Answer:
[tex]A=\left[\begin{array}{ccc}1&1&0\\0&0&0\\\end{array}\right][/tex]
Step-by-step explanation:
We have a linear transformation L mapping R3 into R2 ⇒
[tex]L:[/tex] [tex]IR^{3}[/tex] ⇒ [tex]IR^{2}[/tex]
We need to find a matrix A such that [tex]L(x)=Ax[/tex] for every [tex]x[/tex] in R3
We have the formula of [tex]L(x)[/tex] ⇒
[tex]L(x)=L(\left[\begin{array}{c}x1&x2&x3\end{array}\right])=\left[\begin{array}{c}x1+x2&0\end{array}\right][/tex]
We must notice that [tex]x[/tex] is a vector in R3 and the image of L is a vector in R2.
Given that L is define as [tex]L:[/tex] [tex]IR^{3}[/tex] ⇒ [tex]IR^{2}[/tex] , the matrix that defines the linear transformation L will be a matrix A ∈ [tex]IR^{2x3}[/tex]
How can we find this matrix A? One way is to apply L to a base from the domain of L (R3) and putting that result as the columns of the matrix A.
Let's work with the cannonic base of R3 :
[tex]B_{IR^{3}}=[/tex] { [tex]\left[\begin{array}{c}1&0&0\end{array}\right],\left[\begin{array}{c}0&1&0\end{array}\right],\left[\begin{array}{c}0&0&1\end{array}\right][/tex]}
Now if we transform each vector of the base :
[tex]L(\left[\begin{array}{c}1&0&0\end{array}\right])=\left[\begin{array}{c}1&0\end{array}\right][/tex]
[tex]L(\left[\begin{array}{c}0&1&0\end{array}\right])=\left[\begin{array}{c}1&0\end{array}\right][/tex]
[tex]L(\left[\begin{array}{c}0&0&1\end{array}\right])=\left[\begin{array}{c}0&0\end{array}\right][/tex]
The final step is to put the result of each vector as the columns of the matrix A :
[tex]A=\left[\begin{array}{ccc}1&1&0\\0&0&0\\\end{array}\right][/tex] ⇒
[tex]L(x)=Ax[/tex] being A the matrix calculated. It is important to remark that the transformation L is defined in the cannonic base.
We can verify the matrix A by performing the transformation into any vector of R3. For example, let's calculate :
[tex]L(\left[\begin{array}{c}1&2&3\\\end{array}\right])=\left[\begin{array}{c}3&0\end{array}\right][/tex]
If we use the equation and :
[tex]A\left[\begin{array}{c}1&2&3\end{array}\right]=\left[\begin{array}{ccc}1&1&0\\0&0&0\end{array}\right]\left[\begin{array}{c}1&2&3\end{array}\right]=\left[\begin{array}{c}3&0\end{array}\right][/tex]
If we use the matrix A calculated.
