Question

Find k if the following matrices are singular
(i) $\left[ \begin{matrix}
   7 & 3 \\
   -2 & k \\
\end{matrix} \right]$ (ii) $\left[ \begin{matrix}
   4 & 3 & 1 \\
   7 & k & 1 \\
   10 & 9 & 1 \\
\end{matrix} \right]$ (iii) $\left[ \begin{matrix}
   k-1 & 2 & 3 \\
   3 & 1 & 2 \\
   1 & -2 & 4 \\
\end{matrix} \right]$

Answer 1

Hint: We first discuss the condition of a matrix being singular. We understand that to be a singular determinant value of a matrix has to be 0. All the given matrices are singular. We find their determinant value and equate it with 0 to find the value of k.

Complete step by step answer:
First, we discuss the condition of a matrix being singular.
If the determinant value of a matrix is 0 then the matrix is singular. For example: if A be a matrix and $\det \left( A \right)=\left| A \right|=0$ then we can claim that A is a singular matrix. All the other matrices are non-singular matrices.
Now we find the determinant value of the given matrices. It’s given they are all singular.
For $X=\left[ \begin{matrix}
   7 & 3 \\
   -2 & k \\
\end{matrix} \right]$, the $\det \left( X \right)=\left| X \right|$ value is 0.
$\left| X \right|=7\times k-\left( -2 \right)\times 3=7k+6$.
So, $7k+6=0\Rightarrow k=\dfrac{-6}{7}$.
Therefore, the value of k for $\left[ \begin{matrix}
   7 & 3 \\
   -2 & k \\
\end{matrix} \right]$ is $\dfrac{-6}{7}$.
For $Y=\left[ \begin{matrix}
   4 & 3 & 1 \\
   7 & k & 1 \\
   10 & 9 & 1 \\
\end{matrix} \right]$, the $\det \left( Y \right)=\left| Y \right|$ value is 0. Expanding through third column we get
$\begin{align}
  & \Rightarrow \left| Y \right|=1\times \left| \begin{matrix}
   7 & k \\
   10 & 9 \\
\end{matrix} \right|-1\times \left| \begin{matrix}
   4 & 3 \\
   10 & 9 \\
\end{matrix} \right|+1\times \left| \begin{matrix}
   4 & 3 \\
   7 & k \\
\end{matrix} \right| \\
 & \Rightarrow \left| Y \right|=\left( 63-10k \right)-\left( 36-30 \right)+\left( 4k-21 \right)=36-6k \\
\end{align}$.
So, $36-6k=0\Rightarrow k=\dfrac{36}{6}=6$.
Therefore, the value of k for $\left[ \begin{matrix}
   4 & 3 & 1 \\
   7 & k & 1 \\
   10 & 9 & 1 \\
\end{matrix} \right]$ is 6.
For $Z=\left[ \begin{matrix}
   k-1 & 2 & 3 \\
   3 & 1 & 2 \\
   1 & -2 & 4 \\
\end{matrix} \right]$, the $\det \left( Z \right)=\left| Z \right|$ value is 0. Expanding through third column we get
$\begin{align}
  & \Rightarrow \left| Z \right|=3\times \left| \begin{matrix}
   3 & 1 \\
   1 & -2 \\
\end{matrix} \right|-2\times \left| \begin{matrix}
   k-1 & 2 \\
   1 & -2 \\
\end{matrix} \right|+4\times \left| \begin{matrix}
   k-1 & 2 \\
   3 & 1 \\
\end{matrix} \right| \\
 & \Rightarrow \left| Z \right|=3\left( -6-1 \right)-2\left( 2-2k-2 \right)+4\left( k-1-6 \right)=8k-49 \\
\end{align}$.
So, $8k-49=0\Rightarrow k=\dfrac{49}{8}$.
Therefore, the value of k for $\left[ \begin{matrix}
   k-1 & 2 & 3 \\
   3 & 1 & 2 \\
   1 & -2 & 4 \\
\end{matrix} \right]$ is $\dfrac{49}{8}$.

Note: Even though we are dealing with a singular and non-singular matrix the notion is related to its determinant value. Also, we can’t get confused with the concept of zero matrix and determinant value of a matrix being 0. First one is a matrix and the second one is a value.