0.03 you just times it by ten or if its a decimal take away the 0 infront of it
Answer:
y-determinant = 2
Step-by-step explanation:
Given the following system of equation:
Let's represent it using a matrix:
![\left[\begin{array}{ccc}1&2\\1&-3\end{array}\right] = \left[\begin{array}{ccc}5\\7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%5C%5C1%26-3%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%5C%5C7%5Cend%7Barray%7D%5Cright%5D)
The y‐numerator determinant is formed by taking the constant terms from the system and placing them in the y‐coefficient positions and retaining the x‐coefficients. Then:
![\left[\begin{array}{ccc}1&5\\1&7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%265%5C%5C1%267%5Cend%7Barray%7D%5Cright%5D%20)
y-determinant = (1)(7) - (5)(1) = 2.
Therefore, the y-determinant = 2
The purpose of the tensor-on-tensor regression, which we examine, is to relate tensor responses to tensor covariates with a low Tucker rank parameter tensor/matrix without being aware of its intrinsic rank beforehand.
By examining the impact of rank over-parameterization, we suggest the Riemannian Gradient Descent (RGD) and Riemannian Gauss-Newton (RGN) methods to address the problem of unknown rank. By demonstrating that RGD and RGN, respectively, converge linearly and quadratically to a statistically optimal estimate in both rank correctly-parameterized and over-parameterized scenarios, we offer the first convergence guarantee for the generic tensor-on-tensor regression. According to our theory, Riemannian optimization techniques automatically adjust to over-parameterization without requiring implementation changes.
Learn more about tensor-on-tensor here
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Answer:
could you finish the problem?
Step-by-step explanation:
Also this is in the wrong section
