This transformation is called \(\left[\begin{array}{l} \bar{x} \bar{y} \bar{z} \bar{w} \end{array}\right]=\left[\begin{array}{llll}
2022
This transformation is called
\(\left[\begin{array}{l} \bar{x} \\ \bar{y} \\ \bar{z} \\ \bar{w} \end{array}\right]=\left[\begin{array}{llll} a_{1} & b_{1} & c_{1} & d_{1} \\ a_{2} & b_{2} & c_{2} & d_{2} \\ a_{3} & b_{3} & c_{3} & d_{3} \\ e & f & g & h \end{array}\right]-\left[\begin{array}{l} x \\ y \\ z \\ 1 \end{array}\right]\)
- A.
Scaling
- B.
Shear
- C.
Homography
- D.
Steganography
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Correct answer: C
Answer: Homography (projective transformation).
Why: The expression applies a 4×4 matrix to the vector [x y z 1]^T. Writing coordinates with a final 1 indicates homogeneous coordinates; multiplying by a full 4×4 matrix produces a general projective mapping of 3D points. This is the projective/homography form.
Special cases: If the last row of the matrix is [0 0 0 1], the transform reduces to an affine transformation (which can include translation, rotation, scaling, and shear).
Scaling and shear are specific linear/affine transforms representable by simpler matrices; the given general 4×4 form is more general than either.
Steganography is unrelated: it refers to hiding information inside other data, not to coordinate/matrix transforms.