Numerical Methods & Linear Algebra

Math 2890-003

Write out the augmented matrix corresponding to the linear system. \[ \begin{array}{rrrrrrrrrrrrr} 4x_1 &+& 5x_2 &-& 3x_3 &-& 3x_4 &+& x_5 &+& 7x_6 &=& -2 \\ -7x_1 &+& 2x_2 &+& 9x_3 &+& 8x_4 & & &+& 3x_6 &=& 8 \\ &-& 8x_2 &-& 2x_3 &+& 6x_4 &-& 2x_5 &-& 3x_6 &=& 9 \\ x_1 &-& 3x_2 & & &-& 5x_4 &+& 8x_5 &+& 2x_6 &=& 0 \\ 3x_1 &+& x_2 &-& 3x_3 &+& 5x_4 &+& 2x_5 &+& x_6 &=& 1 \end{array} \]

Write out the linear system corresponding to the augmented matrix. \[ \left[ \begin{array}{rrrrrr|r} 1 & 8 & -2 & 7 & 9 & 0 & 2 \\ 3 & -7 & 8 & 2 & 0 & 2 & 6 \\ 0 & 0 & 0 & 1 &-2 & 2 & 3 \\ -4 & 2 & -1 & 3 & 8 & 1 & 5 \\ 5 & 9 & 5 & -4 & 1 &-9 &-4 \end{array} \right] \]

Let $u =\left(\begin{array}{r} 1 \\ -5 \\ 5 \end{array}\right)$, $v =\left(\begin{array}{r} 0 \\ 4 \\ 2 \end{array}\right)$, $w=\left(\begin{array}{r} 5 \\ -17 \\ 29 \end{array}\right)$ and $x = \left(\begin{array}{r} -1 \\ -7 \\ -11 \end{array}\right).$ Do the given vectors span $\mathbb{R}^3$? Show your work. Explain your answer. answer: The vectors don't span $\mathbb{R}^3$ since (after constructing a matrix using the vectors as the columns) there's no pivot in row 3.
Let $u=\left(\begin{array}{r} -2 \\ 0 \\ 3 \end{array}\right)$ and $v=\left(\begin{array}{r} -6 \\ -2 \\ 18 \end{array}\right)$ and $w = \left(\begin{array}{r} 6 \\ -8 \\ 27 \end{array}\right)$. Are the given vectors linearly independent? Show your work. Explain your answer. answer: The vectors are linearly dependent since (after constructing a matrix using the vectors as the columns) there is no pivot in column 3.

Let $A=\left(\begin{array}{rrrr} 3 & 12 & 3 & 11 \\ -6 & -24 & -11 & 1 \\ 9 & 36 & 4 & 64 \end{array}\right).$
Use Gaussian elimination to reduce the matrix $A$ to row echelon form. Show your work.

Let $A=\left(\begin{array}{rrrr} 6 & -1 & -14 & -40 \\ -12 & 10 & 27 & 27 \\ 18 & -18 & -12 & -30 \end{array}\right).$
Use Gaussian elimination with partial pivoting to reduce the matrix $A$ to row echelon form. Show your work. \answer{Gaussian elimination with partial pivoting reduces $A$ to \[\left(\begin{array}{rrrr} 18 & -18 & -12 & -30 \\ 0 & 5 & -10 & -30 \\ 0 & 0 & 15 & -5 \end{array}\right).\]Here $A=LU$ with $U$ as above and $L=\left(\begin{array}{rrr} 0.3333 & 1 & 0 \\ -0.6667 & -0.4 & 1 \\ 1 & 0 & 0 \end{array}\right)$}

Let $A=\left(\begin{array}{rrrrr} 4 & 3 & -14 & -7 & -2 \\ 3 & -2 & -2 & -1 & 4 \\ -3 & 1 & 4 & 2 & -2 \\ 5 & 2 & -14 & -7 & -3 \end{array}\right).$
Find the reduced row echelon form of $A.$ Show your work. \answer{The reduced row echelon form is $\left(\begin{array}{rrrrr} 1 & 0 & -2 & -1 & 0 \\ 0 & 1 & -2 & -1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right).$}

Let $A=\left(\begin{array}{rrrr} 9 & -7 & -7 & -4 \\ -1 & 4 & -3 & 6 \\ -4 & 3 & 3 & 3 \\ 7 & -6 & -7 & -6 \end{array}\right)$ and $b=\left(\begin{array}{r} -5 \\ 33 \\ 3 \\ -7 \end{array}\right).$
Solve the equation $Ax=b.$ \answer{The solution is $x=\left(\begin{array}{r} 3 \\ 6 \\ -2 \\ 1 \end{array}\right).$}

Let $A=\left(\begin{array}{rrr} -2 & 0 & 1 \\ -2 & -2 & -3 \\ 1 & 3 & -1 \\ -1 & -3 & 2 \end{array}\right)$ and $b=\left(\begin{array}{r} -3 \\ 1 \\ -1 \\ -2 \end{array}\right).$
Solve the equation $Ax=b$ (showing your work) or explain why it doesn't have a solution. \answer{The system is inconsistent because the augmented matrix $(A|b)$ has a pivot in the last column.}

Let $A=\left(\begin{array}{rrr} -1 & 1 & 3 \\ -1 & -1 & -1 \\ 1 & -4 & 1 \\ 3 & -4 & -1 \\ 1 & -3 & 2 \end{array}\right)$ and $b=\left(\begin{array}{r} 12 \\ -8 \\ 8 \\ 4 \\ 13 \end{array}\right).$
Solve the equation $Ax=b$ (showing your work) or explain why it doesn't have a solution. \answer{The solution is $x=\left(\begin{array}{r} 3 \\ 0 \\ 5 \end{array}\right).$}

Let $A=\left(\begin{array}{rrrrrr} 5 & 5 & 20 & 30 & 0 & 1 \\ 2 & -2 & 12 & 8 & 4 & -2 \\ -5 & 4 & -29 & -21 & 4 & 0 \\ -2 & -4 & -6 & -14 & -3 & -2 \end{array}\right)$ and $b=\left(\begin{array}{r} -75 \\ -30 \\ -29 \\ 70 \end{array}\right).$
Find the general solution of the equation $Ax=b.$ Show your work. \textsc{hint:} The augmented matrix $(A|b)$ has reduced row echelon form \[\left(\begin{array}{rrrrrr|r} 1 & 0 & 5 & 5 & 0 & 0 & -7 \\ 0 & 1 & -1 & 1 & 0 & 0 & -8 \\ 0 & 0 & 0 & 0 & 1 & 0 & -8 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{array}\right)\] \answer{The general solution is $x=\left(\begin{array}{r} -7 \\ -8 \\ 0 \\ 0 \\ -8 \\ 0 \end{array}\right)+\left(\begin{array}{rr} -5 & -5 \\ 1 & -1 \\ 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ 0 & 0 \end{array}\right)y$, where $y \in R^{2}.$}

Let $v=\left(\begin{array}{r} 5 \\ -4 \\ 8 \\ -1 \\ 0 \end{array}\right)$ and $w = \left(\begin{array}{r} -7 \\ -7 \\ -3 \\ -2 \\ 2 \end{array}\right)$.
Compute the sum $v+w$ if it is defined; otherwise, explain why it is not defined. \answer{The sum $v+w=\left(\begin{array}{r} -2 \\ -11 \\ 5 \\ -3 \\ 2 \end{array}\right).$}

Let $v=\left(\begin{array}{r} -6 \\ 9 \\ -1 \\ 2 \\ -4 \\ -7 \end{array}\right)$ and $w = \left(\begin{array}{r} 2 \\ 2 \\ -8 \\ 0 \\ 7 \end{array}\right)$.
Compute the sum $v+w$ if it is defined; otherwise, explain why it is not defined. \answer{The sum $v+w$ is not defined because $v$ and $w$ have different dimensions.}

Let $\alpha =-7$, $\beta =5$, $v=\left(\begin{array}{r} 2 \\ 9 \\ -1 \\ 7 \end{array}\right)$ and $w = \left(\begin{array}{r} -1 \\ 7 \\ 6 \\ -1 \end{array}\right).$
Compute the linear combination $v \alpha + w \beta$, or explain why it is impossible. Show your work. \answer{The linear combination $v \alpha + w \beta =\left(\begin{array}{r} -19 \\ -28 \\ 37 \\ -54 \end{array}\right).$}