Question

For a matrix [M] \(= \left[ {\begin{array}{*{20}{c}} {\frac{3}{5}}&{\frac{4}{5}}\\ x&{\frac{3}{5}} \end{array}} \right]\), the transpose of the matrix is equal to the inverse of the matrix [M]^T = [M]^-1. The value of x is given by

1. \(- \frac{4}{5}\)
2. \(- \frac{3}{5}\)
3. \(\frac{3}{5}\)
4. \(\frac{4}{5}\)

Answer 1

Correct Answer - Option 1 : \(- \frac{4}{5}\)

Concept:

Orthogonal matrix:

For an orthogonal matrix, transpose is equal to its inverse. 

[M]^T = [M^-1]

For an orthogonal matrix:

MM^T = I

Calculation:

Given:

[M] \(= \left[ {\begin{array}{*{20}{c}} {\frac{3}{5}}&{\frac{4}{5}}\\ x&{\frac{3}{5}} \end{array}} \right]\;\),

Given M is the orthogonal matrix

 ∴ [M][M]^T = I

 \(\left[ {\begin{array}{*{20}{c}} {\frac{3}{5}}&{\frac{4}{5}}\\ x&{\frac{3}{5}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\frac{3}{5}}&x\\ {\frac{4}{5}}&{\frac{3}{5}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\;\)

\(= \left[ {\begin{array}{*{20}{c}} {1}&{\frac{3x}{5}} +\frac{12}{25}\\ \frac{3x}{5}+\frac{12}{25}&x^2+{\frac{9}{25}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\;\)

so, \(\frac{{12}}{{25}} + \frac{3}{5}x = 0 \Rightarrow x = -\frac{{4}}{5}\;\)

The value of x is \(\frac{{ - 4}}{5}\)