Let $A$ be the following $3 \times 3$ matrix.
\[A=\begin{bmatrix}
1 & 1 & -1 \\
0 &1 &2 \\
1 & 1 & a
\end{bmatrix}.\]
Determine the values of $a$ so that the matrix $A$ is nonsingular.

If $a+1=0$, then the last matrix is in reduced row echelon form.
Thus $A$ is not row equivalent to the identity matrix.

On the other hand, if $a+1 \neq 0$, then we can continue the reduction as follows.
\begin{align*}
\begin{bmatrix}
1 & 0 & -3 \\
0 &1 &2 \\
0 & 0 & a+1
\end{bmatrix}
\xrightarrow{\frac{1}{a+1} R_3}
\begin{bmatrix}
1 & 0 & -3 \\
0 &1 &2 \\
0 & 0 & 1
\end{bmatrix}
\xrightarrow[R_2-2R_3]{R_1+3R_3}
\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & 1
\end{bmatrix}.
\end{align*}
Therefore $A$ is row equivalent to the identity matrix.

We conclude that the matrix $A$ is nonsingular for any values of $a$ except for $a=-1$.

Comment.

If you know how to compute the determinant of a $3 \times 3$ matrix, then you may also solve this using the fact that a matrix is nonsingular if and only if the determinant of it is nonzero.

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