a. [tex]\(A(1,0), B(2,2)\)[/tex], and [tex]\(C(0,3)\)[/tex] are vertices of [tex]\(\triangle ABC\)[/tex]. [tex]\(\triangle ABC\)[/tex] is transformed by the matrix [tex]\(\begin{pmatrix} 2 & -1 \\ 0 & 1 \end{pmatrix}\)[/tex]. Find the images of [tex]\(\triangle ABC\)[/tex].



Answer :

Let's find the images of the vertices [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex] of triangle [tex]\( \triangle ABC \)[/tex] when it undergoes the transformation specified by the matrix [tex]\(\left(\begin{array}{cc}2 & -1 \\ 0 & 1\end{array}\right)\)[/tex]. We will apply this transformation matrix to each vertex.

1. Vertex [tex]\( A(1, 0) \)[/tex]:

To find the image of [tex]\( A \)[/tex], we will multiply the transformation matrix by the coordinates of [tex]\( A \)[/tex]:
[tex]\[ \left(\begin{array}{cc} 2 & -1 \\ 0 & 1 \end{array}\right) \left( \begin{array}{c} 1 \\ 0 \end{array} \right) = \left(\begin{array}{c} 2 \cdot 1 + (-1) \cdot 0 \\ 0 \cdot 1 + 1 \cdot 0 \end{array} \right) = \left( \begin{array}{c} 2 \\ 0 \end{array} \right) \][/tex]

So, the image of [tex]\( A \)[/tex] is [tex]\( A'(2, 0) \)[/tex].

2. Vertex [tex]\( B(2, 2) \)[/tex]:

To find the image of [tex]\( B \)[/tex], we will multiply the transformation matrix by the coordinates of [tex]\( B \)[/tex]:
[tex]\[ \left(\begin{array}{cc} 2 & -1 \\ 0 & 1 \end{array}\right) \left( \begin{array}{c} 2 \\ 2 \end{array} \right) = \left( \begin{array}{c} 2 \cdot 2 + (-1) \cdot 2 \\ 0 \cdot 2 + 1 \cdot 2 \end{array} \right) = \left( \begin{array}{c} 4 - 2 \\ 0 + 2 \end{array} \right) = \left( \begin{array}{c} 2 \\ 2 \end{array} \right) \][/tex]

So, the image of [tex]\( B \)[/tex] is [tex]\( B'(2, 2) \)[/tex].

3. Vertex [tex]\( C(0, 3) \)[/tex]:

To find the image of [tex]\( C \)[/tex], we will multiply the transformation matrix by the coordinates of [tex]\( C \)[/tex]:
[tex]\[ \left(\begin{array}{cc} 2 & -1 \\ 0 & 1 \end{array}\right) \left( \begin{array}{c} 0 \\ 3 \end{array} \right) = \left( \begin{array}{c} 2 \cdot 0 + (-1) \cdot 3 \\ 0 \cdot 0 + 1 \cdot 3 \end{array} \right) = \left( \begin{array}{c} 0 - 3 \\ 0 + 3 \end{array} \right) = \left( \begin{array}{c} -3 \\ 3 \end{array} \right) \][/tex]

So, the image of [tex]\( C \)[/tex] is [tex]\( C'(-3, 3) \)[/tex].

In conclusion, the images of the vertices of [tex]\( \triangle ABC \)[/tex] after the transformation are:
- [tex]\( A'(2, 0) \)[/tex]
- [tex]\( B'(2, 2) \)[/tex]
- [tex]\( C'(-3, 3) \)[/tex]