Respuesta :
Answer:
Let [tex]A = (a_1, ..., a_n)[/tex] and [tex]B = (b_1, ..., b_n)[/tex] bases of V. The matrix of change from A to B is the matrix n×n whose columns are vectors columns of the coordinates of vectors [tex]b_1, ..., b_n[/tex] at base A.
The, we case correspond to find the coordinates of vectors of C,
[tex]\{\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right], \left[\begin{array}{ccc}2\\0\\-1\end{array}\right], \left[\begin{array}{ccc}-3\\1\\2\end{array}\right] \}[/tex]
at base B.
1. We need to find [tex]a,b,c\in\mathbb{R}[/tex] such that
[tex]\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right][/tex]
Then we find these values solving the linear system
[tex]\left[\begin{array}{cccc}1&-2&2&2\\-1&2&-1&-1\\0&-1&1&-1\end{array}\right][/tex]
Using rows operation we obtain the echelon form of the matrix
[tex]\left[\begin{array}{cccc}1&-2&2&2\\0&-1&1&-1\\0&0&1&1\end{array}\right][/tex]
now we use backward substitution
[tex]c=1\\-b+c=-1,\; b=2\\a-2b+2c=2,\; a=4[/tex]
Then the coordinate vector of [tex]\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right][/tex] is [tex]\left[\begin{array}{ccc}4\\2\\1\end{array}\right][/tex]
2. We need to find [tex]a,b,c\in\mathbb{R}[/tex] such that
[tex]\left[\begin{array}{ccc}2\\0\\-1\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right][/tex]
Then we find these values solving the linear system
[tex]\left[\begin{array}{cccc}1&-2&2&2\\-1&2&-1&0\\0&-1&1&-1\end{array}\right][/tex]
Using rows operation we obtain the echelon form of the matrix
[tex]\left[\begin{array}{cccc}1&-2&2&2\\0&-1&1&-1\\0&0&1&2\end{array}\right][/tex]
now we use backward substitution[tex]c=2\\-b+c=-1,\; b=3\\a-2b+2c=2,\; a=4[/tex]
Then the coordinate vector of [tex]\left[\begin{array}{ccc}2\\0\\-1\end{array}\right][/tex] is [tex]\left[\begin{array}{ccc}4\\3\\2\end{array}\right][/tex]
3. We need to find [tex]a,b,c\in\mathbb{R}[/tex] such that
[tex]\left[\begin{array}{ccc}-3\\1\\2\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right][/tex]
Then we find these values solving the linear system
[tex]\left[\begin{array}{cccc}1&-2&2&-3\\-1&2&-1&1\\0&-1&1&2\end{array}\right][/tex]
Using rows operation we obtain the echelon form of the matrix
[tex]\left[\begin{array}{cccc}1&-2&2&-3\\0&-1&1&2\\0&0&1&-2\end{array}\right][/tex]
now we use backward substitution[tex]c=-2\\-b+c=2,\; b=-4\\a-2b+2c=2,\; a=-2[/tex]
Then the coordinate vector of [tex]\left[\begin{array}{ccc}-3\\1\\2\end{array}\right][/tex] is [tex]\left[\begin{array}{ccc}-2\\-4\\-2\end{array}\right][/tex]
Then the change of basis matrix from B to C is
[tex]\left[\begin{array}{ccc}4&4&-2\\2&3&-4\\1&2&-2\end{array}\right][/tex]
