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If $T:\Bbb R^3\to \Bbb R^2$ is a linear transformation, and the matrix of $T$ = $\left(\begin{array}{ccc}0 & 1 & 1 \\0 & 1 & -1\end{array}\right)$.

If you use the basis $\{i,j,k\}$ for $\Bbb R^3$ and and the basis $\{w_1,w_2\}$ for $\Bbb R^2$, where $w_1 = (1,1)$, $w_2 = (1,2)$. What is the matrix of $T$ relative to these bases ?

The answer is supposed to be: $T= \left(\begin{array}{ccc}0 & 1 & 3 \\0 & 0 & -2\end{array}\right)$.

However, the answer I keep getting is: $\left(\begin{array}{ccc}0 & 1 & 1.5 \\0 & 0 & -.5\end{array}\right)$.

What am I doing wrong?

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Since $T(i)=(0,0)=0.w_1+0.w_2, T(j)=(1,1)1.w_1+0.w_2$ and $T(k)=(1,-1)=3w_1-2w_2$, so the columns of matrix of $T$ w.r.t. the new bases are $(0,0)^t,(1,0)^t$ and $(3,-2)^t$ i.e. the matrix of $T$ becomes $\begin{pmatrix} 0&1&3\\ 0&0&-2 \end{pmatrix}.$

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