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In Exercises 11 and $12,$ find the $\mathcal{B}$ -matrix for the transformation $\mathbf{x} \mapsto A \mathbf{x},$ when $\mathcal{B}=\left\{\mathbf{b}_{1}, \mathbf{b}_{2}\right\}$$$A=\left[\begin{array}{rr}{3} & {4} \\ {-1} & {-1}\end{array}\right], \mathbf{b}_{1}=\left[\begin{array}{r}{2} \\ {-1}\end{array}\right], \mathbf{b}_{2}=\left[\begin{array}{l}{1} \\ {2}\end{array}\right]$$

$\left[\begin{array}{ll}{1} & {5} \\ {0} & {1}\end{array}\right]$

Calculus 3

Chapter 5

Eigenvalues and Eigenvectors

Section 4

Eigenvectors and Linear Transformations

Vectors

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were given a Matrix A and column vectors B one B two, which we're told me when in B two form the basis Be and a Is the Matrix for a transformation from X to a X and were asked to find the bee matrix for this transformation. So to do this, we're going to take p to be the column vectors. Be one matrix of column vectors. B one b two. It's The Matrix with Colin Victor's to negative one in 12 Then we have following the example in the book Be Matrix is given by P in verse a p. We can calculate p inverse pretty easily using p, using our formula for the inverse of the two by two matrix. So the determinant of P is four minus negative one, which is five. So we have 1/5 times I'm sorry times and then we have Well, we're going to want to switch D and A so we have to and then negative be so negative. One negative sees a positive one, and then a so too times. And then we have our matrix A which were given us matrix of column vectors. Three negative 14 negative one and finally, p again. So to negative. 112 And so I get 1/5 times to negative 112 and then calculating a P. I get three times two is six minus four is too. Three times one is three plus eight is 11 and negative too. Plus one is negative. One negative one minus two His negative. Three 1/5 times multiplying again. I get four plus one is five. 22 plus three is 25 tu minus two a zero and 11 minus six is five. And so distributing the scaler I get the matrix one five 01 And so this is Thebe matrix for this Trans.

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