Question

If
\[ A=\frac{1}{7} \begin{bmatrix} 3 & -2 & 6 \\ -6 & -3 & 2 \\ -2 & 6 & 3 \end{bmatrix}, \] then

• A. \(A^{-1}=A\)
• B. \(A^{-1}=A^T\)
• C. \(A^{-1}\) does not exist
• D. \(A^{-1}=-A\)

Hint:

If the rows or columns of a matrix are mutually orthogonal and each has unit length, then the matrix is orthogonal and satisfies \(A^{-1}=A^T\).

Answer 1

The Correct Option is B

Step 1: Recall the condition for an orthogonal matrix.
A square matrix \(A\) is orthogonal if
\[ AA^T=I \] where \(I\) is the identity matrix.
For an orthogonal matrix, we have
\[ A^{-1}=A^T \]

Step 2: Write the matrix without the scalar factor.
Let
\[ B= \begin{bmatrix} 3 & -2 & 6 \\ -6 & -3 & 2 \\ -2 & 6 & 3 \end{bmatrix} \] Then,
\[ A=\frac{1}{7}B \]

Step 3: Verify orthogonality by checking row vectors.
The rows of \(B\) are
\[ R_1=(3,-2,6) \] \[ R_2=(-6,-3,2) \] \[ R_3=(-2,6,3) \]
Now compute their dot products.
\[ R_1\cdot R_2 = 3(-6)+(-2)(-3)+6(2) \] \[ =-18+6+12=0 \]
\[ R_1\cdot R_3 = 3(-2)+(-2)(6)+6(3) \] \[ =-6-12+18=0 \]
\[ R_2\cdot R_3 = (-6)(-2)+(-3)(6)+2(3) \] \[ =12-18+6=0 \]
Thus, all rows are mutually orthogonal.
Now find the length of each row:
\[ |R_1|^2=3^2+(-2)^2+6^2 \] \[ =9+4+36=49 \]
Similarly,
\[ |R_2|^2=49 \] and
\[ |R_3|^2=49 \]
Hence, after multiplying by \(\frac{1}{7}\), each row becomes a unit vector.
Therefore, the rows of \(A\) are orthonormal.

Step 4: Conclude the nature of the matrix.
Since the rows are orthonormal,
\[ AA^T=I \] Hence, \(A\) is an orthogonal matrix.
Therefore,
\[ A^{-1}=A^T \]

Step 5: Final conclusion.
Hence,
\[ \boxed{A^{-1}=A^T} \]