Question:

If A=13[122212a2b]A = \frac{1}{3} \begin{bmatrix}1&2&2\\ 2&1&-2\\ a&2&b\end{bmatrix} is an orthogonal matrix, then

Updated On: Jun 18, 2022
  • a = - 2, b = - 1
  • a = 2, b = 1
  • a = 2, b = - 1
  • a = - 2, b = 1
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The Correct Option is A

Solution and Explanation

As A is an orthogonal matrix, AAT=IAA^T = I
13[122212a2b].13[12a21222b]=[100010001]\Rightarrow \frac{1}{3} \begin{bmatrix}1&2&2\\ 2&1&-2\\ a&2&b\end{bmatrix} . \frac{1}{3} \begin{bmatrix}1&2&a\\ 2&1&2\\ 2&-2&b\end{bmatrix} = \begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}
19[122212a2b][12a21222b]=[100010001]\Rightarrow \frac{1}{9} \begin{bmatrix}1&2&2\\ 2&1&-2\\ a&2&b\end{bmatrix} \begin{bmatrix}1&2&a\\ 2&1&2\\ 2&-2&b\end{bmatrix} = \begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}
[90a+4+2b092a+22ba+4+2b2a+22ba2+4+b2]=[900090009]\Rightarrow \begin{bmatrix}9&0&a+4+2b\\ 0&9&2a+2-2b\\ a+4+2b&2a+2-2b&a^{2}+4+b^{2}\end{bmatrix} = \begin{bmatrix}9&0&0\\ 0&9&0\\ 0&0&9\end{bmatrix}
a+4+2b=0,2a+22b=0,a2+4+b2=9\Rightarrow a + 4 + 2b = 0, 2a + 2 - 2b = 0 , a^{2} + 4 + b^{2} = 9
a+2b+4=0,ab+1=0a2+b2=5\Rightarrow a + 2b + 4 = 0, a - b + 1 = 0 a^{2} + b^{2} = 5
a=2,b=1\Rightarrow a = - 2, b = - 1
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Concepts Used:

Transpose of a Matrix

The matrix acquired by interchanging the rows and columns of the parent matrix is called the Transpose matrix. The transpose matrix is also defined as - “A Matrix which is formed by transposing all the rows of a given matrix into columns and vice-versa.”

The transpose matrix of A is represented by A’. It can be better understood by the given example:

A Matrix
A' Matrix
The transpose matrix of A is denoted by A’

Now, in Matrix A, the number of rows was 4 and the number of columns was 3 but, on taking the transpose of A we acquired A’ having 3 rows and 4 columns. Consequently, the vertical Matrix gets converted into Horizontal Matrix.

Hence, we can say if the matrix before transposing was a vertical matrix, it will be transposed to a horizontal matrix and vice-versa.

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