Finding linear transformation from matrix

Question

I tried solving this question by taking given matrix $[A]_{B_{1}}$ = $[T]_{B_{2}}$. The answer that I am getting is $(35,-10)$. Tried solving several times and still reaching at the same answer. Which is the right way to solve this question?

Answer 1

First, you write $(5,5)$ as a linear combination of $(1,2)$ and $(2,-1)$. It turns out that $(5,5)=3(1,2)+(2,-1)$. So, now you know that\begin{align}T(5,5)&=3T(1,2)+T(2,-1)\\&=3(4,2)+(3,-4)\\&=(15,2).\end{align}

Answer 2

Write $\;(5,5)\;$ as a linear combination of $\;B_1\;$ , apply $\;T\;$ and then this last write as a lin. comb. of $\;B_2\;$ (though this last is the usual, canonical basis so you won't have to do anything!):

$$\binom55=3\binom21+1\binom2{-1}\implies \left[\binom 55\right]_{B_1}=\binom31$$

so now

$$T\left[\binom 55\right]_{B_1}=\begin{pmatrix}4&3\\2&\!-4\end{pmatrix}\binom31=\binom{15}2$$

Answer 3

Hey that Matrix $[T]_{\mathcal{B_1}}^{\mathcal{B_2}}$ will take input in Basis $\mathcal{B_1}$ and spit the output in Basis $\mathcal{B_2}$ .Therefore you cannot feed the input in our standard basis. Convert $(5,5)$ into basis $\mathcal{B_1}$ and then feed.

W.K.T $(5,5)=3(1,2)+1(2,-1)$ There fore putting in the matrix $(3,1)$ gives us $(15,2)$ as output convert this into our standard basis which is $\mathcal{B_2}$.

Therefore Option D is correct