College Algebra, Chapter 7, 7.4, Section 7.4, Problem 54

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

For this matrix, we have


$
\begin{equation}
\begin{aligned}

0 =& a \left| \begin{array}{cc}
x + b & x \\
1 & 1
\end{array} \right| - b \left| \begin{array}{cc}
x & x \\
0 & 1
\end{array} \right| + (x - a) \left| \begin{array}{cc}
x & x + b \\
0 & 1
\end{array} \right|
&& \text{Expand}
\\
\\
0 =& a [(x + b)(1) - x \cdot 1] - b (x \cdot 1 - x \cdot 0) + (x - a) [x \cdot 1 - (x + b)(0)]
&& \text{Simplify}
\\
\\
0 =& a(x + b - x) - b(x) + (x - a) (x)
&& \text{Distributive Property}
\\
\\
0 =& ab - bx + x^2 - ax
&& \text{Simplify}
\\
\\
0 =& x^2 - ax - bx + ab
&& \text{Factor}
\\
\\
0 =& (x - a)(x - b)
&& \text{Zero Product Property}
\\
\\
x - a =& 0 \quad x - b = 0
&&
\\
\\
x =& a \qquad x = b
&&

\end{aligned}
\end{equation}
$