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Chapter 1: Q3Q (page 1)

The solutions \(\left( {x,y,z} \right)\) of a single linear equation \(ax + by + cz = d\)

form a plane in \({\mathbb{R}^3}\) when a, b, and c are not all zero. Construct sets of three linear equations whose graphs (a) intersect in a single line, (b) intersect in a single point, and (c) have no points in common. Typical graphs are illustrated in the figure.

Three planes intersecting in a line.

(a)

Three planes intersecting in a point.

(b)

Three planes with no intersection.

(c)

Three planes with no intersection.

(肠鈥�)

Short Answer

Expert verified

(a) The echelon form of the consistent linear system is,,or.

(b) The echelon form of the consistent linear system isthe identity matrix of the order\(3 \times 3\).

(c) The inconsistent linear system of three variables and equations has no common point.

Step by step solution

01

(a) Step 1: Write the condition when the graphs intersect on a single line

Each point on the line is a solution to the given system of equations. And the solution set is infinite if the three planes cross at a single point. As a result, there must be two pivot components in the possible forms.

The echelon form of the consistent linear system is shown below:


Or,

Or,

Here, is the leading entry, and \(\left( * \right)\) can have any value, including 0.

02

(b) Step 2: Write the condition w\(3 \times 3\)hen the graphs intersect at a single point

The system ofthree equations is fulfilled if the three planes cross at a single location. As a consequence, the system is consistent, and it offers a unique solution.

The echelon form that may be produced by solving this system of equations is an identity matrix of the order .

Thus, the echelon form of the consistent linear system is an identity matrix of the order \(3 \times 3\).

03

(c) Step 3: Write the condition when the graphs have no point in common

If there is no common point between the planes, their intersection is not a unique line or point. As a result, there is no way to solve it.

Thus, the inconsistent linear system of three variables and equations has no common point.

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Most popular questions from this chapter

In Exercises 3 and 4, display the following vectors using arrows

on an \(xy\)-graph: u, v, \( - {\bf{v}}\), \( - 2{\bf{v}}\), u + v , u - v, and u - 2v. Notice that u - vis the vertex of a parallelogram whose other vertices are u, 0, and \( - {\bf{v}}\).

4. u and v as in Exercise 2

Let \({{\mathop{\rm v}\nolimits} _1} = \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right],{v_2} = \left[ {\begin{array}{*{20}{c}}{ - 3}\\1\\8\end{array}} \right],\) and \({\rm{y = }}\left[ {\begin{array}{*{20}{c}}h\\{ - 5}\\{ - 3}\end{array}} \right]\). For what values(s) of \(h\) is \(y\) in the plane generated by \({{\mathop{\rm v}\nolimits} _1}\) and \({{\mathop{\rm v}\nolimits} _2}\)

Suppose the system below is consistent for all possible values of \(f\) and \(g\). What can you say about the coefficients \(c\) and \(d\)? Justify your answer.

27. \(\begin{array}{l}{x_1} + 3{x_2} = f\\c{x_1} + d{x_2} = g\end{array}\)

In Exercises 15 and 16, list five vectors in Span \(\left\{ {{v_1},{v_2}} \right\}\). For each vector, show the weights on \({{\mathop{\rm v}\nolimits} _1}\) and \({{\mathop{\rm v}\nolimits} _2}\) used to generate the vector and list the three entries of the vector. Do not make a sketch.

15. \({{\mathop{\rm v}\nolimits} _1} = \left[ {\begin{array}{*{20}{c}}7\\1\\{ - 6}\end{array}} \right],{v_2} = \left[ {\begin{array}{*{20}{c}}{ - 5}\\3\\0\end{array}} \right]\)

In Exercise 23 and 24, make each statement True or False. Justify each answer.

24.

a. Any list of five real numbers is a vector in \({\mathbb{R}^5}\).

b. The vector \({\mathop{\rm u}\nolimits} \) results when a vector \({\mathop{\rm u}\nolimits} - v\) is added to the vector \({\mathop{\rm v}\nolimits} \).

c. The weights \({{\mathop{\rm c}\nolimits} _1},...,{c_p}\) in a linear combination \({c_1}{v_1} + \cdot \cdot \cdot + {c_p}{v_p}\) cannot all be zero.

d. When are \({\mathop{\rm u}\nolimits} \) nonzero vectors, Span \(\left\{ {u,v} \right\}\) contains the line through \({\mathop{\rm u}\nolimits} \) and the origin.

e. Asking whether the linear system corresponding to an augmented matrix \(\left[ {\begin{array}{*{20}{c}}{{{\rm{a}}_{\rm{1}}}}&{{{\rm{a}}_{\rm{2}}}}&{{{\rm{a}}_{\rm{3}}}}&{\rm{b}}\end{array}} \right]\) has a solution amounts to asking whether \({\mathop{\rm b}\nolimits} \) is in Span\(\left\{ {{a_1},{a_2},{a_3}} \right\}\).
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