A Givens rotation is a linear transformation from \({\mathbb{R}^{\bf{n}}}\) to \({\mathbb{R}^{\bf{n}}}\) used in computer programs to create a zero entry in a vector (usually a column of matrix). The standard matrix of a given rotations in \({\mathbb{R}^{\bf{2}}}\) has the form

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

Find \(a\) and \(b\) such that \(\left( {\begin{aligned}{*{20}{c}}4\\3\end{aligned}} \right)\) is rotated into \(\left( {\begin{aligned}{*{20}{c}}5\\0\end{aligned}} \right)\).

As the rotation matrix is \(\left( {\begin{aligned}{*{20}{c}}a&{ - b}\\b&a\end{aligned}} \right)\) and \(\left( {\begin{aligned}{*{20}{c}}4\\3\end{aligned}} \right)\) is rotated through \(\left( {\begin{aligned}{*{20}{c}}5\\0\end{aligned}} \right)\),

\(\left( {\begin{aligned}{*{20}{c}}a&{ - b}\\b&a\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}4\\3\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}5\\0\end{aligned}} \right)\).

The above equation implies that \(4a - 3b = 5\) and \(3a + 4b = 0\).

The augmented matrix for the equation of \(a\) and \(b\) is \(\left( {\begin{aligned}{*{20}{c}}4&{ - 3}&5\\3&4&0\end{aligned}} \right)\).

At row two, multiply row two by 4 and row one by 3 and subtract row one from row two, i.e., \({R_2} \to 4{R_2} - 3{R_1}\).

\(\left( {\begin{aligned}{*{20}{c}}4&{ - 3}&5\\{3 \times 4 - 4 \times 3}&{4 \times 4 - 3\left( { - 3} \right)}&{0 - 5 \times 3}\end{aligned}} \right)\)

After performing the row operations, the matrix becomes

\(\left( {\begin{aligned}{*{20}{c}}4&{ - 3}&5\\0&{25}&{ - 15}\end{aligned}} \right)\).

Divide row two by 25, i.e., \({R_2} \to \frac{{{R_2}}}{{25}}\).

\(\left( {\begin{aligned}{*{20}{c}}4&{ - 3}&5\\0&1&{ - \frac{3}{5}}\end{aligned}} \right)\)

At row one, multiply row two by 3 and add it to row one, i.e., \({R_1} \to {R_1} + 3{R_2}\).

\(\left( {\begin{aligned}{*{20}{c}}{4 + 0}&{ - 3 + 3 \times 1}&{5 - 3 \times \frac{3}{5}}\\0&1&{ - \frac{3}{5}}\end{aligned}} \right)\)

After performing the row operation, the matrix becomes

\(\left( {\begin{aligned}{*{20}{c}}4&0&{\frac{{16}}{5}}\\0&1&{ - \frac{3}{5}}\end{aligned}} \right)\).

Divide row one by \(4\), i.e., \({R_1} \to \frac{{{R_1}}}{4}\).

\(\left( {\begin{aligned}{*{20}{c}}1&0&{\frac{4}{5}}\\0&1&{ - \frac{3}{5}}\end{aligned}} \right)\)

So, \(a = \frac{4}{5}\) and \(b = - \frac{3}{5}\).