Question
If $A=\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 2 & 3\end{array}\right], B=\left[\begin{array}{ccc}1 & -1 & 1 \\ -3 & 2 & -1 \\ -2 & 1 & 0\end{array}\right]$ thenshow that $A B$ and $BA$ are both singular matrices.

Answer

$
\begin{aligned}
& A B=\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 4 & 6 \\
1 & 2 & 3
\end{array}\right]\left[\begin{array}{rrr}
1 & -1 & 1 \\
-3 & 2 & -1 \\
-2 & 1 & 0
\end{array}\right] \\
&=\left[\begin{array}{rrr}
1-6-6 & -1+4+3 & 1-2+0 \\
2-12-12 & -2+8+6 & 2-4+0 \\
1-6-6 & -1+4+3 & 1-2+0
\end{array}\right] \\
&=\left[\begin{array}{rrr}
-11 & 6 & -1 \\
-22 & 12 & -2 \\
-11 & 6 & -1
\end{array}\right] \\
& \therefore| AB |=\left[\begin{array}{rrr}
-11 & 6 & -1 \\
-22 & 12 & -2 \\
-11 & 6 & -1
\end{array}\right]
\end{aligned}
$
By taking -6 common from $C_2$, we get
$
\begin{aligned}
& \therefore| AB |=-6\left|\begin{array}{lll}
-11 & -1 & -1 \\
-22 & -2 & -2 \\
-11 & -1 & -1
\end{array}\right| \\
& =-6 \times 0 \\
& \text {... }\left[\because C_2 \equiv C_3\right] \\
& =0 \\
&
\end{aligned}
$
$\therefore AB$ is a singular matrix.
Also, $BA =\left(\begin{array}{rrr}1 & -1 & 1 \\ -3 & 2 & -1 \\ -2 & 1 & 0\end{array}\right)\left(\begin{array}{lll}1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 2 & 3\end{array}\right)$
$
=\left(\begin{array}{rrr}
1-2+1 & 2-4+2 & 3-6+3 \\
-3+4-1 & -6+8-2 & -9+12-3 \\
-2+2+0 & -4+4+0 & -6+6+0
\end{array}\right)
$
$
=\left(\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right)=0 \quad \therefore| BA |=0
$
$\therefore BA$ is also a singular matrix.
Hence, $AB$ and $BA$ are both singular matrices.

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