# Find All the Values of $x$ so that a Given $3\times 3$ Matrix is Singular ## Problem 169

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

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## Hint.

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 $x=1$. Add to solve later

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