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: Spot the structure.
Write $A=\dfrac{1}{7}B$ where $B=\begin{bmatrix}3&-2&6\\-6&-3&2\\-2&6&3\end{bmatrix}$. The clean factor $\dfrac17$ is a strong hint of an orthogonal matrix.
Step 2: Check the row lengths of $B$.
Row $1$: $3^2+(-2)^2+6^2=9+4+36=49$. The same total $49$ comes out for rows $2$ and $3$. So each row of $A$ has length $\sqrt{49}/7=1$.
Step 3: Check perpendicularity of the rows.
Row$1\cdot$Row$2=-18+6+12=0$, Row$1\cdot$Row$3=-6-12+18=0$, Row$2\cdot$Row$3=12-18+6=0$. All zero.
Step 4: Conclude $A$ is orthogonal.
Unit rows that are mutually perpendicular give $AA^T=I$.
Step 5: Read off the inverse.
Multiplying $AA^T=I$ on the left by $A^{-1}$ shows $A^{-1}=A^T$.
Step 6: State the answer.
Hence the inverse of $A$ is simply its transpose.
\[ \boxed{A^{-1}=A^{T}} \]