Determine the value(s) of \(a\) such that \(\left\{ {\left( {\begin{aligned}{*{20}{c}}1\\a\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}a\\{a + 2}\end{aligned}} \right)} \right\}\) is linearly independent.

Write the vector in the augmented matrix form.

\(\begin{aligned}{l}{x_1}{{\mathop{\rm v}\nolimits} _1} + {x_2}{{\mathop{\rm v}\nolimits} _2} = {{\mathop{\rm v}\nolimits} _3}\\{x_1}\left( {\begin{aligned}{*{20}{c}}1\\a\end{aligned}} \right) + {x_2}\left( {\begin{aligned}{*{20}{c}}a\\{a + 2}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}0\\0\end{aligned}} \right)\,....\left( * \right)\\\left( {\begin{aligned}{*{20}{c}}1&a&0\\a&{a + 2}&0\end{aligned}} \right)\end{aligned}\)

At row two, multiply row one by \(a\) and subtract it from row two.

\(\begin{aligned}{l}\left( {\begin{aligned}{*{20}{c}}1&a&0\\0&{a + 2 - {a^2}}&0\end{aligned}} \right)\\\left( {\begin{aligned}{*{20}{c}}1&a&0\\0&{\left( {2 - a} \right)\left( {1 + a} \right)}&0\end{aligned}} \right)\end{aligned}\)

The columns of matrix \(A\) arelinearly independentif and only if the equation \(Ax = 0\) has only a trivial solution.

There is a non-trivial solution for equation (*) if and only if \(\left( {2 - a} \right)\left( {1 + a} \right) = 0\).

Thus, the vectors are linearly independent for all values of \(a\), except \(a = 2\) and \(a = - 1\).