You try to calm them down, saying "Don’t worry, the matrix also satisfies PA = I, and therefore P is the inverse of A just like you wanted. I’ll prove it to you..." Your proof consists of the following two steps (fill in the details as your answer to this question): Step 1: Argue that your matrix P is the unique Q ∈ R n×n satisfying AQ = I. Step 2: Prove that PA = AP = I. (Hint: consider the matrix A(P+PA−I))

seems like d

step-by-step explanation:

step-by-step explanation:

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.