6. 7 6 12 6 and v = - 7 Let A = -5 1 , b = - 7 Compute Au and Av, and compare them with b. Could u possibly be a least-squares solution of Ax = b? -2 7 6 5 (Answer this without computing a least-squares solution.) Au = (Simplify your answer.) Av = (Simplify your answer.) Compare Au and Av with b. Could u possibly be a least-squares solution of Ax = b? O A. Au is closer to b than Av is. Thus, u could possibly be a least-squares solution of Ax = b. O B. Av is closer to b than Au is. Thus, u cannot possibly be a least-squares solution of Ax = b. O C. Av and Au are equally close to b. Thus, both v and u can be a least-squares solution of Ax b. O D. Av and Au are equally close to b. Thus, neither v nor u can be a least-squares solution of Ax = b.
Correlation
Correlation defines a relationship between two independent variables. It tells the degree to which variables move in relation to each other. When two sets of data are related to each other, there is a correlation between them.
Linear Correlation
A correlation is used to determine the relationships between numerical and categorical variables. In other words, it is an indicator of how things are connected to one another. The correlation analysis is the study of how variables are related.
Regression Analysis
Regression analysis is a statistical method in which it estimates the relationship between a dependent variable and one or more independent variable. In simple terms dependent variable is called as outcome variable and independent variable is called as predictors. Regression analysis is one of the methods to find the trends in data. The independent variable used in Regression analysis is named Predictor variable. It offers data of an associated dependent variable regarding a particular outcome.
![### Least-Squares Solution Comparison
Given the matrix \( A \), vector \( b \), and two potential solution vectors \( u \) and \( v \):
\[ A = \begin{bmatrix} 7 & 6 & 1 \\ -5 & 1 & -1 \\ 7 & 6 & 0 \end{bmatrix}, \]
\[ b = \begin{bmatrix} 12 \\ -7 \\ 5 \end{bmatrix}, \]
\[ u = \begin{bmatrix} 6 \\ -2 \\ 2 \end{bmatrix}, \]
\[ v = \begin{bmatrix} 6 \\ -7 \\ -7 \end{bmatrix}. \]
Compute \( Au \) and \( Av \), and compare them with \( b \). Determine if \( u \) could possibly be a least-squares solution of \( Ax = b \) and compare its suitability relative to \( v \).
(Answer this without computing a least-squares solution.)
#### Step-by-Step Calculations:
1. **Compute \( Au \):**
\[
Au = \begin{bmatrix} 7 & 6 & 1 \\ -5 & 1 & -1 \\ 7 & 6 & 0 \end{bmatrix} \begin{bmatrix} 6 \\ -2 \\ 2 \end{bmatrix}
\]
\[
= \begin{bmatrix} 7(6) + 6(-2) + 1(2) \\ -5(6) + 1(-2) + (-1)(2) \\ 7(6) + 6(-2) + 0(2) \end{bmatrix}
\]
\[
= \begin{bmatrix} 42 - 12 + 2 \\ -30 - 2 - 2 \\ 42 - 12 \end{bmatrix}
\]
\[
= \begin{bmatrix} 32 \\ -34 \\ 30 \end{bmatrix}.
\]
2. **Compute \( Av \):**
\[
Av = \begin{bmatrix} 7 & 6 & 1 \\ -5 & 1 & -1 \\ 7 & 6 & 0 \end{bmatrix} \begin{bmatrix} 6 \\ -7 \\ -7 \end{bmatrix}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F05b5783f-5507-4009-950e-cf9b7485abe1%2F53171e2f-3745-473a-b748-838171654ecf%2Fc50myec_processed.jpeg&w=3840&q=75)
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