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The following equation describes a Givens rotation in \({\mathbb{R}^3}\). Find \(a\) and \(b\).

\(\left( {\begin{aligned}{*{20}{c}}a&0&{ - b}\\0&1&0\\b&0&a\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{\bf{3}}\\{\bf{4}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{{\bf{2}}\sqrt {\bf{5}} }\\{\bf{3}}\\{\bf{0}}\end{aligned}} \right)\), \({a^2} + {b^2} = 1\)

Short Answer

Expert verified

\(a = \frac{1}{{\sqrt 5 }}\), \(b = - \frac{2}{{\sqrt 5 }}\)

Step by step solution

01

Form the equation using the equation of rotation

Upon solving the equation \(\left( {\begin{aligned}{*{20}{c}}a&0&{ - b}\\0&1&0\\b&0&a\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}2\\3\\4\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{2\sqrt 5 }\\3\\0\end{aligned}} \right)\), you get

\(2a - 4b = 2\sqrt 5 \)and \(2b + 4a = 0\).

02

Form the augmented matrix using the two equations

The augmented matrix for the equations \(2a - 4b = 2\sqrt 5 \) and \(2b + 4a = 0\) can be written as \(\left( {\begin{aligned}{*{20}{c}}2&{ - 4}&{2\sqrt 5 }\\4&2&0\end{aligned}} \right)\).

03

Use row operations on the augmented matrix

At row two, multiply row one by two and subtract it from row two, i.e., \({R_2} \to {R_2} - 2{R_1}\).

\(\left( {\begin{aligned}{*{20}{c}}2&{ - 4}&{2\sqrt 5 }\\{4 - 2\left( 2 \right)}&{2 - 2\left( { - 4} \right)}&{0 - 2\left( {2\sqrt 5 } \right)}\end{aligned}} \right)\)

After performing row operations, the matrix becomes

\(\left( {\begin{aligned}{*{20}{c}}2&{ - 4}&{2\sqrt 5 }\\0&{10}&{ - 4\sqrt 5 }\end{aligned}} \right)\).

04

Use row operations on the augmented matrix

Divide row one by 2, i.e., \({R_1} \to \frac{1}{2}{R_1}\).

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 2}&{\sqrt 5 }\\0&{10}&{ - 4\sqrt 5 }\end{aligned}} \right)\)

Divide row two by 10, i.e., \({R_2} \to \frac{1}{{10}}{R_2}\).

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 2}&{\sqrt 5 }\\0&1&{ - \frac{2}{{\sqrt 5 }}}\end{aligned}} \right)\)

05

Use row operations on the augmented matrix

At row one, multiply row two by 2 and add it to row one, i.e., \({R_1} \to {R_1} + 2{R_2}\).

\(\left( {\begin{aligned}{*{20}{c}}{1 + 0}&{ - 2 + 1\left( 2 \right)}&{\sqrt 5 + 2\left( { - \frac{2}{{\sqrt 5 }}} \right)}\\0&1&{ - \frac{2}{{\sqrt 5 }}}\end{aligned}} \right)\)

After performing row operations, the matrix becomes

\(\left( {\begin{aligned}{*{20}{c}}1&0&{\frac{1}{{\sqrt 5 }}}\\0&1&{ - \frac{2}{{\sqrt 5 }}}\end{aligned}} \right)\).

So, the value of \(a\) is \(\frac{1}{{\sqrt 5 }}\) and that of \(b\) is \( - \frac{2}{{\sqrt 5 }}\).

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