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{\bf Question}

Determine whether the following matrices are singular or
non-singular

$$A =\left( \begin{array}{cc} 2 & 3 \\ 4 & 6 \end{array} \right)
\hspace{.2in} B = \left( \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 1
\\ 0 & 1 & 0 \end{array} \right) \hspace{.2in}$$ $$C = \left( \begin{array}{cccc}
 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array}
 \right) \hspace{.2in} D = \left( \begin{array}{ccc} 2 & 1 & 3 \\ 1 & 1 & 2 \\ 2 & 1 & 6
\end{array} \right)$$

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{\bf Answer}

Matrix A

$\left| \begin{array}{cc} 2 & 3 \\ 4 & 6 \end{array} \right| = 2
\times 6 - 3 \times 4 = 12 -12 = 0$

Hence $\left( \begin{array}{cc} 2 & 3 \\ 4 & 6 \end{array}
\right)$ is singular.

Matrix B

$\left| \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0
\end{array} \right| = (-1)^{1+2} \times 1 \times \left|
\begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right| = 0$

Hence $\left( \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 &
0 \end{array} \right)$ is singular.

Matrix C

$\left| \begin{array}{cccc} 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 1 &
1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array} \right| = (-1)^{1+1} \left|
\begin{array}{ccc}0 & 1 & 0 \\ 1 & 0 & 0  \\ 0 & 0 & 0
\end{array} \right|  + (-1)^{1+4}
\left| \begin{array}{ccc} 1 & 0 & 1 \\ 1 & 1 & 0  \\ 1 & 0 & 0
\end{array} \right|  $

First determinant is zero since one row is completely zero.

Hence \begin{eqnarray*} \left| \begin{array}{cccc} 1 & 0 & 0 & 1
\\ 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array}
\right| & = & (-1)^{1+4} \left| \begin{array}{ccc} 1 & 0 & 1 \\ 1
& 1 & 0 \\ 1 & 0 & 0 \end{array} \right|\\ & = & -1 \times \left|
\begin{array} {cc} 1 & 1 \\ 1& 0 \end{array} \right| \\ &  = & -1 \times
-1 \\ & = & 1\end{eqnarray*}


Hence $\left( \begin{array}{cccc} 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0
\\ 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array} \right)$ is not
singular.

Matrix D
\begin{eqnarray*} \left| \begin{array}{ccc} 2 & 1 & 3 \\ 1 & 1 & 2 \\ 2 & 1 & 6
\end{array} \right| & = & \left| \begin{array}{ccc} 0 & 0 & -3 \\ 1 & 1 & 2 \\ 2 & 1 & 6
\end{array} \right| \ \ {\rm row\ 1\ \rightarrow row\ 1\ -\ row\ 3} \\ & = & \left|
\begin{array}{ccc} 0 & 0 & -3 \\ 0 & 1 & 2 \\ 1 & 1 & 6
\end{array} \right|\ \   {\rm column\ 1\ \rightarrow column\ 1\ -\ column\ 2}
 \\ & = & 3  \end{eqnarray*}

Hence $\left( \begin{array}{ccc} 2 & 1 & 3 \\ 1 & 1 & 2 \\ 2 & 1 &
6 \end{array} \right)$ is not singular.



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