverse exists and is said to be non-invertible if its determinant is equal to zero.
$\therefore$A square matrix that is not invertible is called singular/ degenerate. Non-square matrices $\left( m\times n \right)$ where $m\ne n$ do not have an inverse. In some cases there are left inverse and right inverse.
Given the matrix $\left( \begin{matrix} 2-k & 4 \\ -5 & 1 \\ \end{matrix} \right)$
Letâ€™s write it as $A=\left( \begin{matrix} 2-k & 4 \\ -5 & 1 \\ \end{matrix} \right)$
As the matrix is non-invertible, the determinant of A is zero.
$\left| A \right|=0$
$\left| \left( \begin{matrix} 2-k & 3 \\ -5 & 1 \\ \end{matrix} \right) \right|=0$
To work out the determinant, multiply A by the determinant of the matrix that is not in a row or column and then sum up.
\begin{align} & \left[ \left( 2-k \right)\left( 1 \right) \right]-\left[ \left( 3 \right)\left( -5 \right) \right] \\ & =\left( 2-k \right)-\left( -15 \right) \\ & =2-k+15=0 \\ & \therefore k=17 \\ \end{align}
$\therefore$Value of k = 17.

Note:
For the case of a non-invertible matrix, we took$\left| A \right|=0$.
If the matrix was invertible, then $\left| A \right|\ne 0$but${{A}^{-1}}$.