English

Let F ( α ) = ⎡ ⎢ ⎣ Cos α − Sin α 0 Sin α Cos α 0 0 0 1 ⎤ ⎥ ⎦ and G ( β ) = ⎡ ⎢ ⎣ Cos β 0 Sin β 0 1 0 − Sin β 0 Cos β ⎤ ⎥ ⎦ Show that [ F ( α ) ] − 1 = F ( − α ) - Mathematics

Advertisements
Advertisements

Question

Let
\[F \left( \alpha \right) = \begin{bmatrix}\cos \alpha & - \sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{bmatrix}\text{ and }G\left( \beta \right) = \begin{bmatrix}\cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ - \sin \beta & 0 & \cos \beta\end{bmatrix}\]

Show that

(i) \[\left[ F \left( \alpha \right) \right]^{- 1} = F \left( - \alpha \right)\]
(ii) \[\left[ G \left( \beta \right) \right]^{- 1} = G \left( - \beta \right)\]
(iii) \[\left[ F \left( \alpha \right)G \left( \beta \right) \right]^{- 1} = G \left( - \beta \right)F \left( - \alpha \right)\]
Sum
Advertisements

Solution

(i) \[ F(\alpha) = \begin{bmatrix}\cos\alpha & - \sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix}\]
\[ \Rightarrow F( - \alpha) = \begin{bmatrix}\cos\left( - \alpha \right) & - \sin\left( - \alpha \right) & 0 \\ \sin\left( - \alpha \right) & \cos\left( - \alpha \right) & 0 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}\cos\alpha & \sin\alpha & 0 \\ - \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix}\]
\[Now, \]
\[ C_{11} = \begin{vmatrix}\cos\alpha & 0 \\ 0 & 1\end{vmatrix} = \cos\alpha, C_{12} = - \begin{vmatrix}\sin\alpha & 0 \\ 0 & 1\end{vmatrix} = - \sin\alpha\text{ and }C_{13} = \begin{vmatrix}\sin\alpha & \cos\alpha \\ 0 & 0\end{vmatrix} = 0\]
\[ C_{21} = - \begin{vmatrix}- \sin\alpha & 0 \\ 0 & 1\end{vmatrix} = \sin\alpha, C_{22} = \begin{vmatrix}\cos\alpha & 0 \\ 0 & 1\end{vmatrix} = \cos\alpha\text{ and }C_{23} = - \begin{vmatrix}\cos\alpha & - \sin\alpha \\ 0 & 0\end{vmatrix} = 0\]
\[ C_{31} = \begin{vmatrix}- \sin\alpha & 0 \\ \cos\alpha & 0\end{vmatrix} = 0, C_{32} = - \begin{vmatrix}\cos\alpha & 0 \\ \sin\alpha & 0\end{vmatrix} = 0\text{ and }C_{33} = \begin{vmatrix}\cos\alpha & - \sin\alpha \\ \sin\alpha & \cos\alpha\end{vmatrix} = 1\]
\[ \Rightarrow adj\left\{ F(\alpha) \right\} = \begin{bmatrix}\cos\alpha & - \sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix}^T = \begin{bmatrix}\cos\alpha & \sin\alpha & 0 \\ - \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix}\]
\[ \Rightarrow \left| F(\alpha) \right| = 1\]
\[ \therefore \left[ F\left( \alpha \right) \right]^{- 1} = \begin{bmatrix}\cos\alpha & \sin\alpha & 0 \\ - \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix} . . . \left( 1 \right)\]
\[ \Rightarrow \left[ F\left( \alpha \right) \right]^{- 1} = F( - \alpha) \]
(ii) \[ G(\beta) = \begin{bmatrix}\cos\beta & 0 & \sin\beta \\ 0 & 1 & 0 \\ - \sin\beta & 0 & \cos\beta\end{bmatrix}\]
\[ \Rightarrow G( - \beta) = \begin{bmatrix}\cos\left( - \beta \right) & 0 & \sin\left( - \beta \right) \\ 0 & 1 & 0 \\ - \sin\left( - \beta \right) & 0 & \cos\left( - \beta \right)\end{bmatrix} = \begin{bmatrix}\cos\beta & 0 & - \sin\beta \\ 0 & 1 & 0 \\ \sin\beta & 0 & \cos\beta\end{bmatrix}\]
\[Now, \]
\[ C_{11} = \begin{vmatrix}1 & 0 \\ 0 & \cos\beta\end{vmatrix} = \cos\beta, C_{12} = - \begin{vmatrix}0 & 0 \\ - \sin\beta & \cos\beta\end{vmatrix} = 0\text{ and }C_{13} = \begin{vmatrix}0 & 1 \\ - \sin\beta & 0\end{vmatrix} = \sin\beta\]
\[ C_{21} = - \begin{vmatrix}0 & \sin\beta \\ 0 & \cos\beta\end{vmatrix} = 0, C_{22} = \begin{vmatrix}\cos\beta & \sin\beta \\ - \sin\beta & \cos\beta\end{vmatrix} = 1\text{ and }C_{23} = - \begin{vmatrix}\cos\beta & 0 \\ - \sin\beta & 0\end{vmatrix} = 0\]
\[ C_{31} = \begin{vmatrix}0 & \sin\beta \\ 1 & 0\end{vmatrix} = - \sin\beta, C_{32} = - \begin{vmatrix}\cos\beta & \sin\beta \\ 0 & 0\end{vmatrix} = 0\text{ and }C_{33} = \begin{vmatrix}\cos\beta & 0 \\ 0 & 1\end{vmatrix} = \cos\beta\]
\[adj\left\{ G(\beta) \right\} = \begin{bmatrix}\cos\beta & 0 & \sin\beta \\ 0 & 1 & 0 \\ - \sin\beta & 0 & \cos\beta\end{bmatrix}^T = \begin{bmatrix}\cos\beta & 0 & - \sin\beta \\ 0 & 1 & 0 \\ \sin\beta & 0 & \cos\beta\end{bmatrix}\]
\[\left| G(\beta) \right| = 1\]
\[ \therefore G(\beta )^{- 1} = \begin{bmatrix}\cos\beta & 0 & - \sin\beta \\ 0 & 1 & 0 \\ \sin\beta & 0 & \cos\beta\end{bmatrix} . . . \left( 2 \right) \]
\[ \Rightarrow G(\beta )^{- 1} = = G( - \beta) \]
(iii) \[ F(\alpha) = \begin{bmatrix}\cos\alpha & - \sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix}\]
\[ \Rightarrow F( - \alpha) = \begin{bmatrix}\cos\left( - \alpha \right) & - \sin\left( - \alpha \right) & 0 \\ \sin\left( - \alpha \right) & \cos\left( - \alpha \right) & 0 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}\cos\alpha & \sin\alpha & 0 \\ - \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix} . . . \left( 3 \right)\]
\[G(\beta) = \begin{bmatrix}\cos\beta & 0 & \sin\beta \\ 0 & 1 & 0 \\ - \sin\beta & 0 & \cos\beta\end{bmatrix}\]
\[ \Rightarrow G( - \beta) = \begin{bmatrix}\cos\left( - \beta \right) & 0 & \sin\left( - \beta \right) \\ 0 & 1 & 0 \\ - \sin\left( - \beta \right) & 0 & \cos\left( - \beta \right)\end{bmatrix} = \begin{bmatrix}\cos\beta & 0 & - \sin\beta \\ 0 & 1 & 0 \\ \sin\beta & 0 & \cos\beta\end{bmatrix} . . . \left( 4 \right)\]
\[ \left[ F(\alpha)G(\beta) \right]^{- 1} = \left[ G(\beta) \right]^{- 1} \left[ F(\alpha) \right]^{- 1} \]
\[ = \begin{bmatrix}\cos\beta & 0 & - \sin\beta \\ 0 & 1 & 0 \\ \sin\beta & 0 & \cos\beta\end{bmatrix}\begin{bmatrix}\cos\alpha & \sin\alpha & 0 \\ - \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix} \]               [Using equation (1) and (2)]
\[ = G( - \beta)F( - \alpha) \]                 [Using equatio (3) and (4)]

shaalaa.com
  Is there an error in this question or solution?
Chapter 7: Adjoint and Inverse of a Matrix - Exercise 7.1 [Page 23]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 7 Adjoint and Inverse of a Matrix
Exercise 7.1 | Q 16 | Page 23

RELATED QUESTIONS

Find the inverse of the matrices (if it exists).

`[(1,0,0),(3,3,0),(5,2,-1)]`


For the matrix A = `[(3,2),(1,1)]` find the numbers a and b such that A2 + aA + bI = 0.


For the matrix A = `[(1,1,1),(1,2,-3),(2,-1,3)]` show that A3 − 6A2 + 5A + 11 I = 0. Hence, find A−1.


Let A = `[(1,2,1),(2,3,1),(1,1,5)]` verify that

  1. [adj A]–1 = adj(A–1)
  2. (A–1)–1 = A

Compute the adjoint of the following matrix:

\[\begin{bmatrix}2 & - 1 & 3 \\ 4 & 2 & 5 \\ 0 & 4 & - 1\end{bmatrix}\]

Verify that (adj A) A = |A| I = A (adj A) for the above matrix.


For the matrix 

\[A = \begin{bmatrix}1 & - 1 & 1 \\ 2 & 3 & 0 \\ 18 & 2 & 10\end{bmatrix}\] , show that A (adj A) = O.

Find the inverse of the following matrix.
\[\begin{bmatrix}1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2\end{bmatrix}\]


For the following pair of matrix verify that \[\left( AB \right)^{- 1} = B^{- 1} A^{- 1} :\]

\[A = \begin{bmatrix}2 & 1 \\ 5 & 3\end{bmatrix}\text{ and }B \begin{bmatrix}4 & 5 \\ 3 & 4\end{bmatrix}\]


If \[A = \begin{bmatrix}2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2\end{bmatrix}\].
Verify that \[A^3 - 6 A^2 + 9A - 4I = O\]  and hence find A−1.

If \[A = \frac{1}{9}\begin{bmatrix}- 8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & - 8 & 4\end{bmatrix}\],
prove that  \[A^{- 1} = A^3\]

If \[A = \begin{bmatrix}- 1 & 2 & 0 \\ - 1 & 1 & 1 \\ 0 & 1 & 0\end{bmatrix}\] , show that  \[A^2 = A^{- 1} .\]


Find the matrix X for which 

\[\begin{bmatrix}3 & 2 \\ 7 & 5\end{bmatrix} X \begin{bmatrix}- 1 & 1 \\ - 2 & 1\end{bmatrix} = \begin{bmatrix}2 & - 1 \\ 0 & 4\end{bmatrix}\]

 


Find the matrix X satisfying the equation 

\[\begin{bmatrix}2 & 1 \\ 5 & 3\end{bmatrix} X \begin{bmatrix}5 & 3 \\ 3 & 2\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} .\]

\[\text{ If }A^{- 1} = \begin{bmatrix}3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2\end{bmatrix}\text{ and }B = \begin{bmatrix}1 & 2 & - 2 \\ - 1 & 3 & 0 \\ 0 & - 2 & 1\end{bmatrix},\text{ find }\left( AB \right)^{- 1} .\]

If \[A = \begin{bmatrix}1 & - 2 & 3 \\ 0 & - 1 & 4 \\ - 2 & 2 & 1\end{bmatrix},\text{ find }\left( A^T \right)^{- 1} .\]


Find the adjoint of the matrix \[A = \begin{bmatrix}- 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1\end{bmatrix}\]  and hence show that \[A\left( adj A \right) = \left| A \right| I_3\]. 


Find the inverse by using elementary row transformations:

\[\begin{bmatrix}7 & 1 \\ 4 & - 3\end{bmatrix}\]


Find the inverse by using elementary row transformations:

\[\begin{bmatrix}1 & 6 \\ - 3 & 5\end{bmatrix}\]


Find the inverse by using elementary row transformations:

\[\begin{bmatrix}2 & - 1 & 3 \\ 1 & 2 & 4 \\ 3 & 1 & 1\end{bmatrix}\]


Find the inverse by using elementary row transformations:

\[\begin{bmatrix}1 & 3 & - 2 \\ - 3 & 0 & - 1 \\ 2 & 1 & 0\end{bmatrix}\]


Find the inverse by using elementary row transformations:

\[\begin{bmatrix}- 1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{bmatrix}\]


If A is a non-singular symmetric matrix, write whether A−1 is symmetric or skew-symmetric.


If A is an invertible matrix such that |A−1| = 2, find the value of |A|.


Find the inverse of the matrix \[\begin{bmatrix}3 & - 2 \\ - 7 & 5\end{bmatrix} .\]


If A is a singular matrix, then adj A is ______.


If for the matrix A, A3 = I, then A−1 = _____________ .


If A and B are invertible matrices, which of the following statement is not correct.


If a matrix A is such that \[3A^3 + 2 A^2 + 5 A + I = 0,\text{ then }A^{- 1}\] equal to _______________ .


If A is an invertible matrix, then det (A1) is equal to ____________ .


A square matrix A is invertible if det A is equal to ____________.


The value of `abs (("cos" (alpha + beta),-"sin" (alpha + beta),"cos"  2 beta),("sin" alpha, "cos" alpha, "sin" beta),(-"cos" alpha, "sin" alpha, "cos" beta))` is independent of ____________.


A and B are invertible matrices of the same order such that |(AB)-1| = 8, If |A| = 2, then |B| is ____________.


If A is a square matrix of order 3 and |A| = 5, then |adj A| = ______.


If A = `[(2, -3, 5),(3, 2, -4),(1, 1, -2)]`, find A–1. Use A–1 to solve the following system of equations 2x − 3y + 5z = 11, 3x + 2y – 4z = –5, x + y – 2z = –3


A furniture factory uses three types of wood namely, teakwood, rosewood and satinwood for manufacturing three types of furniture, that are, table, chair and cot.

The wood requirements (in tonnes) for each type of furniture are given below:

  Table Chair Cot
Teakwood 2 3 4
Rosewood 1 1 2
Satinwood 3 2 1

It is found that 29 tonnes of teakwood, 13 tonnes of rosewood and 16 tonnes of satinwood are available to make all three types of furniture.

Using the above information, answer the following questions:

  1. Express the data given in the table above in the form of a set of simultaneous equations.
  2. Solve the set of simultaneous equations formed in subpart (i) by matrix method.
  3. Hence, find the number of table(s), chair(s) and cot(s) produced.

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×