Science:Math Exam Resources/Courses/MATH221/December 2011/Question 05 (b)

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MATH221 December 2011

• Q1 (a) • Q1 (b) • Q2 (a) • Q2 (b) • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) • Q6 (a) • Q6 (b) • Q7 • Q8 (a) • Q8 (b) •

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Question 05 (b)
Let $b_{1}=\left[\!\!{\begin{array}{c}1\\-1\end{array}}\!\!\right]\quad b_{2}=\left[{\begin{array}{c}1\\0\end{array}}\right]\quad b_{3}=\left[{\begin{array}{c}3\\4\end{array}}\right]$ Let T denote a linear transformation of R² such that $T(b_{1})=\left[{\begin{array}{c}1\\1\end{array}}\right]\quad T(b_{2})=\left[{\begin{array}{c}1\\2\end{array}}\right]$ Find T(b₃) and give the matrix of T with respect to the standard basis of R²

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
To find T(b₃) express b₃ as a linear combination of b₁ and b₂. Then use the linearity of T.

Hint 2
To find the matrix representation of T you need to calculate T(e₁) and T(e₂), where e₁ = [1,0] and e₂ = [0,1]. Again, linear combinations and linearity of T will be your friends.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
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Solution
Since $\displaystyle b_{3}=7b_{2}-4b_{1}$ we have that ${\begin{aligned}T(b_{3})&=T(7b_{2}-4b_{1})\\&=7T(b_{2})-4T(b_{1})\\&=7\left[{\begin{array}{c}1\\2\end{array}}\right]-4\left[{\begin{array}{c}1\\1\end{array}}\right]\\&=\left[{\begin{array}{c}3\\10\end{array}}\right]\end{aligned}}$ To find the matrix of T with respect to the standard basis, we need to compute the image of each of these standard vectors. ${\begin{aligned}T(e_{1})&=T\left(\left[{\begin{array}{c}1\\0\end{array}}\right]\right)\\&=T(b_{2})\\&=\left[{\begin{array}{c}1\\2\end{array}}\right]\end{aligned}}$ and ${\begin{aligned}T(e_{2})&=T\left(\left[{\begin{array}{c}0\\1\end{array}}\right]\right)\\&=T(b_{2}-b_{1})\\&=T(b_{2})-T(b_{1})\\&=\left[{\begin{array}{c}1\\2\end{array}}\right]-\left[{\begin{array}{c}1\\1\end{array}}\right]&=\left[{\begin{array}{c}0\\1\end{array}}\right]\end{aligned}}$ And so the matrix of T in the standard basis is $\left[{\begin{array}{cc}1&0\\2&1\end{array}}\right]$ Note that you can check that your result makes sense by verifying this matrix on the vector b₃, indeed we have that $\left[{\begin{array}{cc}1&0\\2&1\end{array}}\right]\left[{\begin{array}{c}3\\4\end{array}}\right]=\left[{\begin{array}{c}3\\10\end{array}}\right]$