Question

Find which of the given matrices is not invertible.
A. $\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]$
B. $\left[ \begin{matrix}
   -1 & -1 \\
   -1 & 2 \\
\end{matrix} \right]$
C. $\left[ \begin{matrix}
   2 & 3 \\
   4 & 6 \\
\end{matrix} \right]$
D. $\left[ \begin{matrix}
   2 & -2 \\
   1 & 1 \\
\end{matrix} \right]$

Answer 1

Hint: We first define what is the condition for a matrix to be invertible. We establish the concept of a non-singular matrix. Then we find the determinant value of the given matrices. If the matrix is nonsingular then we find the inverse of the matrix.

Complete step-by-step solution
We know that for a matrix A to be invertible, it has to be non-singular which means the determinant value is non-zero. So, we have to find the determinant value of every given matrix.
Let’s assume $A=\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]$. So, $\det \left( A \right)=\left| \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right|$.
Using expansion method, we get $\det \left( A \right)=\left| A \right|=1\times 1-0\times 1=1$. So, A is invertible.
Now we find its inverse. If ${{A}^{-1}}$ be the inverse of A then ${{A}^{-1}}=\dfrac{adj\left( A \right)}{\left| A \right|}$. Here $adj\left( A \right)$ defines the cofactors of elements. If $A=\left[ {{a}_{ij}} \right]$, then $adj\left( A \right)=\left[ {{A}_{ji}} \right]$. Here ${{a}_{ij}}$ represents the element of ${{i}^{th}}$ row and ${{j}^{th}}$ column. ${{A}_{ji}}$ represents the cofactor of the element ${{a}_{ij}}$.
$adj\left( A \right)=\left[ \begin{matrix}
   1 & -1 \\
   0 & 1 \\
\end{matrix} \right]$ and ${{A}^{-1}}=\left[ \begin{matrix}
   1 & -1 \\
   0 & 1 \\
\end{matrix} \right]$ as $\det \left( A \right)=1$.
Let’s assume $B=\left[ \begin{matrix}
   -1 & -1 \\
   -1 & 2 \\
\end{matrix} \right]$. So, $\det \left( B \right)=\left| \begin{matrix}
   -1 & -1 \\
   -1 & 2 \\
\end{matrix} \right|$.
Using expansion method, we get $\det \left( B \right)=\left| B \right|=2\times \left( -1 \right)-\left( -1 \right)\times \left( -1 \right)=-2-1=-3$. So, B is invertible.
Now we find its inverse. If ${{B}^{-1}}$ be the inverse of B then ${{B}^{-1}}=\dfrac{adj\left( B \right)}{\left| B \right|}$. Here $adj\left( B \right)$ defines the cofactors of elements. If $B=\left[ {{b}_{ij}} \right]$, then $adj\left( B \right)=\left[ {{B}_{ji}} \right]$. Here ${{b}_{ij}}$ represents the element of ${{i}^{th}}$ row and ${{j}^{th}}$ column. ${{B}_{ji}}$ represents the cofactor of the element ${{b}_{ij}}$.
$adj\left( B \right)=\left[ \begin{matrix}
   2 & 1 \\
   1 & -1 \\
\end{matrix} \right]$ and ${{B}^{-1}}=\dfrac{-1}{3}\left[ \begin{matrix}
   2 & 1 \\
   1 & -1 \\
\end{matrix} \right]=\dfrac{1}{3}\left[ \begin{matrix}
   -2 & -1 \\
   -1 & 1 \\
\end{matrix} \right]$ as $\det \left( B \right)=-3$.
Let’s assume $C=\left[ \begin{matrix}
   2 & 3 \\
   4 & 6 \\
\end{matrix} \right]$. So, $\det \left( C \right)=\left| \begin{matrix}
   2 & 3 \\
   4 & 6 \\
\end{matrix} \right|$.
Using expansion method, we get $\det \left( C \right)=\left| C \right|=2\times 6-3\times 4=12-12=0$. So, C is not invertible.
Let’s assume $D=\left[ \begin{matrix}
   2 & -2 \\
   1 & 1 \\
\end{matrix} \right]$. So, $\det \left( D \right)=\left| \begin{matrix}
   2 & -2 \\
   1 & 1 \\
\end{matrix} \right|$.
Using expansion method, we get $\det \left( D \right)=\left| D \right|=2\times 1-\left( -2 \right)\times 1=2+2=4$. So, D is invertible.
Now we find its inverse. If ${{D}^{-1}}$ be the inverse of D then ${{D}^{-1}}=\dfrac{adj\left( D \right)}{\left| D \right|}$. Here $adj\left( D \right)$ defines the cofactors of elements. If $D=\left[ {{d}_{ij}} \right]$, then $adj\left( D \right)=\left[ {{D}_{ji}} \right]$. Here ${{d}_{ij}}$ represents the element of ${{i}^{th}}$ row and ${{j}^{th}}$ column. ${{D}_{ji}}$ represents the cofactor of the element ${{d}_{ij}}$.
$adj\left( D \right)=\left[ \begin{matrix}
   1 & 2 \\
   -1 & 2 \\
\end{matrix} \right]$ and ${{D}^{-1}}=\dfrac{1}{4}\left[ \begin{matrix}
   1 & 2 \\
   -1 & 2 \\
\end{matrix} \right]=\dfrac{1}{2}\left[ \begin{matrix}
   1 & 1 \\
   -1 & 1 \\
\end{matrix} \right]$ as $\det \left( A \right)=4$.
Therefore, the matrix in option C isn’t invertible.

Note: We need to remember that another trick way to find the adjoint of only $2\times 2$ matrix A is that we interchange the elements of ${{a}_{ij}},i=j$. There are two such elements ${{a}_{11}},{{a}_{22}}$. Then we just change the sign of the element of ${{a}_{ij}},i\ne j$ where we have two such elements ${{a}_{12}},{{a}_{21}}$. To find the inverse we follow the normal procedure of ${{A}^{-1}}=\dfrac{adj\left( A \right)}{\left| A \right|}$..