Find all the values of $x$ so that the following matrix $A$ is a singular matrix.
\[A=\begin{bmatrix}
x & x^2 & 1 \\
2 &3 &1 \\
0 & -1 & 1
\end{bmatrix}.\]

Use the fact that a matrix is singular if and only if its determinant is zero.

Solution.

Note that a matrix is singular if and only if its determinant is zero.
So we compute the determinant of the matrix $A$ as follows.
\begin{align*}
&\det(A)=\begin{vmatrix}
x & x^2 & 1 \\
2 &3 &1 \\
0 & -1 & 1
\end{vmatrix}\\
&=(-1)^{3+1}\cdot 0 \cdot \begin{vmatrix}
x^2 & 1\\
3& 1
\end{vmatrix}
+(-1)^{3+2}\cdot(-1)\cdot \begin{vmatrix}
x & 1\\
2& 1
\end{vmatrix}
+(-1)^{3+3}\cdot 1\cdot \begin{vmatrix}
x & x^2\\
2& 3
\end{vmatrix}\\
&\text{by the third row cofactor expansion}\\
&= 0+(x-2)+(3x-2x^2)\\
&=-2x^2+4x-2.
\end{align*}

Thus the determinant of $A$ is zero if
\[\det(A)=-2x^2+4x-2=0,\]
equivalently,
\[x^2-2x+1=(x-1)^2=0.\]
Thus, the determinant of the matrix $A$ is zero if and only if $x=1$.
Hence the matrix $A$ is singular if and only if $x=1$.

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