Question
If $A=\left[\begin{array}{lll}1 & -1 & 2 \\ 0 & -1 & 3\end{array}\right], B=\left[\begin{array}{cc}-2 & 1 \\ 3 & -1 \\ 0 & 2\end{array}\right]$, show that matrix $\mathrm{AB}$ is non singular.
✓
Answer
$
\text { let } \begin{aligned}
A B & =\left[\begin{array}{ccc}
1 & -1 & 2 \\
0 & -1 & 3
\end{array}\right]\left[\begin{array}{cc}
-2 & 1 \\
3 & -1 \\
0 & 2
\end{array}\right] \\
& =\left[\begin{array}{cc}
-2-3+0 & 1+1+4 \\
0-3+0 & 0+1+6
\end{array}\right] \\
& =\left[\begin{array}{ll}
-5 & 6 \\
-3 & 7
\end{array}\right] \\
\therefore|A B| & =\left[\begin{array}{ll}
-5 & 6 \\
-3 & 7
\end{array} \mid\right. \\
& =-35+18=-17 \neq 0
\end{aligned}
$
$\therefore$ matrix $\mathrm{AB}$ is nonsingular.
\text { let } \begin{aligned}
A B & =\left[\begin{array}{ccc}
1 & -1 & 2 \\
0 & -1 & 3
\end{array}\right]\left[\begin{array}{cc}
-2 & 1 \\
3 & -1 \\
0 & 2
\end{array}\right] \\
& =\left[\begin{array}{cc}
-2-3+0 & 1+1+4 \\
0-3+0 & 0+1+6
\end{array}\right] \\
& =\left[\begin{array}{ll}
-5 & 6 \\
-3 & 7
\end{array}\right] \\
\therefore|A B| & =\left[\begin{array}{ll}
-5 & 6 \\
-3 & 7
\end{array} \mid\right. \\
& =-35+18=-17 \neq 0
\end{aligned}
$
$\therefore$ matrix $\mathrm{AB}$ is nonsingular.
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