Determine linear transformation using matrix representation

Problem 324

Let $T$ be the linear transformation from the $3$-dimensional vector space $\R^3$ to $\R^3$ itself satisfying the following relations.
\begin{align*}
T\left(\, \begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix} \,\right)
=\begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}, \qquad T\left(\, \begin{bmatrix}
2 \\
3 \\
5
\end{bmatrix} \, \right) =
\begin{bmatrix}
0 \\
2 \\
-1
\end{bmatrix}, \qquad
T \left( \, \begin{bmatrix}
0 \\
1 \\
2
\end{bmatrix} \, \right)=
\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}.
\end{align*}
Then for any vector
\[\mathbf{x}=\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}\in \R^3,\] find the formula for $T(\mathbf{x})$.

 
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We give two solutions using two different methods.

Solution 1 using the matrix representation.

The first solution uses the matrix representation of $T$.
Let $A$ be the matrix representation of the linear transformation $T$ with respect to the standard basis of $\R^3$.
Then we have $T(\mathbf{x})=A\mathbf{x}$ by definition.
We determine the matrix $A$ as follows.
We have
\begin{align*}
A\begin{bmatrix}
1 & 2 & 0 \\
1 &3 &1 \\
1 & 5 & 2
\end{bmatrix}&= \begin{bmatrix}
A\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}, & A\begin{bmatrix}
2 \\
3 \\
5
\end{bmatrix}, & A\begin{bmatrix}
0 \\
1 \\
2
\end{bmatrix} \\
\end{bmatrix}\\[6 pt] &=\begin{bmatrix}
T\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}, & T\begin{bmatrix}
2 \\
3 \\
5
\end{bmatrix}, & T\begin{bmatrix}
0 \\
1 \\
2
\end{bmatrix} \\
\end{bmatrix}
=\begin{bmatrix}
1 & 0 & 1 \\
0 &2 &0 \\
1 & -1 & 0
\end{bmatrix}.
\end{align*}
It follows that we have
\begin{align*}
A=\begin{bmatrix}
1 & 0 & 1 \\
0 &2 &0 \\
1 & -1 & 0
\end{bmatrix}
\begin{bmatrix}
1 & 2 & 0 \\
1 &3 &1 \\
1 & 5 & 2
\end{bmatrix}^{-1}.
\end{align*}

Let us find the inverse matrix by using augmented matrix.
\begin{align*}
\left[\begin{array}{rrr|rrr}
1 & 2 & 0 & 1 &0 & 0 \\
1 & 3 & 1 & 0 & 1 & 0 \\
1 & 5 & 2 & 0 & 0 & 1 \\
\end{array} \right] \xrightarrow{\substack{R_2-R_1\\R_3-R_1}}
\left[\begin{array}{rrr|rrr}
1 & 2 & 0 & 1 &0 & 0 \\
0 & 1 & 1 & -1 & 1 & 0 \\
0 & 3 & 2 & -1 & 0 & 1 \\
\end{array} \right]\\[6 pt] \xrightarrow{\substack{R_1-2R_2 \\R_3-3R_2}}
\left[\begin{array}{rrr|rrr}
1 & 0 & -2 & 3 &-2 & 0 \\
0 & 1 & 1 & -1 & 1 & 0 \\
0 & 0 & -1 & 2 & -3 & 1 \\
\end{array} \right] \xrightarrow{-R_3}
\left[\begin{array}{rrr|rrr}
1 & 0 & -2 & 3 &-2 & 0 \\
0 & 1 & 1 & -1 & 1 & 0 \\
0 & 0 & 1 & -2 & 3 & -1 \\
\end{array} \right]\\[6 pt] \xrightarrow{\substack{R_1+2R_3\\ R_2-R_3}}
\left[\begin{array}{rrr|rrr}
1 & 0 & 0 & -1 &4 & -2 \\
0 & 1 & 0 & 1 & -2 & 1 \\
0 & 0 & 1 & -2 & 3 & -1 \\
\end{array} \right].
\end{align*}
Thus we obtain the inverse matrix
\[\begin{bmatrix}
1 & 2 & 0 \\
1 &3 &1 \\
1 & 5 & 2
\end{bmatrix}^{-1}=\begin{bmatrix}
-1 & 4 & -2 \\
1 &-2 &1 \\
-2 & 3 & -1
\end{bmatrix},\] and hence we have
\begin{align*}
A=\begin{bmatrix}
1 & 0 & 1 \\
0 &2 &0 \\
1 & -1 & 0
\end{bmatrix}
\begin{bmatrix}
-1 & 4 & -2 \\
1 &-2 &1 \\
-2 & 3 & -1
\end{bmatrix}
=\begin{bmatrix}
-3 & 7 & -3 \\
2 &-4 &2 \\
-2 & 6 & -3
\end{bmatrix}.
\end{align*}
Using the relation $T(\mathbf{x})=A\mathbf{x}$, the formula for $T(\mathbf{x})$ is obtained as follows.
\begin{align*}
T(\mathbf{x})&=A\mathbf{x}\\
&=\begin{bmatrix}
-3 & 7 & -3 \\
2 &-4 &2 \\
-2 & 6 & -3
\end{bmatrix}\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}\\[6 pt] &=\begin{bmatrix}
-3x+7y-3z \\
2x-4y+2z \\
-2x+6y-3z
\end{bmatrix}.
\end{align*}

Solution 2 using a linear combination and linearity.

The second method is to find the linear combination
\[\mathbf{x}=c_1\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}+c_2\begin{bmatrix}
2 \\
3 \\
5
\end{bmatrix}+c_3\begin{bmatrix}
0 \\
1 \\
2
\end{bmatrix}\] and use the linearity of the linear transformation.
To find the coefficients $c_1, c_2, c_3$, we consider the augmented matrix
\[ \left[\begin{array}{rrr|r}
1 & 2 & 0 & x \\
1 &3 & 1 & y \\
1 & 5 & 2 & z
\end{array} \right]\] and we reduce this matrix by elementary row operations.
The reduction operations are exactly the same as in solution 1 and we obtain
\begin{align*}
c_1&=-x+4y-2z\\
c_2&=x-2y+z\\
c_3&=-2x+3y-z.
\end{align*}

Therefore, we have by the linearity of $T$
\begin{align*}
T(\mathbf{x})&=T\left(\,c_1\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}+c_2\begin{bmatrix}
2 \\
3 \\
5
\end{bmatrix}+c_3\begin{bmatrix}
0 \\
1 \\
2
\end{bmatrix} \, \right)\\[6 pt] &=c_1T\left(\,\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}\, \right)+c_2T\left(\,\begin{bmatrix}
2 \\
3 \\
5
\end{bmatrix}\, \right)+c_3T\left(\,\begin{bmatrix}
0 \\
1 \\
2
\end{bmatrix} \, \right)\\[6 pt] &=(-x+4y-2z)\begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}+(x-2y+z)\begin{bmatrix}
0 \\
2 \\
-1
\end{bmatrix} +(-2x+3y-z)\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}\\[6 pt] &=\begin{bmatrix}
-3x+7y-3z \\
2x-4y+2z \\
-2x+6y-3z
\end{bmatrix}.
\end{align*}

Related Question.

A similar problem with a linear transformation from $\R^2$ to $\R^3$ is given in the post
“Give a formula for a linear transformation from $\R^2$ to $\R^3$“.

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