Question
If $A=\left[\begin{array}{ccc}4 & 3 & 2 \\ -1 & 2 & 0\end{array}\right], B=\left[\begin{array}{cc}1 & 2 \\ -1 & 0 \\ 1 & -2\end{array}\right]$, show that matrix $A B$ is non singular.
$\begin{aligned} & \therefore \quad\left[\begin{array}{cc}\alpha^2+0 & 0+0 \\ \alpha+1 & 0+1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right] \\ & \therefore \quad\left[\begin{array}{cc}\alpha^2 & 0 \\ \alpha+1 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right]\end{aligned}$
∴ By equality of matrices, we get
$\alpha^2=1$ and $\alpha+1=2$
∴ α = ± 1 and α = 1 ∴ α = 1
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$\left[\begin{array}{ccc}1 & 2 & -5 \\ 2 & -3 & 4 \\ -5 & 4 & 9\end{array}\right]$