Answer
See below
Work Step by Step
Given: $A=\begin{bmatrix}
2 & k-3 & k^2\\
2 & 1 & 4\\
1 & k & 0
\end{bmatrix}$
a) For $A$ to be invertible, the determinant should not be equal to zero.
Find $\det(A)=4(k-3)+2k^3-k^2-8k=2k^3-k^2-4k-12$
then $\det(A)=2k^3-k^2-4k-12 \ne 0 \rightarrow k\ne 2.389, k\ne -0.944 \pm 1.273i$
b) In terms of $k$, we will determine the volume of the parallelepiped determined by the row vectors of the matrix $A$
$V_1=\begin{vmatrix}
2& k-3& k^2\\
2 & 1& 4\\
1 & k & 0
\end{vmatrix}=2k^3-k^2-4k-12$
and we have $V_2=\det(A)=2k^3-k^2-4k-12$
We can see that $V_1=V_2$.
Without calculation, we still can know the volumes are the same because according to Property P4, we know $V_1=|A^T|,|A|=|A^T|$
