College Algebra, Chapter 7, 7.4, Section 7.4, Problem 52

Solve for $x$ if the matrix $\displaystyle \left| \begin{array}{ccc}
x & 1 & 1 \\
1 & 1 & x \\
x & 1 & x
\end{array} \right| = 0$.

For this matrix we have


$
\begin{equation}
\begin{aligned}

0 =& x \left| \begin{array}{cc}
1 & x \\
1 & x
\end{array} \right| - 1 \left| \begin{array}{cc}
1 & x \\
x & x
\end{array} \right| + 1 \left| \begin{array}{cc}
1 & 1 \\
x & 1
\end{array} \right|
&& \text{Expand}
\\
\\
0 =& x (1 \cdot x - x \cdot 1) - 1 (1 \cdot x - x \cdot x) + 1 (1 \cdot 1 - 1 \cdot x)
&& \text{Simplify}
\\
\\
0 =& x(0) - (x - x^2) + 1 -x
&& \text{Distributive Property}
\\
\\
0 =& -x + x^2 + 1 - x
&& \text{Combine like terms}
\\
\\
0 =& x^2 - 2x + 1
&& \text{Factor}
\\
\\
0 =& (x - 1)^2
&& \text{Take the square root of both sides}
\\
\\
0 =& x - 1
&& \text{Add } 1
\\
\\
x =& 1
&&

\end{aligned}
\end{equation}
$