Linear Algebra, Linear Transformation problem

Question

My task is this

I am wondering how to go about doing this.

Anyone have ant idea?

Thanks!

Answer 1

$\text{(a)}$ We are looking for a matrix $P = (p_{i j})_{1 \leq i, j \leq 3}$ such that $$\left \{\begin{array}{c c c c c c c} p_{1 1} R &+& p_{1 2} G &+& p_{1 3} B &=& I \\ p_{2 1} R &+& p_{2 2} G &+& p_{2 3} B &=& L \\ p_{3 1} R &+& p_{3 2} G &+& p_{3 3} B &=& S \end{array} \right.$$

Using the formulae given, we deduce that $$P = \begin{pmatrix} \dfrac{1}{3} & \dfrac{1}{3} & \dfrac{1}{3} \\ 1 & -1 & 0 \\ -\dfrac{1}{2} & -\dfrac{1}{2} & 1 \end{pmatrix}$$

$\text{(b)}$ Let $R', G', B'$ be the amount of red light, green light and blue light after the light has passed through the sunglasses. We have $$\left \{\begin{array}{c c c c c c c} R' &=& R \\ G' &=& G \\ B' &=& 0 \end{array} \right.$$

By doing exactly the same thing as in the previous question, we find $$A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$

$\text{(c)}$ The matrix for the composition of the transformations is simply the product $P A$ (Notice that the matrix corresponding to the first transformation is to the right of the matrix corresponding to the second transformation).

$$P A = \begin{pmatrix} \dfrac{1}{3} & \dfrac{1}{3} & \dfrac{1}{3} \\ 1 & -1 & 0 \\ -\dfrac{1}{2} & -\dfrac{1}{2} & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} \dfrac{1}{3} & \dfrac{1}{3} & 0 \\ 1 & -1 & 0 \\ -\dfrac{1}{2} & -\dfrac{1}{2} & 0 \end{pmatrix} $$