Maths > Matrices and Determinants > 3.0 Special Matrices

  Matrices and Determinants
    1.0 Introduction
    2.0 Algebra of Matrices
    3.0 Special Matrices
    4.0 Determinant of a square matrix
    5.0 Adjoint of a square Matrix
    6.0 Inverse of a Matrix
    7.0 Types of Equations Homogenous & Non-Homogenous
    8.0 Cramer's rule
    9.0 Types of Linear Equations

3.6 (f) Orthogonal Matrix :
Any square matrix $A$ of order $n$ is said to be orthogonal if $AA' = A'A = {I_n}$.


Question 5.

Prove that $A = \frac{1}{3}\left[ {\begin{array}{c} 1&{ - 2}&2 \\ { - 2}&1&2 \\ { - 2}&{ - 2}&{ - 1} \end{array}} \right]$ is an orthogonal matrix.

Solution:

Given $A = \frac{1}{3}\left[ {\begin{array}{c} 1&{ - 2}&2 \\ { - 2}&1&2 \\ { - 2}&{ - 2}&{ - 1} \end{array}} \right]$

$\therefore {A^T} = \frac{1}{3}\left[ {\begin{array}{c} 1&{ - 2}&{ - 2} \\ { - 2}&1&{ - 2} \\ { - 2}&{ - 2}&{ - 1} \end{array}} \right]$

$\therefore A{A^T} = \frac{1}{3}\left[ {\begin{array}{c} 1&{ - 2}&2 \\ { - 2}&1&2 \\ { - 2}&{ - 2}&{ - 1} \end{array}} \right] \times \frac{1}{3}\left[ {\begin{array}{c} 1&{ - 2}&{ - 2} \\ { - 2}&1&{ - 2} \\ { - 2}&{ - 2}&{ - 1} \end{array}} \right]$

$ = \frac{1}{9}\left[ {\begin{array}{c} {1 + 4 + 4}&{ - 2 - 2 + 4}&{ - 2 + 4 - 2} \\ { - 2 - 2 + 4}&{4 + 1 + 4}&{4 - 2 - 2} \\ { - 2 + 4 - 2}&{4 - 2 - 2}&{4 + 4 + 1} \end{array}} \right]$

$ = \frac{1}{9}\left[ {\begin{array}{c} 9&0&0 \\ 0&9&0 \\ 0&0&9 \end{array}} \right]$

$ = \left[ {\begin{array}{c} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right]$

$ = I$

Hence $A$ is an orthogonal matrix.

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