$\displaystyle A =\left(

\begin{array}{cc}

1 & 3 \\

2 & 4 \\

-1 & -1 \\

0 & 1

\end{array}

\right)

$

$\displaystyle Q = \left(

\begin{array}{cc}

1/\sqrt{6} & 1/\sqrt{3}\\

2/\sqrt{6} & 0\\

-1/\sqrt{6} & 1/\sqrt{3}\\

0 & 1/\sqrt{3}

\end{array}

\right)

$

The columns of Q were obtained by applying the Gram-Schmidt Process to the columns of A. Find the upper triangular matrix $\displaystyle R$ such that $\displaystyle A=QR$

Now I'm pretty sure that $\displaystyle R=Q^TA$ could someone please verify this and show what they got for $\displaystyle R$ thanks