Chapter 9

Solve each linear system of equations. In addition, for each system, graph the two lines corresponding to the two equations in a single coordinate system and use your graph to explain your solution. $$ \begin{array}{l} x-y=1 \\ x-2 y=-2 \end{array} $$

Let \(\mathbf{x}=[1,4,-1]^{\prime}\) and \(\mathbf{y}=[-2,1,0]^{\prime}\). (a) Find \(\mathbf{x}+\mathbf{y}\). (b) Find \(2 \mathbf{x}\). (c) Find \(-3 \mathbf{y}\).

Let $$ A=\left[\begin{array}{ll} 2 & 1 \\ 3 & 4 \end{array}\right], \quad \mathbf{x}=\left[\begin{array}{l} x_{1} \\ x_{2} \end{array}\right], \quad \text { and } \quad \mathbf{y}=\left[\begin{array}{l} y_{1} \\ y_{2} \end{array}\right] $$ (a) Show by direct calculation that \(A(\mathbf{x}+\mathbf{y})=A \mathbf{x}+A \mathbf{y}\). (b) Show by direct calculation that \(A(\lambda \mathbf{x})=\lambda(A \mathbf{x})\).

Solve each linear system of equations. In addition, for each system, graph the two lines corresponding to the two equations in a single coordinate system and use your graph to explain your solution. $$ \begin{array}{r} 2 x+3 y=6 \\ x-4 y=-4 \end{array} $$

Let \(\mathbf{x}=[-4,3,1]^{\prime}\) and \(\mathbf{y}=[0,-2,3]^{\prime}\). (a) Find \(\mathbf{x}-\mathbf{y}\). (b) Find \(2 \mathbf{x}+3 \mathbf{y}\). (c) Find \(-\mathbf{x}-2 \mathbf{y}\).

Let $$ A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right], \quad \mathbf{x}=\left[\begin{array}{l} x_{1} \\ x_{2} \end{array}\right], \quad \text { and } \quad \mathbf{y}=\left[\begin{array}{l} y_{1} \\ y_{2} \end{array}\right] $$ (a) Show by direct calculation that \(A(\mathbf{x}+\mathbf{y})=A \mathbf{x}+A \mathbf{y}\). $$ \text { (b) Show by direct calculation that } A(\lambda \mathbf{x})=\lambda(A \mathbf{x}) \text { . } $$

Solve each linear system of equations. In addition, for each system, graph the two lines corresponding to the two equations in a single coordinate system and use your graph to explain your solution. $$ \begin{aligned} x-3 y &=6 \\ y &=3+\frac{1}{3} x \end{aligned} $$

Let \(A=(2,3)\) and \(B=(4,1)\). Find the vector representation of \(A B\)

Solve each linear system of equations. In addition, for each system, graph the two lines corresponding to the two equations in a single coordinate system and use your graph to explain your solution. $$ \begin{array}{l} 2 x+y=\frac{1}{3} \\ 6 x+3 y=1 \end{array} $$

Let \(A=(-1,0)\) and \(B=(2,-4)\). Find the vector representation of \(\overrightarrow{A B}\).

In Problems , represent each given vector \(\mathbf{x}=\left[\begin{array}{l}x_{1} \\ x_{2}\end{array}\right]\) in the \(x_{1}-x_{2}\) plane, and determine its length and the angle that it forms with the positive \(x_{1}-\) axis (measured counterclockwise). $$ \mathbf{x}=\left[\begin{array}{r} -2 \\ 0 \end{array}\right] $$

Determine \(c\) such that $$ \begin{array}{l} 2 x-3 y=5 \\ 4 x-6 y=c \end{array} $$ has (a) infinitely many solutions and (b) no solutions. (c) Is it possible to choose a number for \(c\) so that the system has exactly one solution? Explain your answer.

Let \(A=(0,1,-3)\) and \(B=(-1,-1,2) .\) Find the vector representation of \(\overrightarrow{A B}\)

$$ \text { Show that } 2(A+B)=2 A+2 B \text { . } $$

(a) Determine the solution of $$ \begin{array}{r} -2 x+3 y=5 \\ a x-y=1 \end{array} $$ in terms of \(a\). (b) For which values of \(a\) are there no solutions, exactly one solution, and infinitely many solutions?

6\. Let \(A=(1,3,-2)\) and \(B=(0,-1,0)\). Find the vector representation of \(\overrightarrow{A B}\).

In Problems , represent each given vector \(\mathbf{x}=\left[\begin{array}{l}x_{1} \\ x_{2}\end{array}\right]\) in the \(x_{1}-x_{2}\) plane, and determine its length and the angle that it forms with the positive \(x_{1}-\) axis (measured counterclockwise).$$ \mathbf{x}=\left[\begin{array}{l} -1 \\ -1 \end{array}\right] $$

$$ \begin{array}{r} A=\left[\begin{array}{rrr} 1 & 3 & -1 \\ 2 & 4 & 1 \\ 0 & -2 & 2 \end{array}\right], \quad B=\left[\begin{array}{rrr} 5 & -1 & 4 \\ 2 & 0 & 1 \\ 1 & -3 & -3 \end{array}\right], \\ \quad C=\left[\begin{array}{rrr} -2 & 0 & 4 \\ 1 & -3 & 1 \\ 0 & 0 & 2 \end{array}\right] \end{array} $$ $$ \text { Find } 2 A+3 B-C . $$

Show that the solution of $$ \begin{array}{l} a_{11} x_{1}+a_{12} x_{2}=b_{1} \\ a_{21} x_{1}+a_{22} x_{2}=b_{2} \end{array} $$ is given by $$ x_{1}=\frac{a_{22} b_{1}-a_{12} b_{2}}{a_{11} a_{22}-a_{21} a_{12}} $$ and $$ x_{2}=\frac{-a_{21} b_{1}+a_{11} b_{2}}{a_{11} a_{22}-a_{21} a_{12}} $$

Find the length of \(\mathbf{x}=[1,3]^{\prime}\).

In Problems , represent each given vector \(\mathbf{x}=\left[\begin{array}{l}x_{1} \\ x_{2}\end{array}\right]\) in the \(x_{1}-x_{2}\) plane, and determine its length and the angle that it forms with the positive \(x_{1}-\) axis (measured counterclockwise). $$ \mathbf{x}=\left[\begin{array}{r} -\sqrt{3} \\ 1 \end{array}\right] $$

$$ \begin{array}{r} A=\left[\begin{array}{rrr} 1 & 3 & -1 \\ 2 & 4 & 1 \\ 0 & -2 & 2 \end{array}\right], \quad B=\left[\begin{array}{rrr} 5 & -1 & 4 \\ 2 & 0 & 1 \\ 1 & -3 & -3 \end{array}\right], \\ \quad C=\left[\begin{array}{rrr} -2 & 0 & 4 \\ 1 & -3 & 1 \\ 0 & 0 & 2 \end{array}\right] \end{array} $$ $$ \text { Find } 3 C-B+\frac{1}{2} A $$

Find the length of \(\mathbf{x}=[-2,7]^{\prime}\).

$$ \begin{array}{r} A=\left[\begin{array}{rrr} 1 & 3 & -1 \\ 2 & 4 & 1 \\ 0 & -2 & 2 \end{array}\right], \quad B=\left[\begin{array}{rrr} 5 & -1 & 4 \\ 2 & 0 & 1 \\ 1 & -3 & -3 \end{array}\right], \\ \quad C=\left[\begin{array}{rrr} -2 & 0 & 4 \\ 1 & -3 & 1 \\ 0 & 0 & 2 \end{array}\right] \end{array} $$ $$ \text { Determine } D \text { so that } A+B+C+D=\mathbf{0} \text { . } $$

In Problems 9-16, reduce the system of linear equations to upper triangular form and solve. $$ \begin{array}{r} 2 x-y=3 \\ x-3 y=7 \end{array} $$

Find the length of \(\mathbf{x}=[0,1,5]^{\prime}\).

In Problems 9-12, vectors are given in their polar coordinate representation (length \(r\), and angle \(\alpha\) measured counterclockwise from the positive \(x_{1}-\) axis \() .\) Find the representation of the vector \(\left[\begin{array}{l}x_{1} \\ x_{2}\end{array}\right]\)\text { in Cartesian coordinates. } $$ r=2, \alpha=30^{\circ} $$

$$ \begin{array}{r} A=\left[\begin{array}{rrr} 1 & 3 & -1 \\ 2 & 4 & 1 \\ 0 & -2 & 2 \end{array}\right], \quad B=\left[\begin{array}{rrr} 5 & -1 & 4 \\ 2 & 0 & 1 \\ 1 & -3 & -3 \end{array}\right], \\ \quad C=\left[\begin{array}{rrr} -2 & 0 & 4 \\ 1 & -3 & 1 \\ 0 & 0 & 2 \end{array}\right] \end{array} $$ $$ \text { Determine } D \text { so that } A+4 B=2(A+B)+D \text { . } $$

Reduce the system of linear equations to upper triangular form and solve. $$ \begin{array}{l} 5 x-3 y=2 \\ 2 x+7 y=3 \end{array} $$

Find the length of \(\mathbf{x}=[-2,1,-3]^{\prime}\).

In Problems , vectors are given in their polar coordinate representation (length \(r\), and angle \(\alpha\) measured counterclockwise from the positive \(x_{1}-\) axis \() .\) Find the representation of the vector \(\left[\begin{array}{l}x_{1} \\ x_{2}\end{array}\right]\)\text { in Cartesian coordinates. } $$ r=3, \alpha=150^{\circ} $$

$$ \begin{array}{r} A=\left[\begin{array}{rrr} 1 & 3 & -1 \\ 2 & 4 & 1 \\ 0 & -2 & 2 \end{array}\right], \quad B=\left[\begin{array}{rrr} 5 & -1 & 4 \\ 2 & 0 & 1 \\ 1 & -3 & -3 \end{array}\right], \\ \quad C=\left[\begin{array}{rrr} -2 & 0 & 4 \\ 1 & -3 & 1 \\ 0 & 0 & 2 \end{array}\right] \end{array} $$ $$ \text { Show that } A+B=B+A \text { . } $$

Reduce the system of linear equations to upper triangular form and solve. $$ \begin{array}{l} 7 x-y=4 \\ 3 x+2 y=1 \end{array} $$

Normalize \([1,3,-1]^{\prime}\).

In Problems , vectors are given in their polar coordinate representation (length \(r\), and angle \(\alpha\) measured counterclockwise from the positive \(x_{1}-\) axis \() .\) Find the representation of the vector \(\left[\begin{array}{l}x_{1} \\ x_{2}\end{array}\right]\)\text { in Cartesian coordinates. } $$ r=1, \alpha=120^{\circ} $$

$$ \begin{array}{r} A=\left[\begin{array}{rrr} 1 & 3 & -1 \\ 2 & 4 & 1 \\ 0 & -2 & 2 \end{array}\right], \quad B=\left[\begin{array}{rrr} 5 & -1 & 4 \\ 2 & 0 & 1 \\ 1 & -3 & -3 \end{array}\right], \\ \quad C=\left[\begin{array}{rrr} -2 & 0 & 4 \\ 1 & -3 & 1 \\ 0 & 0 & 2 \end{array}\right] \end{array} $$ $$ \text { Show that }(A+B)+C=A+(B+C) \text { . } $$

Reduce the system of linear equations to upper triangular form and solve. $$ \begin{array}{l} 5 x+2 y=8 \\ -x+3 y=9 \end{array} $$

Normalize \([2,0,-4]^{\prime}\).

$$ \begin{array}{r} A=\left[\begin{array}{rrr} 1 & 3 & -1 \\ 2 & 4 & 1 \\ 0 & -2 & 2 \end{array}\right], \quad B=\left[\begin{array}{rrr} 5 & -1 & 4 \\ 2 & 0 & 1 \\ 1 & -3 & -3 \end{array}\right], \\ \quad C=\left[\begin{array}{rrr} -2 & 0 & 4 \\ 1 & -3 & 1 \\ 0 & 0 & 2 \end{array}\right] \end{array} $$ $$ \text { Show that if } A+B=C, \text { then } A=C-B $$

Reduce the system of linear equations to upper triangular form and solve. $$ \begin{array}{r} 3 x-y=1 \\ -3 x+y=4 \end{array} $$

Suppose a vector \(\left[\begin{array}{l}x_{1} \\ x_{2}\end{array}\right]\) has length 3 and is \(15^{\circ}\) clockwise from the positive \(x_{1}\) -axis. Find \(x_{1}\) and \(x_{2}\).

Find the transpose of $$ A=\left[\begin{array}{r} 2 \\ -3 \\ 5 \end{array}\right] $$

Reduce the system of linear equations to upper triangular form and solve. $$ \begin{aligned} 2 x+3 y &=5 \\ -\quad y &=-2+\frac{2}{3} x \end{aligned} $$

Normalize \([0,-3,1,3]^{\prime}\).

. Suppose a vector \(\left[\begin{array}{l}x_{1} \\ x_{2}\end{array}\right]\) has length 2 and is \(140^{\circ}\) clockwise from the positive \(x_{1}\) -axis. Find \(x_{1}\) and \(x_{2}\).

Find the transpose of $$ A=\left[\begin{array}{rrr} -1 & 0 & 3 \\ 2 & 1 & -4 \end{array}\right] $$

Reduce the system of linear equations to upper triangular form and solve. $$ \begin{array}{r} x+2 y=3 \\ 4 y+2 x=6 \end{array} $$

Find the dot product of \(\mathbf{x}=[1,2]^{\prime}\) and \(\mathbf{y}=[3,-1]^{\prime}\).

Suppose a vector \(\left[\begin{array}{l}x_{1} \\ x_{2}\end{array}\right]\) has length 5 and is \(25^{\circ}\) counterclockwise from the positive \(x_{2}\) -axis. Find \(x_{1}\) and \(x_{2}\).

Suppose \(A\) is a \(2 \times 2\) matrix. Find conditions on the entries of \(A\) such that $$ A+A^{\prime}=\mathbf{0} $$