Any arbitrary vector in \(\displaystyle{\mathbb{{{P}_{{{2}}}}}}\) can be written as,
\(\displaystyle{\left[{\mathbf{{{p}}}}\ {\left({t}\right)}\ =\ {a}\ {\left(\ {1}\ +\ {t}^{{{2}}}\ \right)}\ +\ {b}\ {\left(\ {t}\ +\ {t}^{{{2}}}\ \right)}\ +\ {c}\ {\left(\ {1}\ +\ {2}\ {t}\ +\ {t}^{{{2}}}\ \right)}\right]}\)
Thus, for the vector \(\displaystyle{\left[{\mathbf{{{p}}}}\ {\left(\ {t}\ \right)}={6}\ +\ {3}\ {t}\ -\ {t}^{{{2}}}\right]}\) can be written as,
1) \(\displaystyle{\left[{a}\ {\left(\ {1}\ +\ {t}\ \right)}\ +\ {b}\ {\left(\ {1}\ +\ {t}^{{{2}}}\ \right)}\ +\ {c}\ {\left(\ {t}\ +\ {t}^{{{2}}}\ \right)}={6}\ +\ {3}{t}\ -\ {t}^{{{2}}}\right]}\)
On comparing the terms of both the sides in the above equation (1) gives the following equations.
\(\displaystyle{a}\ +\ {b}={6}\)
\(\displaystyle{a}\ +\ {c}={3}\)
\(\displaystyle{b}\ +\ {c}=-{1}\)
The representation of the linear system in a matrix is,
\(\begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} a \\ c \\ d \end{bmatrix}=\begin{bmatrix} 6 \\ 3 \\ -1 \end{bmatrix}\)
The solution of the above system gives as by solving the augmented matrix
\(\begin{bmatrix} 1 & 1 & 0 & 6 \\ 1 & 0 & 1 & 3 \\ 0 & 1 & 1 & -1 \end{bmatrix}\)
Use Elementary row transformations to reduce the augmented matrix to row reduced Echelon form.
Step 1:
\(\begin{bmatrix} 1 & 1 & 0 &\ \ 6 \\ 1 & 0 & 1 &\ \ 3 \\ 0 & 1 & 1 & -1 \end{bmatrix}\underrightarrow{R_{2}\ \rightarrow\ R_{2}\ -\ R_{1}} \begin{bmatrix} 1 & 1 & 0 &\ \ \ 6 \\ 0 & -1\ \ \ & 1 & -3 \\ 0 & 1 & 1 & -1 \end{bmatrix}\)
Step 2
\(\begin{bmatrix} 1 & 1 & 0 &\ \ \ 6 \\ 0 & -1\ \ \ & 1 & -3 \\ 0 & 1 & 1 & -1 \end{bmatrix} \underrightarrow{R_{3}\ \rightarrow\ R_{3}\ +\ R_{2}} \begin{bmatrix} 1 & 1 & 0 &\ \ \ 6 \\ 0 & -1\ \ \ & 1 & -3 \\ 0 & 0 & 2 & -4 \end{bmatrix}\)
Step 3:
\(\begin{bmatrix} 1 & 1 & 0 &\ \ \ 6 \\ 0 & -1\ \ \ & 1 & -3 \\ 0 & 0 & 2 & -4 \end{bmatrix}\underrightarrow {R_{1}\ \rightarrow\ R_{1}\ +\ R_{2}\ - \frac{1}{2}\ R_{3}} \begin{bmatrix} 1 & 0 & 0 &\ \ \ 5 \\ 0 & -1\ \ \ & 1 & -3 \\ 0 & 0 & 2 & -4 \end{bmatrix}\)
Step 4:
\(\begin{bmatrix} 1 & 0 & 0 &\ \ \ 5 \\ 0 & -1\ \ \ & 1 & -3 \\ 0 & 0 & 2 & -4 \end{bmatrix}\underrightarrow {R_{2}\ \rightarrow\ R_{2}\ - \frac{1}{2}\ R_{3}} \begin{bmatrix} 1 & 0 & 0 &\ \ \ 5 \\ 0 & 1 & 0 &\ \ \ 1 \\ 0 & 0 & 1 & -2 \end{bmatrix}\)
Step 5:
\(\begin{bmatrix} 1 & 0 & 0 &\ \ \ 5 \\ 0 & -1\ \ \ & 1 & -1 \\ 0 & 0 & 2 & -4 \end{bmatrix}\underrightarrow {R_{2}\ \rightarrow\ -1\ \times\ R_{2}}_{R_{3}\rightarrow \frac{1}{2}\ R_{3}} \begin{bmatrix} 1 & 0 & 0 &\ \ \ 5 \\ 0 & 1 & 0 &\ \ \ 1 \\ 0 & 0 & 1 & -2 \end{bmatrix}\)
Thus, the solution is \(\begin{bmatrix} a \\ c \\ d \end{bmatrix}=\begin{bmatrix} \ \ \ \ 5 \\ \ \ \ \ 1 \\ -2 \end{bmatrix}\)
Therefore, the coordinate vector \(\mathbf {[p]_{\mathscr {B}}}\ is \begin{bmatrix} \ \ \ \ 5 \\ \ \ \ \ 1 \\ -2 \end{bmatrix}\)
The coordinate vector of \displaystyle{\left[{\mathbf{{{p}}}}
The coordinate vector of \displaystyle{\left[{\mathbf{{{p}}}}
Answers (1)
Relevant Questions
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