Science:Math Exam Resources/Courses/MATH307/April 2005/Question 06 (b)

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MATH307 April 2005

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Question 06 (b)
Let B be the basis $\{v_{1},v_{2},v_{3}\}$ of $\mathbb {R} ^{3}$ , where $v_{1}=\left[{\begin{array}{c}1\\0\\0\end{array}}\right],\quad v_{2}=\left[{\begin{array}{c}1\\1\\0\end{array}}\right],\quad v_{3}=\left[{\begin{array}{c}1\\1\\1\end{array}}\right]$ and let $T:\mathbb {R} ^{3}\to \mathbb {R} ^{3}$ be the linear transformation such that $M_{B}^{B}(T)=\left[{\begin{array}{ccc}2&-3&3\\-2&1&-7\\5&-5&7\end{array}}\right]$ (b) Calculate the matrix T with respect to the standard basis {e₁, e₂, e₃} of $\mathbb {R} ^{3}$ . You may leave your answer unsimplified.

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Hint
Science:Math Exam Resources/Courses/MATH307/April 2005/Question 06 (b)/Hint 1

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Solution
The transformations are linear so we can use the formula $Tv_{1}=T(e_{1})$ $Tv_{2}=T(e_{1}+e_{2})=Te_{1}+Te_{2}$ and Tv₃=T(e₁+e₂+e₃) = Te₁+Te₂+Te₃ where e is the standard basis. We also know that $Tv_{n}$ are the columns of $M_{B}^{B}$ $T(v_{1})=T(e_{1})=\left[{\begin{array}{c}2\\-2\\5\end{array}}\right]$ $T(v_{2})=T(v_{1})+T(e_{2})$ $T(e_{2})=\left[{\begin{array}{c}-3\\1\\-5\end{array}}\right]-\left[{\begin{array}{c}2\\-2\\5\end{array}}\right]=\left[{\begin{array}{c}5\\-3\\10\end{array}}\right]$ $T(v_{3})=T(v_{2})+T(e_{3})=\left[{\begin{array}{c}3\\-7\\7\end{array}}\right]$ $T(e_{3})=\left[{\begin{array}{c}3\\-7\\7\end{array}}\right]-\left[{\begin{array}{c}5\\-3\\10\end{array}}\right]=\left[{\begin{array}{c}-2\\-4\\-3\end{array}}\right]$ So we have our transformation matrix $T=\left[{\begin{array}{ccc}2&5&-2\\-2&-3&-4\\5&10&-3\end{array}}\right]$