Question 1

Find a QR factorization of the matrix A.
-$$A = \left[ \begin{array} { r r } 
- 4 & 6 \
2 & - 18 \
0 & 6
\end{array} \right]$$

Accepted Answer

A) 
$$Q = \left[ \begin{array} { r r } 
- 4 & - 6 \
2 & - 12 \
0 & 6
\end{array} \right] , R = \left[ \begin{array} { c c } 
\frac { - 10 } { \sqrt { 5 } } & \frac { 30 } { \sqrt { 5 } } \
0 & \frac { 36 } { \sqrt { 6 } }
\end{array} \right]$$ 
B) 
$$Q = \left[ \begin{array} { c c } 
\frac { 2 } { \sqrt { 5 } } & - \frac { 1 } { \sqrt { 6 } } \
- \frac { 1 } { \sqrt { 5 } } & - \frac { 2 } { \sqrt { 6 } } \
0 & \frac { 1 } { \sqrt { 6 } }
\end{array} \right] , R = \left[ \begin{array} { c c } 
\frac { - 10 } { \sqrt { 5 } } & \frac { 30 } { \sqrt { 5 } } \
0 & \frac { 36 } { \sqrt { 6 } }
\end{array} \right]$$ 
C) 
$Q=\left[\begin{array}{rr}
-4 & 6 \
2 & -18 \
0 & 6
\end{array}\right], R=\left[\begin{array}{ll}
\frac{30}{\sqrt{5}} & 0 \
\frac{36}{\sqrt{6}} & \frac{-10}{\sqrt{5}}
\end{array}\right]$ 
D) 
   
A) 
$$Q = \left[ \begin{array} { r r } 
- 4 & - 6 \
2 & - 12 \
0 & 6
\end{array} \right] , R = \left[ \begin{array} { c c } 
\frac { - 10 } { \sqrt { 5 } } & \frac { 30 } { \sqrt { 5 } } \
0 & \frac { 36 } { \sqrt { 6 } }
\end{array} \right]$$ 
B) 
$$Q = \left[ \begin{array} { c c } 
\frac { 2 } { \sqrt { 5 } } & - \frac { 1 } { \sqrt { 6 } } \
- \frac { 1 } { \sqrt { 5 } } & - \frac { 2 } { \sqrt { 6 } } \
0 & \frac { 1 } { \sqrt { 6 } }
\end{array} \right] , R = \left[ \begin{array} { c c } 
\frac { - 10 } { \sqrt { 5 } } & \frac { 30 } { \sqrt { 5 } } \
0 & \frac { 36 } { \sqrt { 6 } }
\end{array} \right]$$ 
C) 
$Q=\left[\begin{array}{rr}
-4 & 6 \
2 & -18 \
0 & 6
\end{array}\right], R=\left[\begin{array}{ll}
\frac{30}{\sqrt{5}} & 0 \
\frac{36}{\sqrt{6}} & \frac{-10}{\sqrt{5}}
\end{array}\right]$ 
D)

Question 2

Express the vector x as a linear combination of the uʹs.
-$\mathbf { u } _ { 1 } = \left[ \begin{array} { r } - 2 \ 0 \ 1 \end{array} \right] , \mathbf { u } _ { 2 } = \left[ \begin{array} { l } 3 \ 5 \ 6 \end{array} \right] , \mathbf { u } _ { 3 } = \left[ \begin{array} { r } - 2 \ 6 \ - 4 \end{array} \right] , \mathbf { x } = \left[ \begin{array} { r } 9 \ 39 \ - 2 \end{array} \right]$

Accepted Answer

A)  $\mathbf { x } = 4 \mathbf { u } _ { 1 } + 6 \mathbf { u } _ { 2 } - 4 \mathbf { u } _ { 3 }$ 
B)  $\mathbf { x } = 4 \mathbf { u } _ { 1 } - 3 \mathbf { u } _ { 2 } - 4 \mathbf { u } _ { 3 }$ 
C)  $\mathbf { x } = - 8 \mathbf { u } _ { 1 } + 6 \mathbf { u } _ { 2 } + 8 \mathbf { u } _ { 3 }$ 
D)  $\mathbf { x } = - 4 \mathbf { u } _ { 1 } + 3 \mathbf { u } _ { 2 } + 4 \mathbf { u } _ { 3 }$ 
A)  $\mathbf { x } = 4 \mathbf { u } _ { 1 } + 6 \mathbf { u } _ { 2 } - 4 \mathbf { u } _ { 3 }$ 
B)  $\mathbf { x } = 4 \mathbf { u } _ { 1 } - 3 \mathbf { u } _ { 2 } - 4 \mathbf { u } _ { 3 }$ 
C)  $\mathbf { x } = - 8 \mathbf { u } _ { 1 } + 6 \mathbf { u } _ { 2 } + 8 \mathbf { u } _ { 3 }$ 
D)  $\mathbf { x } = - 4 \mathbf { u } _ { 1 } + 3 \mathbf { u } _ { 2 } + 4 \mathbf { u } _ { 3 }$

Question 3

$$\text { Find the equation } y = \beta _ { 0 } + \beta _ { 1 } x \text { of the least-squares line that best fits the given data points. }$$
-Data points: (2, 1), (3, 2), (7, 3), (8, 1)33) $$\mathrm { X } = \left[ \begin{array} { l l } 
1 & 2 \
1 & 3 \
1 & 7 \
1 & 8
\end{array} \right] , \mathrm { y } = \left[ \begin{array} { l } 
1 \
2 \
3 \
1
\end{array} \right]$$

Accepted Answer

A)  $y = - \frac { 9 } { 52 } + \frac { 1 } { 13 } x$ 
B)  $y = \frac { 71 } { 52 } - \frac { 1 } { 26 } x$ 
C)  $y = \frac { 71 } { 52 } + \frac { 1 } { 13 } x$ 
D)  $y = - \frac { 105 } { 52 } + \frac { 1 } { 13 } x$ 
A)  $y = - \frac { 9 } { 52 } + \frac { 1 } { 13 } x$ 
B)  $y = \frac { 71 } { 52 } - \frac { 1 } { 26 } x$ 
C)  $y = \frac { 71 } { 52 } + \frac { 1 } { 13 } x$ 
D)  $y = - \frac { 105 } { 52 } + \frac { 1 } { 13 } x$

Question 4

Find the distance between the two vectors.
-$$\mathbf { u } = ( 0,0,0 ) , \mathbf { v } = ( 9,9,6 )$$

Accepted Answer

A)  198 
B)  $2 \sqrt { 6 }$ 
C)  $3 \sqrt { 22 }$ 
D)  24 
A)  198 
B)  $2 \sqrt { 6 }$ 
C)  $3 \sqrt { 22 }$ 
D)  24

Find a QR factorization of the matrix A. - $A = \left[ \begin{array} { r r } - 4 & 6 \\2 & - 18 \\0 & 6\end{array} \right]$

Find the distance between the two vectors. - $\mathbf { u } = ( 0,0,0 ) , \mathbf { v } = ( 9,9,6 )$

Linear Equations in Linear Algebra

Matrix Algebra

Determinants

Vector Spaces

Eigenvalues and Eigenvectors

Symmetric Matrices and Quadratic Forms

The Geometry of Vector Spaces

Optimization Online Only

Filters

Exam 6: Orthogonality and Least Squares

Find a QR factorization of the matrix A. - A=[−462−1806]A = \left[ \begin{array} { r r } - 4 & 6 \\2 & - 18 \\0 & 6\end{array} \right]A=​−420​6−186​​

Find the distance between the two vectors. - u=(0,0,0),v=(9,9,6)\mathbf { u } = ( 0,0,0 ) , \mathbf { v } = ( 9,9,6 )u=(0,0,0),v=(9,9,6)