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Question
For the multiplication of matrices as a binary operation on the set of all matrices of the form \[\begin{bmatrix}a & b \\ - b & a\end{bmatrix}\] a, b ∈ R the inverse of \[\begin{bmatrix}2 & 3 \\ - 3 & 2\end{bmatrix}\] is ___________________ .
Options
\[\begin{bmatrix}- 2 & 3 \\ - 3 & - 2\end{bmatrix}\]
\[\begin{bmatrix}2 & 3 \\ - 3 & 2\end{bmatrix}\]
\[\begin{bmatrix}2/13 & - 3/13 \\ 3/13 & 2/13\end{bmatrix}\]
\[\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\]
Solution
\[\begin{bmatrix}2/13 & - 3/13 \\ 3/13 & 2/13\end{bmatrix}\]
To find the identity element,
\[\text{ Let }A = \begin{bmatrix}a & b \\ - b & a\end{bmatrix} \text{ and }I = \begin{bmatrix}x & y \\ - y & x\end{bmatrix} \text{ such that }\]
\[A . I = I . A = A\]
\[A . I = A\]
\[\begin{bmatrix}a & b \\ - b & a\end{bmatrix}\begin{bmatrix}x & y \\ - y & x\end{bmatrix} = \begin{bmatrix}x & y \\ - y & x\end{bmatrix}\]
\[\begin{bmatrix}ax - by & ay + bx \\ - \left( ay + bx \right) & ax - by\end{bmatrix} = \begin{bmatrix}x & y \\ - y & x\end{bmatrix}\]
\[ \Rightarrow ax - by = x . . . \left( 1 \right)\]
\[ \Rightarrow ay + bx = y . . . \left( 2 \right)\]
\[\text{ Solving these two equations, we get }\]
\[x = 1 \text{ and }y = 0\]
\[\text{ Thus },\]
\[I = \begin{bmatrix}x & y \\ - y & x\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} (\text{ which is usually an identity matrix })\]
\[\text{ Let} \begin{bmatrix}m & n \\ - n & m\end{bmatrix} \text{ be the inverse of} \begin{bmatrix}2 & 3 \\ - 3 & 2\end{bmatrix} . \]
\[ \therefore \begin{bmatrix}2 & 3 \\ - 3 & 2\end{bmatrix} \begin{bmatrix}m & n \\ - n & m\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}2m - 3n & 2n + 3m \\ - 3m - 2n & - 3n + 2m\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\]
\[ \Rightarrow 2m - 3n = 1 . . . (3) \]
\[2n + 3m = 0 . . . (4) \]
\[ - 3m - 2n = 0 . . . (5) \]
\[ - 3n + 2m = 1 . . . (6) \]
From eq. (4)
\[n = \frac{- 3m}{2} . . . (7) \]
\[\text{ Substituting the value of n in eq } . (3) \]
\[2m - 3\left( \frac{- 3m}{2} \right) = 1\]
\[ \Rightarrow 2m + \frac{9m}{2} = 1\]
\[ \Rightarrow \frac{13m}{2} = 1\]
\[ \Rightarrow m = \frac{2}{13}\]
\[\text{ Substituting the value of m in eq } . (7)\]
\[ \Rightarrow n = \frac{- 3}{2} \times \frac{2}{13} = \frac{- 3}{13}\]
\[\text{ Hence, the inverse of } \begin{bmatrix}2 & 3 \\ - 3 & 2\end{bmatrix} is \begin{bmatrix}\frac{2}{13} & \frac{- 3}{13} \\ \frac{3}{13} & \frac{2}{13}\end{bmatrix} .\]
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