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Wize University Linear Algebra Textbook > Systems of Linear Equations (SLEs) (Linear Systems)

Geometric Interpretation of Linear Systems

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Geometric Interpretation of Solutions to SLEs

Linear systems are made up of equations of hyperplanes (e.g. lines in R2\reals^2; planes in R3\reals^3).
A solution is the intersection of all the hyperplanes.

R2\colorOne{\reals^2}: Intersection of Lines

  • No intersection
  • Unique point of intersection (2 lines cross)
  • Infinitely many points of intersection
  • Line →\rightarrow 1-parameter family of solutions (same line)
Example
{−x+y=2y=1  ⟹  [−112011]  ⟹  x⃗=[−11]POI\begin{cases} -x&+&y&=&2\\ &&y&=&1 \end{cases} \quad \implies \quad \left[ \begin{array}{rr|r} -1 & 1 & 2\\ 0 & 1 & 1\\ \end{array} \right] \quad \implies \quad \overset{\normalsize POI}{ \boxed{ \vec x = \left[ \begin{array}{r} -1\\ 1 \end{array} \right] }}



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R3\colorOne{\reals^3}: Intersection of Planes

  • No intersection
  • Unique point of intersection (requires 3 planes)
  • Infinitely many points of intersection
  • Line →\rightarrow 1-parameter family of solutions (2 planes intersect)
  • Plane →\rightarrow 2-parameter family of solutions (2 of the same plane)
Example
{−x+y=2y=1  ⟹  [−11020101]  ⟹  x⃗=[−110]+t[001]line of intersection\begin{cases} -x&+&y&=&2\\ &&y&=&1 \end{cases} \quad \implies \quad \left[ \begin{array}{rrr|r} -1 & 1 & 0 & 2\\ 0 & 1 & 0 & 1\\ \end{array} \right] \quad \implies \quad \overset{\text{\normalsize line of intersection}}{ \boxed{ \vec x = \left[ \begin{array}{r} -1\\ 1\\ 0 \end{array} \right] +t \left[ \begin{array}{r} 0\\ 0\\ 1 \end{array} \right] }}


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Example: Geometric Interpretation of a Linear System

Solve the system of linear equations and interpret the solution geometrically:
2x+2y=6−x−y=−3\begin{array}{rl} 2x+2y&=6\\ -x-y&=-3\\ \end{array}
Steps
  1. Let's first write this in augmented matrix form:
2x+2y=6−x−y=−3→[226−1−1−3]\begin{array}{rcr} 2x+2y&=&6\\[0.2em] -x-y&=&-3\\[0.2em] \end{array} \quad \rightarrow \quad \left[\begin{array}{rr|r} 2&2&6\\ -1&-1&-3\\ \end{array}\right]
  1. Turn the matrix into RREF using EROs:
[226−1−1−3]12R1⟶[113−1−1−3]R2+R1⟶[113000]\begin{aligned} &\left[\begin{array}{rr|r} 2&2&6\\ -1&-1&-3\\ \end{array}\right] \begin{array}{l} \dfrac{1}{2}R_1\\ \\ \\ \end{array}\\[2.5em] \longrightarrow &\left[\begin{array}{rr|r} 1&1&3\\ -1&-1&-3\\ \end{array}\right] \begin{array}{l} \\ R_2 + R_1\\ \end{array}\\[2.5em] \longrightarrow &\left[\begin{array}{rr|r} 1&1&3\\ 0&0&0\\ \end{array}\right] \end{aligned}
  1. Find all solutions to the SLE:
The coefficient matrix has just one leading 1, but 2 columns/variables: rank(A)=1  <   n=2{\rm rank}(A)=1 \ \ \bm< \ \ \ n=2
Thus, there is 2−1=12-1 = 1 free variable (yy). Let y=ty=t.
Rewrite the augmented matrix back into a system of linear equations:
{x+t=3y=t  ⟹  {x=3−ty=t\begin{cases} x+t=3\\ y=t \end{cases} \quad\implies\quad \begin{cases} x=3-t\\ y=t \end{cases}
Therefore, the solution is x⃗=[30]+t[−11]\vec x = \begin{bmatrix} 3\\ 0\\ \end{bmatrix} +t \begin{bmatrix} -1\\ 1\\ \end{bmatrix}.
  1. Interpret the solution geometrically:
We were given a system of linear equations in 2 variables   ⟹  R2\implies \reals^2.
Each equation is a hyperplane: in R2\reals^2, hyperplanes are simply lines (in general, a hyperplane in Rn\reals^n is n−1n-1 dimensional).
So we have 2 lines, but both lines are identical (hence reducing to a row of 0s).
This means the lines intersect everywhere on the line defined by x⃗=[30]+t[−11]\vec x = \begin{bmatrix} 3\\ 0\\ \end{bmatrix} +t \begin{bmatrix} -1\\ 1\\ \end{bmatrix}.

Consider the following augmented matrix in RREF:
[1−100−540010−2200014−3]\left[\begin{array}{rrrrr|r} 1&-1&0&0&-5&4\\[0.5em] 0&0&1&0&-2&2\\[0.5em] 0&0&0&1&4&-3 \end{array}\right]
If the unknowns in this system are x1, x2, x3, x4, x5x_1,\ x_2,\ x_3,\ x_4,\ x_5, find the solution(s) to the linear system.
Then, determine which of the following is the correct geometrical interpretation of the solution.
Extra Practice
What is the reduced row echelon form of an augmented matrix with three equations and three unknowns,
a) If all solutions lie on the line[x y z]=(3,1,0)+s(1,2,3)[x\ y\ z] =(3,1,0)+s(1,2,3)
b) If all solutions lie on the plane x+y+2z=1x+y+2z=1
For what value of p does the following system of equations have no solution? What is the geometric interpretation of these equations?
x+3y=1x+3y=1
px−6y=−4px-6y=-4
Practice Question 6: Solving Systems of Linear Equations
Practice Question: Solving Systems of Linear Equations
The following equations represent three planes in ℝ3. Describe the regions of intersection.
a) x1−x2+2x3=0x_1-x_2+2x_3=0
Solving Linear Systems
Practice: Geometric Interpretation of a Linear System
Consider the following augmented matrix in RREF:
[1−100−540010−2200014−3]\left[\begin{array}{rrrrr|r} 1&-1&0&0&-5&4\\[0.5em] 0&0&1&0&-2&2\\[0.5em] 0&0&0&1&4&-3 \end{array}\right]
Geometric Interpretation of Linear Systems
Example: Geometric Interpretation of a Linear System
Solve the system of linear equations and interpret the solution geometrically:
2x+2y=6−x−y=−3\begin{array}{rl} 2x+2y&=6\\ -x-y&=-3\\ \end{array}
Homogeneous equations
Find the solution(s) of Ax=0, if the solution of Ax=b is given as follows:
xp=[1030]+s[−1100]+t[00−23]x_p=\begin{bmatrix} 1\\ 0\\ 3\\ 0 \end{bmatrix}+s\begin{bmatrix} -1\\ 1\\ 0\\ 0 \end{bmatrix}+t\begin{bmatrix} 0\\ 0\\ -2\\ 3 \end{bmatrix}
Interpretation of Linear Systems
Which of the following describes the intersection between the following 3 hyperplanes in R4\reals^4?
2x1+3x3+x4=1x2−x4=0x2=−3\begin{array}{rcl} 2x_1+3x_3+x_4&=&1\\ x_2-x_4&=&0\\ x_2&=&-3 \end{array}
The following linear equations represent planes in R3R^3. Describe the intersection of these planes (if any).
2x+y−z=3x−y−z=33x+z=6\begin{array}{c} 2x+y-z=3\\ x-y-z=3\\ 3x+z=6 \end{array}
Homogeneous equations
Find the solution(s) of Ax=0, if the solution of Ax=b is given as follows:
xp=[1030]+s[−1100]+t[00−23]x_p=\begin{bmatrix} 1\\ 0\\ 3\\ 0 \end{bmatrix}+s\begin{bmatrix} -1\\ 1\\ 0\\ 0 \end{bmatrix}+t\begin{bmatrix} 0\\ 0\\ -2\\ 3 \end{bmatrix}
Practice: Solutions to a System of Equations (Step 3)
Practice: Solutions to a System of Equations (Step 3)
Find the solution(s) to the systems of linear equations whose augmented matrices are given:
a.) [1−1030012]\left[ \begin{array}{ccc|c} 1&-1&0&3\\ 0&0&1&2\\ \end{array}\right]
System of Equations
e.g. True or False: if a linear system has more unknowns than equations, then it must have infinitely many solutions. Justify your answer with a proof or counter-example.
System of Equations
True or False: if a linear system has more unknowns than equations, then it must have infinitely many solutions. Justify your answer with a proof or counter-example.
Linear combinations
If u⃗=[5−23]\vec{u}=\begin{bmatrix}5\\-2\\3\end{bmatrix}is a linear combination of v⃗=[−2k3]\vec{v}=\begin{bmatrix}-2\\k\\3\end{bmatrix}and w⃗=[32−1]\vec{w}=\begin{bmatrix}3\\2\\-1\end{bmatrix}, what is the value of kk?
Solving Linear Systems
Consider the following augmented matrix in RREF:
[1−100−540010−2200014−3]\left[\begin{array}{rrrrr|r} 1&-1&0&0&-5&4\\[0.5em] 0&0&1&0&-2&2\\[0.5em] 0&0&0&1&4&-3 \end{array}\right]
If the unknowns in this system are x1, x2, x3, x4, x5x_1,\ x_2,\ x_3,\ x_4,\ x_5, find the solution(s) to the linear system.
Solving Linear Systems
Consider the following augmented matrix in RREF:
[1−100−540010−2200014−3]\left[\begin{array}{rrrrr|r} 1&-1&0&0&-5&4\\[0.5em] 0&0&1&0&-2&2\\[0.5em] 0&0&0&1&4&-3 \end{array}\right]
If the unknowns in this system are x1, x2, x3, x4, x5x_1,\ x_2,\ x_3,\ x_4,\ x_5, find the solution(s) to the linear system.
Basics of Systems of Linear Equations
Determine which vectors are solutions to the following SLE [select all that apply]:
x1 − 5x2 +x3=2−x1 +5x2 −3x3= −6\begin{alignedat}{8} x_1\ &-&\ 5x_2\ &+& x_3 &=& 2\\ -x_1\ &+& 5x_2\ &-& 3x_3 &=&\ -6 \end{alignedat}
Interpretation of Linear Systems
Which of the following describes the intersection between the following 3 hyperplanes in R4\reals^4?
2x1+3x3+x4=1x2−x4=0x2=−3\begin{array}{rcl} 2x_1+3x_3+x_4&=&1\\ x_2-x_4&=&0\\ x_2&=&-3 \end{array}