# Determine a Value of Linear Transformation From $\R^3$ to $\R^2$

## Problem 368

Let $T$ be a linear transformation from $\R^3$ to $\R^2$ such that
$T\left(\, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}\,\right) =\begin{bmatrix} 1 \\ 2 \end{bmatrix} \text{ and }T\left(\, \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}\,\right)=\begin{bmatrix} 0 \\ 1 \end{bmatrix}.$ Then find $T\left(\, \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} \,\right)$.

(The Ohio State University, Linear Algebra Exam Problem)

## Solution.

We first express the vector $\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}$ as a linear combination
$\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}=c_1\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}+c_2\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}.$ Then we find that $c_1=-1$ and $c_2=2$. Hence we obtain
$\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}=-\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}+2\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}.$

We now compute
\begin{align*}
T\left(\, \begin{bmatrix}
0 \\
1 \\
2
\end{bmatrix} \,\right)
&=T\left(\, -\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix}+2\begin{bmatrix}
0 \\
1 \\
1
\end{bmatrix} \,\right)\\
&=-T\left(\, \begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix} \,\right)+2\left(\, \begin{bmatrix}
0 \\
1 \\
1
\end{bmatrix} \,\right) && \text{by linearity of $T$}\\
&=-\begin{bmatrix}
1 \\
2
\end{bmatrix}+2\begin{bmatrix}
0 \\
1
\end{bmatrix}\\
&=\begin{bmatrix}
-1 \\
0
\end{bmatrix}.
\end{align*}
Therefore we have found that
$T\left(\, \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} \,\right)=\begin{bmatrix} -1 \\ 0 \end{bmatrix}$

## Linear Algebra Midterm Exam 2 Problems and Solutions

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