Solved Problems

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Solved Problems for Chapter 2

You have a system of $k$ equations in two variables, $k\geq 2$ . Explain the geometric significance of

(a): No solution.

The $k$ lines do not have a point common to all of them.
(b): A unique solution.

All $k$ lines intersect at a single point.
(c): An infinite number of solutions.

The $k$ lines coincide.

Click the arrow to see answer.

Consider the following augmented matrix in which $\ast$ denotes an arbitrary number and $\blacksquare$ denotes a nonzero number. Determine whether the given augmented matrix corresponds to a consistent system. If consistent, is the solution unique? $\begin{equation*} \left [ \begin{array}{ccccc|c} \blacksquare & \ast & \ast & \ast & \ast & \ast \\ 0 & \blacksquare & \ast & \ast & 0 & \ast \\ 0 & 0 & 0 & 0 & \blacksquare & 0 \\ 0 & 0 & 0 & 0 & \blacksquare & \blacksquare \end{array} \right ] \end{equation*}$ Click the arrow to see answer.

The third equation implies that $x_5 = 0$ . The fourth equation implies that $x_5 \ne 0$ . We conclude that the system is inconsistent.

Suppose a system of equations has fewer equations than variables. Will such a system necessarily be consistent? If so, explain why and if not, give an example which is not consistent.

Click the arrow to see answer.

No. Consider $x+y+z=2$ and $x+y+z=1.$

If a system of equations has more equations than variables, can it have a solution? If so, give an example and if not, explain why not.

Click the arrow to see answer.

These can have a solution. For example, $x+y=1,2x+2y=2,3x+3y=3$ even has an infinite set of solutions.

Find $h$ such that $\begin{equation*} \left [ \begin{array}{rr|r} 2 & h & 4 \\ 3 & 6 & 7 \end{array} \right ] \end{equation*}$ is the augmented matrix of an inconsistent system.

$\left [ \begin {array}{cc|c} 2 & h & 4 \\ 3 & 6 & 7 \end {array} \right ]\rightsquigarrow \left [ \begin {array}{cc|c} 3 & (3/2)h & 6 \\ 3 & 6 & 7 \end {array} \right ]$ . The system is inconsistent for $h=4$ .

Find $h$ such that $\begin{equation*} \left [ \begin{array}{rr|r} 1 & h & 3 \\ 2 & 4 & 6 \end{array} \right ] \end{equation*}$ is the augmented matrix of a consistent system.

Click the arrow to see answer.

The reduced row-echelon form will never have a row of the form $[0\, 0|1]$ . The system is consistent for all $h$ .

Choose $h$ and $k$ such that the augmented matrix shown has each of the following:

(a): one solution
(b): no solution
(c): infinitely many solutions

$\begin{equation*} \left [ \begin{array}{rr|r} 1 & 2 & 2 \\ 2 & h & k \end{array} \right ] \end{equation*}$ Click the arrow to see answer.

If $h\neq 4,$ then there is exactly one solution. If $h=4$ and $k\neq 4,$ then there are no solutions. If $h=4$ and $k=4,$ then there are infinitely many solutions.

Determine if the system is consistent. If so, is the solution unique? $\begin {array}{ccccccccc} x & +&2y&+&z&-&w&= &2 \\ x& -&y&+&z&+&w&=&1\\ 2x& +&y&-&z&&&=&1\\ 4x&+&2y&+&z&&&=&5 \end {array}$

Click the arrow to see answer.

There is no solution because the rref of the augmented matrix contains a row of the form $\begin {bmatrix}0 & 0 & 0 & 0 & | & 1\end {bmatrix}$ $\mbox {rref}\left (\left [ \begin {array}{rrrr|r} 1 & 2 & 1 & -1 & 2 \\ 1 & -1 & 1 & 1 & 1 \\ 2 & 1 & -1 & 0 & 1 \\ 4 & 2 & 1 & 0 & 5 \end {array} \right ]\right ) = \left [ \begin {array}{rrrr|r} 1 & 0 & 0 & \frac {1}{3} & 0 \\ 0 & 1 & 0 & - \frac {2}{3} & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end {array} \right ] .$

Determine if the system is consistent. If so, is the solution unique?

$\begin {array}{ccccccccc} x & +&2y&+&z&-&w&= &2 \\ x& -&y&+&z&+&w&=&0\\ 2x& +&y&-&z&&&=&1\\ 4x&+&2y&+&z&&&=&3 \end {array}$

Click the arrow to see answer.

$\text {rref}\left (\left [ \begin {array}{rrrr|r} 1 & 2 & 1 & -1 & 2 \\ 1 & -1 & 1 & 1 & 0 \\ 2 & 1 & -1 & 0 & 1 \\ 4 & 2 & 1 & 0 & 3 \end {array} \right ]\right )=\left [ \begin {array}{rrrr|r} 1 & 0 & 0 & 1/3 & 1/3 \\ 0 & 1 & 0 & -2/3 & 2/3 \\ 0 & 0 & 1 & 0 & 1/3 \\ 0 & 0 & 0 & 0 & 0 \end {array} \right ]$ There are infinitely many solutions: $w=t$ , $z=\frac {1}{3}$ , $y=\frac {2}{3}+\frac {2}{3}t$ , $x=\frac {1}{3}-\frac {1}{3}t$ .

Find the solution of the system whose augmented matrix is $\begin{equation*} \left [ \begin{array}{rrr|r} 1 & 2 & 0 & 2 \\ 1 & 3 & 4 & 2 \\ 1 & 0 & 2 & 1 \end{array} \right ] \end{equation*}$

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$\text {rref}\left (\left [ \begin {array}{rrr|r} 1 & 2 & 0 & 2 \\ 1 & 3 & 4 & 2 \\ 1 & 0 & 2 & 1 \end {array} \right ]\right )=\left [ \begin {array}{rrr|r} 1 & 0 & 0 & 6/5 \\ 0 & 1 & 0 & 2/5 \\ 0 & 0 & 1 & -1/10 \end {array} \right ]$ $x=\frac {6}{5}, y=\frac {2}{5}, z=-\frac {1}{10}$

Solve the system whose augmented matrix is $\begin{equation*} \left [ \begin{array}{rrr|r} 1 & 1 & 0 & 1 \\ 1 & 0 & 4 & 2 \end{array} \right ] \end{equation*}$ Click the arrow to see answer.

The reduced row echelon form is $\left [ \begin {array}{rrr|r} 1 & 0 & 4 & 2 \\ 0 & 1 & -4 & -1 \end {array} \right ]$ and so the solution is $z=t,y=4t,x=2-4t.$

Solve the system if the rref of its augmented matrix is $\left [ \begin {array}{rrrrr|r} 1 & 0 & 0 & 0 & 9 & 3 \\ 0 & 1 & 0 & 0 & -4 & 0 \\ 0 & 0 & 1 & 0 & -7 & -1 \\ 0 & 0 & 0 & 1 & 6 & 1 \end {array} \right ]$ Click the arrow to see answer.

$x_{5}=t,x_{4}=1-6t,x_{3}=-1+7t,x_{2}=4t,x_{1}=3-9t$ .

Solve the system if the rref of its augmented matrix is $\left [ \begin {array}{rrrrr|r} 1 & 0 & 2 & 0 & -1/2 & 5/2 \\ 0 & 1 & 0 & 0 & 1/2 & 3/2 \\ 0 & 0 & 0 & 1 & 3/2 & -1/2 \\ 0 & 0 & 0 & 0 & 0 & 0 \end {array} \right ]$ Click the arrow to see answer.

The free variables are $x_{5}=t,x_{3}=s$ . The other variables are given by $x_{4}=-\frac {1}{2}-\frac {3}{2}t$ , $x_{2}=\frac {3}{2}-\frac {1}{2}t$ , $x_{1}=\frac {5}{2}+\frac {1}{2}t-2s$ .

Suppose a system of equations has fewer equations than variables and you have found a solution to this system of equations. Is it possible that your solution is the only one? Explain.

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No. The rank of the coefficient matrix in this case is smaller than the number of columns (variables). So, there has to be a free variable. The parameter ( $t$ is a typical choice) assigned to the free variable will guarantee infinitely many solutions.

Suppose $A$ is an $m\times n$ matrix. Explain why the rank of $A$ is always no larger than $\min \left ( m,n\right ).$

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Find the rank of the following matrix. $\begin {bmatrix} 4 & 4 & 20 & -1 & 17 \\ 1 & 1 & 5 & 0 & 5 \\ 1 & 1 & 5 & -1 & 2 \\ 3 & 3 & 15 & -3 & 6 \end {bmatrix}$

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$\text {rref}\left (\begin {bmatrix}4 & 4 & 20 & -1 & 17 \\ 1 & 1 & 5 & 0 & 5 \\ 1 & 1 & 5 & -1 & 2 \\ 3 & 3 & 15 & -3 & 6\end {bmatrix}\right )=\begin {bmatrix}1 &1 &5 &0 &5\\ 0 & 0 &0 &1 &3\\ 0 &0 &0 &0 &0\\ 0& 0& 0& 0& 0\end {bmatrix}$

The rank is the number of leading $1$ ’s. The rank of the given matrix is $2$ .

Bibliography

These problems come from the end of Chapter 1 of Ken Kuttler’s A First Course in Linear Algebra. (CC-BY)

Ken Kuttler, A First Course in Linear Algebra, Lyryx 2017, Open Edition, pp. 42–49.

2024-09-06 23:26:39

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Solved Problems for Chapter 2

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