Answer
$d=10$ forces a row exchange and the triangular system after row exchange is
$
\begin{bmatrix}
2 && 5 && 1 && 0\\
0 && 1 && -1 && 3\\
0 && 0 && -1 && 2\\
\end{bmatrix}$
For $d=11$, the system becomes singular.
Work Step by Step
$
2x + 5y + z = 0\\
4x + dy + z = 2\\
y − z = 3\\
$
Converting the above equation in matrix form
$\begin{bmatrix}
2 && 5 && 1 && 0\\
4 && d && 1 && 2\\
0 && 1 && -1 && 3\\
\end{bmatrix}$
Subtract 2 times row 1 from row 2
$\sim
\begin{bmatrix}
2 && 5 && 1 && 0\\
0 && d-10 && -1 && 2\\
0 && 1 && -1 && 3\\
\end{bmatrix}$
$d=10$ forces a row exchange. Hence the system after $d=10$ and after row exchange would be
$\sim
\begin{bmatrix}
2 && 5 && 1 && 0\\
0 && 1 && -1 && 3\\
0 && 0 && -1 && 2\\
\end{bmatrix}$
Now for $d\neq 10$
Subtract $\frac{1}{d-10}$ times row 2 from row 3
$\sim
\begin{bmatrix}
2 && 5 && 1 && 0\\
0 && d-10 && -1 && 2\\
0 && 0 && \frac{d-11}{10-d} && \frac{3d-32}{d-10}\\
\end{bmatrix}$
For no third pivot $\frac{d-11}{10-d}=0$
Hence, $d=11$
The matrix finally changes to
$\sim
\begin{bmatrix}
2 && 5 && 1 && 0\\
0 && 1 && -1 && 2\\
0 && 0 && 0 && 1\\
\end{bmatrix}$
