Answer
See below
Work Step by Step
Given: $A=\begin{bmatrix}
4 &-1 &0\\5 &1 &4
\end{bmatrix}$
Adding one row $\begin{vmatrix}
a&b&c
\end{vmatrix}$ to matrix we have
$\det(B)=\begin{vmatrix}
4 &-1 & 0 \\5 &1 &4\\ a &b &c
\end{vmatrix}=10\\
\rightarrow a\begin{vmatrix}
-1 &0\\1 &4
\end{vmatrix}-b\begin{vmatrix}
4&0 \\5&4
\end{vmatrix}+c\begin{vmatrix}
4 &-1\\5 &1
\end{vmatrix}=-4a-16b+9c=10$
Assume that $a=-\frac{5}{2},b=0,c=0$
Substitute: $-4(-\frac{5}{2})-16(0)+9(0)=10$
Consequently, matrix (B) can be $B=\begin{bmatrix}
4 &-1 &0\\5 &1 &4\\-\frac{2}{5}& 0 & 0
\end{bmatrix}$
