Question

If a matrix \(A=\dfrac{1}{3}\begin{bmatrix}1& 2& 2\\2& 1&-2-2& 2&-1\end{bmatrix}\), then

• A. \(|A|\ne 1\)
• B. \(A\) is a symmetric matrix
• C. \(A\) is a skew symmetric matrix
• D. \(A\) is an orthogonal matrix

Hint:

To check an orthogonal matrix, verify that rows or columns are mutually perpendicular and each has unit length.

Answer 1

The Correct Option is D

Concept:
A square matrix \(A\) is called orthogonal if: \[ A^TA=I \] This means the rows or columns of the matrix are mutually perpendicular and have unit length.

Step 1: Write the given matrix. \[ A=\frac{1}{3}\begin{bmatrix} 1& 2& 2\\ 2& 1&-2\\ -2& 2&-1 \end{bmatrix} \]

Step 2: Check row lengths before multiplication by \(\frac{1}{3}\).
First row: \[ 1^2+2^2+2^2=1+4+4=9 \] Second row: \[ 2^2+1^2+(-2)^2=4+1+4=9 \] Third row: \[ (-2)^2+2^2+(-1)^2=4+4+1=9 \] After multiplying by \(\frac{1}{3}\), each row has length: \[ \sqrt{\frac{9}{9}}=1 \]

Step 3: Check perpendicularity of rows.
Dot product of first and second rows: \[ (1)(2)+(2)(1)+(2)(-2)=2+2-4=0 \] Dot product of first and third rows: \[ (1)(-2)+(2)(2)+(2)(-1)=-2+4-2=0 \] Dot product of second and third rows: \[ (2)(-2)+(1)(2)+(-2)(-1)=-4+2+2=0 \]

Step 4: Conclusion.
All rows are unit vectors and mutually perpendicular. Therefore: \[ A^TA=I \] Hence, \(A\) is an orthogonal matrix. \[ \therefore \text{Correct Answer is (D)} \]