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If $A=\large\frac{1}{3}$$\begin{bmatrix} 2 & 2 & 1 \\-2 & 1 & 2 \\1 & -2 & 2 \end{bmatrix}$ prove that $A^{-1}=A^T$.

\[\begin{array} {1 1}(1) u \; is \; a \; unit \; vector & (2) \overrightarrow{u}=\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\\(3)\overrightarrow{u}=\overrightarrow{0}& (4) \overrightarrow{u}\neq \overrightarrow{0}\end{array}\]

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1 Answer

We know $\overrightarrow{u}=\overrightarrow{0}$
Hence 3 is the correct answer.
answered May 5, 2014 by meena.p
 

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