Answer to Question #102725 in Linear Algebra for Hamna

Question

Consider the basis S = {v1, v2, v3} for R3, where v1 = (1, 2, 1), v2 = (2, 9, 0), and v3 = (3, 3, 4), and let T:R3 rightarrow R2 be the linear operator such that T(V1) = (1, 0), T(v2) = (- 1, 1), T(v3) = (0, 1) Find a formula for T(x1, x2, x3), and use that formula to find

Answer 1

In case we want to transform R3 into R2 we need a Matrix of size 3x2.$\begin{pmatrix} x1 & x4 \\ x2 & x5 \\ x3 & x6 \end{pmatrix}$ The Matrix gives us 6 variables which we can find by solving 6 corresponding equations 

$\begin{cases} x1+2*x2+x3=1\\ x4+2*x5+x6=0\\ 2*x1+9*x2=-1\\ 2*x4+9*x5=1\\ 3*x1+3*x2+4*x3=0\\ 3*x4+3*x5+4*x6=1\\ \end{cases}$ $\implies$ $\begin{cases} x1+2*x2+x3=1\\ 3*x1+3*x2+4*x3=0\\ 2*x1+9*x2=-1\\ \end{cases} \begin{cases} 2*x4+9*x5=1\\ 3*x4+3*x5+4*x6=1\\ x4+2*x5+x6=0\\ \end{cases}$ $\implies$

$\begin{cases} x1+5*x2=4\\ 3*x1+3*x2+4*x3=0\\ 2*x1+9*x2=-1\\ \end{cases} \begin{cases} 2*x4+9*x5=1\\ 3*x4+3*x5+4*x6=1\\ x4+5*x5=-1\\ \end{cases}$ $\implies$

$\begin{cases} x2=9\\ x1=-41\\ x3=24\\ \end{cases} \begin{cases} x5=-3\\ x4=14\\ x6=-8\\ \end{cases}$

Solution is $\begin{pmatrix} -41 & 14 \\ 9 & -3 \\ 24 & -8 \end{pmatrix}$.

Then we can get the formula, which is T(x,y,z)= (-41*x+9*y+24*z, 14*x-3*y-8*z)