Answer:
![D=\left[\begin{array}{cc}0&0\\1&1\end{array}\right]](https://tex.z-dn.net/?f=D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0%260%5C%5C1%261%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
The matrix equation is already set up as ...
PD = Q
The normal method of solution is to multiply the equation by P^-1, but that matrix does not exist. Here, we'll solve by considering what the equation means, element by element.
Suppose matrix D looks like ...
![D=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]](https://tex.z-dn.net/?f=D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D)
We require the product PD = Q, so that means ...
0·a +0·c = 0
0·b +0·d = 0
1·a + 0·c = 0
1·b +0·d = 0
We can see from these equations that a=0 and b=0 are required. The values of c and d can be anything you like.
A suitable matrix for D could be ...
![D=\left[\begin{array}{cc}0&0\\1&1\end{array}\right]](https://tex.z-dn.net/?f=D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0%260%5C%5C1%261%5Cend%7Barray%7D%5Cright%5D)