document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1168921>Question 1168921</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Let vector p=2−x+x^2 and let B be the basis for P2:<BR>\n" );
document.write( "{1+x,1+x^2,x+x^2}. <BR>\n" );
document.write( "Find [p]B. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #851470 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=851470\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Solution:<BR>\n" );
document.write( "Let the basis $B = \{v_1, v_2, v_3\}$, where $v_1 = 1+x$, $v_2 = 1+x^2$, and $v_3 = x+x^2$.<BR>\n" );
document.write( "We are given the vector $p = 2 - x + x^2$.<BR>\n" );
document.write( "We need to find the coordinates of $p$ with respect to the basis $B$, which means finding scalars $c_1, c_2, c_3$ such that:<BR>\n" );
document.write( "$p = c_1 v_1 + c_2 v_2 + c_3 v_3$<BR>\n" );
document.write( "$2 - x + x^2 = c_1(1+x) + c_2(1+x^2) + c_3(x+x^2)$<BR>\n" );
document.write( "$2 - x + x^2 = c_1 + c_1 x + c_2 + c_2 x^2 + c_3 x + c_3 x^2$\r<BR>\n" );
document.write( "\n" );
document.write( "Group the terms by powers of $x$:<BR>\n" );
document.write( "$2 - x + x^2 = (c_1 + c_2) + (c_1 + c_3)x + (c_2 + c_3)x^2$\r<BR>\n" );
document.write( "\n" );
document.write( "For this equation to hold for all $x$, the coefficients of the corresponding powers of $x$ on both sides must be equal. This gives us a system of linear equations:<BR>\n" );
document.write( "1. Coefficient of $x^0$ (constant term): $c_1 + c_2 = 2$<BR>\n" );
document.write( "2. Coefficient of $x^1$ (term with $x$): $c_1 + c_3 = -1$<BR>\n" );
document.write( "3. Coefficient of $x^2$ (term with $x^2$): $c_2 + c_3 = 1$\r<BR>\n" );
document.write( "\n" );
document.write( "We need to solve this system for $c_1, c_2, c_3$.<BR>\n" );
document.write( "From equation (1), $c_2 = 2 - c_1$.<BR>\n" );
document.write( "From equation (2), $c_3 = -1 - c_1$.\r<BR>\n" );
document.write( "\n" );
document.write( "Substitute these expressions for $c_2$ and $c_3$ into equation (3):<BR>\n" );
document.write( "$(2 - c_1) + (-1 - c_1) = 1$<BR>\n" );
document.write( "$1 - 2c_1 = 1$<BR>\n" );
document.write( "$-2c_1 = 0$<BR>\n" );
document.write( "$c_1 = 0$\r<BR>\n" );
document.write( "\n" );
document.write( "Now, substitute the value of $c_1$ back into the expressions for $c_2$ and $c_3$:<BR>\n" );
document.write( "$c_2 = 2 - c_1 = 2 - 0 = 2$<BR>\n" );
document.write( "$c_3 = -1 - c_1 = -1 - 0 = -1$\r<BR>\n" );
document.write( "\n" );
document.write( "So, the coordinates of $p$ with respect to the basis $B$ are $c_1 = 0$, $c_2 = 2$, and $c_3 = -1$.<BR>\n" );
document.write( "The coordinate vector $[p]_B$ is given by:<BR>\n" );
document.write( "$[p]_B = \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 2 \\ -1 \end{bmatrix}$\r<BR>\n" );
document.write( "\n" );
document.write( "To verify, we can substitute these values back into the linear combination:<BR>\n" );
document.write( "$0(1+x) + 2(1+x^2) + (-1)(x+x^2) = 0 + 2 + 2x^2 - x - x^2 = 2 - x + x^2 = p$.\r<BR>\n" );
document.write( "\n" );
document.write( "Final Answer: The final answer is $\boxed{\begin{bmatrix} 0 \\ 2 \\ -1 \end{bmatrix}}$ </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );