High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize University Linear Algebra Textbook > Systems of Linear Equations (SLEs) (Linear Systems)

Solving Linear Systems

Rank of a MatrixPrevious Section
Geometric Interpretation of Linear SystemsNext Section
Popular Courses
Find My Course
0:00 / 0:00

Solutions of a Linear System

Matrix Form of a Linear System

Given a linear system of equations, we have seen how to write the augmented matrix [ A ∣ b⃗ ][\ A\ |\ \vec b\ ].
We may also write the SLE as a matrix equation, using the same coefficient matrix AA and constant vector b⃗\vec b:
Ax⃗=b⃗\boxed{\quad A\vec x = \vec b \quad}
Here, x⃗=[x1x2⋮xn]\vec x = \begin{bmatrix} x_1\\ x_2\\ \vdots\\ x_n \end{bmatrix} is the unknown solution vector.
PAGE BREAK

Procedure to Solve Ax⃗=b⃗\colorOne{A\vec x = \vec b}

  1. Write the linear system as an augmented matrix
  2. Turn the matrix into RREF using EROs
  3. Find all solutions
  4. Use rank to determine the number of solutions
  5. Turn the augmented matrix back into a system of linear equations (assign parameters if needed)
  6. Solve for any missing variables
Example 1
Find the general solution to the system represented by the augmented matrix:
[100a010b001c]\left[\begin{array}{rrr|r} 1&0&0&a\\[0.25em] 0&1&0&b\\[0.25em] 0&0&1&c \end{array}\right]
  ⟹  {x1=ax2=bx3=c  ⟹  x⃗=[abc]\begin{aligned} &\implies \begin{cases} x_1=a\\[0.25em] x_2=b\\[0.25em] x_3=c \end{cases}\\[3em] &\implies \boxed{ \vec x = \left[ \begin{array}{l} a\\[0.25em] b\\[0.25em] c \end{array} \right] } \end{aligned}
PAGE BREAK
Example 2
Find the general solution to the system represented by the augmented matrix:
[102a013b]\left[\begin{array}{rrr|r} 1&0&2&a\\[0.25em] 0&1&3&b \end{array}\right]
n−rank(A) = 3−2 = 1n - {\rm rank}(A) \ =\ 3-2 \ =\ 1 free variable.
Note thatx1x_1 and x2x_2 have leading 1s in their columns, but x3x_3 does not.
  ⟹  x3\implies x_3 is a free variable, and we assign it a parameter: let x3=t\colorOne{x_3=t}.
[102a013b]  ⟹  {x1+2x3=ax2+3x3=bx3=t  ⟹  {x1+2t=ax2+3t=bx3=t  ⟹  {x1=a−2tx2=b−3tx3=t  ⟹  x⃗=[a−2tb−3tt]  ⟹  x⃗=[ab0]+t[−2−31]   is the 1-parameter family of solutions\begin{aligned} \left[\begin{array}{rrr|r} 1&0&2&a\\[0.25em] 0&1&3&b \end{array}\right] &\implies \left\{\begin{array}{l} x_1+2 \colorOne{x_3}=a\\[0.25em] x_2+3 \colorOne{x_3}=b\\[0.25em] \colorOne{x_3 = t} \end{array}\right.\\[2em] &\implies \left\{\begin{array}{l} x_1+2 \colorOne{t}=a\\[0.25em] x_2+3 \colorOne{t}=b\\[0.25em] \colorOne{x_3 = t} \end{array}\right.\\[2em] &\implies \left\{\begin{array}{l} x_1=a-2 \colorOne{t}\\[0.25em] x_2=b-3 \colorOne{t}\\[0.25em] \colorOne{x_3 = t} \end{array}\right.\\[2em] &\implies \vec x = \left[ \begin{array}{c} a-2 \colorOne{t}\\[0.25em] b-3 \colorOne{t}\\[0.25em] \colorOne{t} \end{array} \right]\\[2em] &\implies \boxed{ \vec x = \left[ \begin{array}{c} a\\ b\\ 0 \end{array} \right] +\colorOne{t} \left[ \begin{array}{c} -2\\ -3\\ 1 \end{array} \right]} \ \ \text{ is the 1-parameter family of solutions} \end{aligned}
0:00 / 0:00

Homogeneous Linear Systems

A system of linear equations is said to be homogeneous if the augmented column is all 0s:
[a11a12…a1n0a21a22…a2n0⋮⋮⋮⋮am1am2…amn0]\left[ \begin{array}{cccc|c} a_{11} & a_{12} & \dots & a_{1n} & \colorThree{0}\\ a_{21} & a_{22} & \dots & a_{2n} & \colorThree{0}\\ \vdots & \vdots &&\vdots&\vdots\\ a_{m1} & a_{m2} & \dots & a_{mn} & \colorThree{0}\\ \end{array} \right]
As a matrix equation, a homogeneous system is written:
Ax⃗=0⃗\boxed{\quad A\vec x = \vec 0 \quad}
Notes
  • A homogeneous system of equations is always consistent!
  • unique solution →\rightarrow trivial solution (x⃗=0⃗\vec x = \vec 0)
  • infinitely many solutions
  • The general solution is a linear combination of basic solutions (vectors that appear next to parameters)
Example
[1101000120]\left[\begin{array}{rrrr|r} 1&1&0&1&0\\[0.25em] 0&0&1&2&0 \end{array}\right] represents a system with general solution:
x⃗=s[−1100]⏟basic solution+t[−10−21]⏟basic solution\vec x = s \underbrace{ \begin{bmatrix} -1\\ 1\\ 0\\ 0\\ \end{bmatrix} }_{\text{basic solution}} + t \underbrace{ \begin{bmatrix} -1\\ 0\\ -2\\ 1\\ \end{bmatrix} }_{\text{basic solution}}
The vectors being multiplied by parameters are non-unique basic solutions.
Wize Concept
Any linear combination of basic solutions is another basic solution.

0:00 / 0:00

Example: Solving a Linear System (Unique Solution)

Solve the system of linear equations:
2x+2y−2z=2x+2z−7=03y=−7−2z\begin{array}{ccccccl} 2x&+&2y&-&2z&=&2\\ x&+&2z&-&7&=&0\\ &&&&3y &=&-7-2z \end{array}
Steps
  1. Let's first write this in augmented matrix form:
2x+2y−2z=21x+0y+2z=70x+3y+2z=−7→[22−221027032−7]\begin{array}{rcr} 2x+2y-2z&=&2\\[0.2em] 1x+0y+2z&=&7\\[0.2em] 0x+3y+2z&=&-7 \end{array} \quad \rightarrow \quad \left[\begin{array}{rrr|r} 2&2&-2&2\\ 1&0&2&7\\ 0&3&2&-7 \end{array}\right]
  1. Turn the matrix into RREF using EROs:
[22−221027032−7]R1↔R2⟶[102722−22032−7]R2−2R1⟶[102702−6−12032−7]12R2⟶[102701−3−6032−7]R3−3R2⟶[102701−3−6001111]111R3⟶[102701−3−60011]R1−2R3R2+3R3⟶[1005010−30011]\begin{aligned} &\left[ \begin{array}{rrr|r} 2 & 2 & -2 & 2 \\ 1 & 0 & 2 & 7\\ 0 & 3 & 2 & -7 \\ \end{array} \right] \begin{array}{l} R_1 \leftrightarrow R_2\\ \\ \\ \end{array}\\[2.5em] \longrightarrow &\left[ \begin{array}{rrr|r} 1 & 0 & 2 & 7\\ 2 & 2 & -2 & 2 \\ 0 & 3 & 2 & -7 \\ \end{array} \right] \begin{array}{l} \\ R_2 - 2R_1\\ \\ \end{array}\\[2.5em] \longrightarrow &\left[ \begin{array}{rrr|r} 1 & 0 & 2 & 7\\ 0 & 2 & -6 & -12 \\ 0 & 3 & 2 & -7 \\ \end{array} \right] \begin{array}{l} \\ \dfrac{1}{2}R_2\\ \\ \end{array}\\[2.5em] \longrightarrow &\left[ \begin{array}{rrr|r} 1 & 0 & 2 & 7\\ 0 & 1 & -3 & -6 \\ 0 & 3 & 2 & -7 \\ \end{array} \right] \begin{array}{l} \\ \\ R_3 - 3R_2\\ \end{array}\\[2.5em] \longrightarrow &\left[ \begin{array}{rrr|r} 1 & 0 & 2 & 7\\ 0 & 1 & -3 & -6 \\ 0 & 0 & 11 & 11 \\ \end{array} \right] \begin{array}{l} \\ \\ \dfrac{1}{11}R_3\\ \end{array}\\[2.5em] \longrightarrow &\left[ \begin{array}{rrr|r} 1 & 0 & 2 & 7\\ 0 & 1 & -3 & -6 \\ 0 & 0 & 1 & 1 \\ \end{array} \right] \begin{array}{l} R_1 - 2 R_3\\ R_2 + 3 R_3\\ \\ \end{array}\\[2.5em] \longrightarrow &\left[ \begin{array}{rrr|r} 1 & 0 & 0 & 5\\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 1 \\ \end{array} \right] \end{aligned}
  1. Find all solutions to the SLE:
The coefficient matrix has 3 leading 1s and 3 columns: rank(A)=n{\rm rank}(A) = n   ⟹  \implies there is a unique solution.
Rewrite the augmented matrix back into a system of linear equations:
x=5y=−3z=1\begin{array}{c} x&=&5\\ y&=&-3\\ z&=&1 \end{array}
Therefore, the unique solution is
x⃗=[5−31]\boxed{\vec x = \begin{bmatrix} 5\\ −3\\ 1 \end{bmatrix}}
0:00 / 0:00

Example: Solving a Linear System (Infinitely Many Solutions)

Solve the following system of linear equations:
x+2y=4y+z=3z+w=1\begin{array}{c} x+2y&=&4\\ y+z&=&3\\ z+w&=&1 \end{array}
The standard form of the linear system is 1x+2y+0z+0w=40x+1y+1z+0w=30x+0y+1z+1w=1\begin{array}{c} 1x+2y+0z+0w&=&4\\ 0x+1y+1z+0w&=&3\\ 0x+0y+1z+1w&=&1 \end{array} and the augmented matrix is [120040110300111]\left[\begin{array}{rrrr|r} 1&2&0&0&4\\ 0&1&1&0&3\\ 0&0&1&1&1 \end{array} \right].
(Note that the order of the variables does not matter, but be consistent!)
We need to reduce the matrix to RREF:
[120040110300111]R1−2R2⟶[10−20−20110300111]R1+2R3R2−R3⟶[10020010−1200111]\begin{aligned} &\left[ \begin{array}{rrrr|r} 1 & 2 & 0 & 0 & 4\\ 0 & 1 & 1 & 0 & 3\\ 0 & 0 & 1 & 1 & 1\\ \end{array} \right] \begin{array}{l} R_1 - 2R_2\\ \\ \\ \end{array}\\[2.5em] \longrightarrow &\left[ \begin{array}{rrrr|r} 1 & 0 & -2 & 0 & -2\\ 0 & 1 & 1 & 0 & 3\\ 0 & 0 & 1 & 1 & 1\\ \end{array} \right] \begin{array}{l} R_1 + 2R_3\\ R_2 - R_3\\ \\ \end{array}\\[2.5em] \longrightarrow &\left[ \begin{array}{rrrr|r} 1 & 0 & 0 & 2 & 0\\ 0 & 1 & 0 & -1 & 2\\ 0 & 0 & 1 & 1 & 1\\ \end{array} \right] \end{aligned}
There are 3 leading 1s and 4 columns in the RREF, so rank(A)=3  <  n=4{\rm rank}(A)=3 \ \ \bm<\ \ n=4, so there are infinitely many solutions.
n−rank(A) = 4−3 = 1n-{\rm rank}(A) \ =\ 4-3 \ =\ 1 so there is one free variable (1-parameter family of solutions).
Let's assign a parameter to the free variable (the column with a missing leading 1, ww) and rewrite the SLE using the RREF.
Let w=t\colorTwo{w=t}. (Remember to include this equation in the list of equations we obtain from the RREF. We need a solution for all variables!)
{x+2w=0y−w=2z+w=1w=t  ⟹  {x+2t=0y−t=2z+t=1w=t  ⟹  {x=−2ty=2+tz=1−tw=t\begin{cases} x+2\colorTwo{w}&=&0\\ y-\colorTwo{w}&=&2\\ z+\colorTwo{w}&=&1\\ \colorTwo{w}&=&\colorTwo{t} \end{cases} \quad\implies\quad \begin{cases} x+2\colorTwo{t}&=&0\\ y-\colorTwo{t}&=&2\\ z+\colorTwo{t}&=&1\\ \colorTwo{w}&=&\colorTwo{t} \end{cases} \quad\implies\quad \begin{cases} x&=&-2\colorTwo{t}\\ y&=&2+\colorTwo{t}\\ z&=&1-\colorTwo{t}\\ \colorTwo{w}&=&\colorTwo{t} \end{cases}
Therefore, the 1-parameter family of solutions is x⃗=[0210]+t[−21−11]\boxed{ \vec x = \begin{bmatrix} 0\\ 2\\ 1\\ 0\\ \end{bmatrix} + \colorTwo{t} \begin{bmatrix} -2\\ 1\\ -1\\ 1\\ \end{bmatrix} }.
Solve the following system of linear equations. What type of solution is it?
3x1+6x2=12x1+4x2=2x2+x3=3\begin{aligned} 3x_1 + 6x_2&= 1\\ 2x_1 + 4x_2 &=2\\ x_2 + x_3 &= 3 \end{aligned}

Solve the following homogeneous system of linear equations. What type of solution is it?
2x1−x2+2x4=02x2+4x3−x4=0\begin{aligned} 2x_1-x_2+2x_4 &= 0\\ 2x_2 + 4x_3 − x_4 &= 0 \end{aligned}
Extra Practice
Find size of soln set to nonhom linear system
Find all solutions to the system of linear equations 5x+y+z=0x+2z=−12x+y−5z=1\begin{array}{rcl} 5x+y+z=0\\ x+2z=-1\\ 2x+y-5z=1 \end{array}.
Given the augmented matrix [1201010200k2−9k−3]\left[\begin{array}{rrr|r} 1 & 2 &0 &1 \\ 0 & 1 &0&2\\ 0&0&k^2-9&k-3 \end{array}\right], answer the following questions.
$\tkcth{Moved }\bcth{\to} \key{ Mid} \tkco{ S } \tkcth{ch 2 quiz ✓}$ | 133 - FML 1 - 18.1W e.g. 44.4
Interpret the following matrix as the RREF of a matrix of coefficients augmented by the vector of constants and determine whether the systems has a unique solution.
[100001000010] \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix}
Find the all values of p and q such that the system has at least one solution.
[12401300p∣15q]\left[\begin{array}{l} 1&2&4\\ 0&1&3\\ 0&0&p\\ \end{array}\right.\left|\begin{array}{l}1 \\5\\q\end{array}\right]
Find the value of p such that there is a unique solution to this system of equations:
L:{         2x1+x3=4x1+3x2+px3=−10     x1−x2+3x3=5L:\begin{cases} \ \ \ \ \ \ \ \ \ 2x_1+x_3=4\\ x_1+3x_2+px_3=-10\\ \ \ \ \ \ x_1-x_2+3x_3=5 \end{cases}
a) For what values of c does the system have one solution?
{2𝑥+𝒄𝑦=0𝒄𝑥+𝑦=3\begin{cases} 2𝑥 + 𝒄𝑦 = 0\\ 𝒄𝑥 + 𝑦 = 3 \end{cases}
b) For what values of c does it have infinitely many solutions?
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: Solving a System of Linear Equations (Infinitely Many Solutions)
Practice Question: Solving a SLE
Solve the following system of linear equations: x+2y=4y+z=3z+w=1\begin{array}{rcl} x+2y&=&4\\ y+z&=&3\\ z+w&=&1 \end{array}.
The row echelon form of the augmented matrix for a linear system is:
[A∣b][10111012400000100000∣1258].[A|b]\left[\begin{array}{l} 1&0&1&1&1\\ 0&1&2&4&0\\ 0&0&0&0&1\\ 0&0&0&0&0 \end{array}\right.\left|\begin{array}{l}1 \\2\\5\\8\end{array}\right]_.
Which of the following options is the general solution of the following system of linear equations?
x−y−z=2x+y+z=0\begin{array}{l}x-y-z=2\\x+y+z=0\end{array}
Find the basic solutions of the following homogeneous system of linear equations:
x1−2x2+3x3+x4+2x5=0x1−x2+3x3−x4+3x5=0−x1+2x2−4x3+x4+x5=0\begin{array}{l}x_1 - 2x_2 + 3x_3 + x_4 + 2x_5 = 0\\x_1 - x_2 + 3x_3 - x_4 + 3x_5 = 0\\-x_1 + 2x_2 - 4x_3 + x_4 + x_5 = 0\end{array}
What can you say about the following systems of linear equations?
x−2y−3z=42x+2y−6z=−1\begin{array}{l}x-2y-3z=4\\2x+2y-6z=-1\end{array}
What can you say about the following systems of linear equations?
x−y+3z=52x−2y+6z=−10\begin{array}{l}x-y+3z=5\\2x-2y+6z=-10\end{array}
Write the general solution to the system of linear equations
x−y−3z=−42x−3y+z=0\begin{array}{l} x - y - 3z = -4\\ 2x - 3y + z = 0 \end{array}
Wize Concept Clarifier--Gaussian Jordan Elimination
Wize Concept Clarifier
Solve the system x+y−z=24y−4z=12x+3y−3z=−1\begin{array}{} x+y-z=2\\ 4y-4z=1\\ 2x+3y-3z=-1 \end{array}.
The row echelon form of the augmented matrix for a linear system is:
[A∣b][10111012400000100000∣1258].[A|b]\left[\begin{array}{l} 1&0&1&1&1\\ 0&1&2&4&0\\ 0&0&0&0&1\\ 0&0&0&0&0 \end{array}\right.\left|\begin{array}{l}1 \\2\\5\\8\end{array}\right]_.
Linear Systems - Mixing
You are the cook on a celebrity yacht, and you have run out of flour, sugar, and baking powder -
except for that contained in a large quantity of 3 different pancake mixes:
Mix A: 8 parts flour to 8 parts sugar to 1 part baking powder
Basics of Systems of Linear Equations
Practice: Solutions to Linear Systems
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}
123 - FML #1 - 18.1W - e.g. 20
Consider the following system of linear equations:
{x−2y+4z=22x+y−2z=−13x−y+2z=12x+6y−12z=−6\bcb{ \begin{cases} x - 2y + 4z = 2& \\ 2x + y - 2z = -1 &\\ 3x - y + 2z = 1 &\\ 2x + 6y - 12z = -6& \end{cases}}
a) Construct the augmented matrix A\bcb{A} corresponding to this system.
Solve linear systems
Solve the following systems of linear equations:
a) x+2y=−3−x+3y=1\begin{array}{r}x+2y=-3\\-x+3y=1\end{array}
b) x+y−2z=1−4x+3y+2z=1−3x+11x−6z=0\begin{array}{r} x+y-2z=1\\ -4x+3y+2z=1\\ -3x+11x-6z=0 \end{array}
123 - FML #1 - 18.1W - e.g. 21.1
Solve the following system of linear equations:
{2x+y+z+3w=13x−2z+2w=−15x+y−z+5w=07x+2y+8w=1\bcb{ \begin{cases} 2x + y + z + 3w = 1 \\ 3x - 2z + 2w = -1 \\ 5x + y - z + 5w = 0 \\ 7x + 2y + 8w = 1 \end{cases} }
Solve the following linear system of equations by Gaussian Elimination.
2x−y+5z=14x+y+z=4−x−3y+3z=−2\begin{array}{c} 2x - y + 5z = 14\\ x + y + z = 4\\ -x - 3y + 3z = - 2 \end{array}
Solve the following linear system of equations.
3x+2y−2z=185x+2y−z=27x+y+z=9\begin{array}{c} 3x + 2y - 2z = 18\\ 5x + 2y - z =27\\ x + y + z = 9 \end{array}
Find soln to nonhom linear system
Find all solutions to the following system of linear equations[10210131]\left[\begin{array}{rrr|r} 1&0&2&1\\ 0&1&3&1 \end{array}\right].
Balancing Chemical Equations
The chemical equation for the combustion of octane (the primary component of gasoline) is:
C8H18+O2=H2O+CO2C_8H_{18}+O_2 = H_2O+CO_2
Balance this equation using Gauss-Jordan elimination
Find all possible solutions to the following linear system of equations.
x+2y−z=9x−y+2z=3\begin{array}{c} x + 2y - z = 9\\ x - y + 2z = 3 \end{array}
Solving Linear Systems
Practice: Solving a Linear System
Solve the following system of linear equations. What type of solution is it?
3x1+6x2=12x1+4x2=2x2+x3=3\begin{aligned} 3x_1 + 6x_2&= 1\\ 2x_1 + 4x_2 &=2\\ x_2 + x_3 &= 3 \end{aligned}
Solving Linear Systems
Practice: Solving a Homogeneous Linear System
Solve the following homogeneous system of linear equations. What type of solution is it?
2x1−x2+2x4=02x2+4x3−x4=0\begin{aligned} 2x_1-x_2+2x_4 &= 0\\ 2x_2 + 4x_3 − x_4 &= 0 \end{aligned}
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]
Find I0 in the following circuit.
Solving Linear Systems
Example: Solving a Linear System (Unique Solution)
Solve the system of linear equations:
2x+2y−2z=2x+2z−7=03y=−7−2z\begin{array}{ccccccl} 2x&+&2y&-&2z&=&2\\ x&+&2z&-&7&=&0\\ &&&&3y &=&-7-2z \end{array}
Solving Linear Systems
Example: Solving a Linear System (Infinitely Many Solutions)
Solve the following system of linear equations:
x+2y=4y+z=3z+w=1\begin{array}{c} x+2y&=&4\\ y+z&=&3\\ z+w&=&1 \end{array}
19.4F_Final_Builder_Ch_2.13_Algebra_SLE_$\tkco{eg22}$_$\key{Final}$_Builder_$\tkcth{2.13.}\tkcf{22}$_
Given the system of equations:
{2x1−4x2=−1−4x1+8x2=2 \left\{ \begin{array}{ll} 2x_1 - 4x_2 & = \quad -1 \\ -4x_1 + 8x_2 & = \quad 2 \end{array} \right.
Show that x⃗g=x⃗p+tx⃗h\vx_g = \vx_p + t \vx_h, where x⃗p=[−120]\vx_p = \colvec{-\frac12}{0} and x⃗h=[21]\vx_h = \colvec{2}{1} is a solution to the system when t=1,2,3t = 1, 2, 3, i.e. show that the vectors:
Find (if possible) conditions on aa and bb so that the following system has
(a) no solution
(b) one solution
Which of the following options is the general solution of the following system of linear equations?
x−2y+z=3x−y−z=0\begin{array}{l}x-2y+z=3\\x-y-z=0\end{array}
What can you say about the following systems of linear equations?
x+y−2z=22x+2y−6z=3\begin{array}{l}x+y-2z=2\\2x+2y-6z=3\end{array}
What can you say about the following systems of linear equations?
x+y−2z=−32x+2y−4z=−6\begin{array}{l}x+y-2z=-3\\2x+2y-4z=-6\end{array}
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}
Matrix Multiplication
e.g. Find a 3 × 3 matrix M such that
 ⁣M[   0−2   0]=[−1   1−1] ⁣, ⁣M[001]=[   0−2   3] ⁣,and M[1300]=[−2   2−1]\!M\begin{bmatrix} \ \ \ 0\\-2\\\ \ \ 0 \end{bmatrix}= \begin{bmatrix} -1\\\ \ \ 1\\-1 \end{bmatrix}\!, \hspace{0.7cm} \!M\begin{bmatrix} 0\\0\\1 \end{bmatrix}= \begin{bmatrix} \ \ \ 0\\-2\\\ \ \ 3 \end{bmatrix}\!, \hspace{0.7cm} \text{\small{and}}\ M\begin{bmatrix} \frac{1}{3}\\0\\0 \end{bmatrix}= \begin{bmatrix} -2\\\ \ \ 2\\-1 \end{bmatrix}
System of Linear Equations
Find (if possible) conditions on aa and bb so that the following system has
(a) no solution
(b) one solution
Homogeneous System of Equations
Find basic solutions of a homogeneous system of linear equations with reduced row echelon form of:
[1203000150]\left[\begin{array}{cccc|c}1 & 2 & 0 & 3 & 0 \\0 & 0 & 1 & 5 & 0\end{array}\right]
Practice Question: Solving a System of Linear Equations (Unique Solution)
Practice Question: Solving a SLE--Unique Solution
Solve the following systems of linear equations.
𝑥−𝑧=5𝑥−𝑧=5
Find size of soln set to nonhom linear system
Find all solutions to the system of linear equations 5x+y+z=0x+2z=−12x+y−5z=1\begin{array}{rcl} 5x+y+z=0\\ x+2z=-1\\ 2x+y-5z=1 \end{array}.
What can you say about the following systems of linear equations?
x+y−2z=22x+2y−6z=3\begin{array}{l}x+y-2z=2\\2x+2y-6z=3\end{array}
Write the general solution (all solutions) to the system of linear equations
x−y−3z=−42x−3y+z=0\begin{array}{l}x-y-3z=-4\\2x-3y+z=0\end{array}
Find (if possible) conditions on aa and bb so that the following system has
(a) no solution
(b) one solution
What can you say about the following systems of linear equations?
x+y−2z=−32x+2y−4z=−6\begin{array}{l}x+y-2z=-3\\2x+2y-4z=-6\end{array}
Which of the following options is the general solution of the following system of linear equations?
x−2y+z=3x−y−z=0\begin{array}{l}x-2y+z=3\\x-y-z=0\end{array}
Find all solutions to the system of linear equations 5x+y+z=0x+2z=−12x+y−5z=1\begin{array}{c} 5x+y+z=0\\ x+2z=-1\\ 2x+y-5z=1 \end{array}.
Practice Question: Solving a System of Linear Equations (Infinitely Many Solutions)
Practice Question: Solving a SLE
Solve the following system of linear equations: x+2y=4y+z=3z+w=1\begin{array}{rcl} x+2y&=&4\\ y+z&=&3\\ z+w&=&1 \end{array}.
Solve the homogeneous linear system
When the right hand side of every equation in a linear system is 0, we say that the system is homogeneous
Find the general solution of the following homogeneous linear system:
{3x+2y=0−6x+4y=0\left\{\begin{array}{rcl} 3x+2y&=&0\\ -6x+4y&=&0 \end{array} \right.
Solving a system of linear equations
Solve the following system of linear equations:
−2x1+3x2+13x3=5−x1+2x2+5x3=14x1−14x2−2x3=14\begin{array}{rcr} -2x_1+3x_2+13x_3&=&5\\ -x_1+2x_2+5x_3&=&1\\ 4x_1-14x_2-2x_3&=&14\\ \end{array}
Solving a system of linear equations
Solve the following system of linear equations:
−2x1+3x2+13x3=5−x1+2x2+5x3=14x1−14x2−2x3=14\begin{array}{rcr} -2x_1+3x_2+13x_3&=&5\\ -x_1+2x_2+5x_3&=&1\\ 4x_1-14x_2-2x_3&=&14\\ \end{array}
Solving a system of linear equations
Solve the following system of linear equations:
−2x1−4x2−6x3−19x4=−153x1+10x2−30x3+9x4=24x1+3x2−6x3+4x4=87x1+17x2+45x4=55\begin{array}{rcr} -2x_1-4x_2-6x_3-19x_4&=&-15\\ 3x_1+10x_2-30x_3+9x_4&=&24\\ x_1+3x_2-6x_3+4x_4&=&8\\ 7x_1+17x_2+45x_4&=&55 \end{array}
where we wish to solve for the quadruple (x1,x2,x3,x4)(x_1,x_2,x_3,x_4) of real numbers that satisfy this system
Practice: Solutions to a System of Equations
Practice: Solutions to a System of Equations
Given the system of linear equations x+y=0x−cy=1\begin{array}{c} x+y=0\\ x-cy=1 \end{array}, find the values of cc such that,
Solving a system of linear equations
Solve the following system of linear equations:
−2x1+3x2+13x3=5−x1+2x2+5x3=14x1−14x2−2x3=14\begin{array}{rcr} -2x_1+3x_2+13x_3&=&5\\ -x_1+2x_2+5x_3&=&1\\ 4x_1-14x_2-2x_3&=&14\\ \end{array}
where we wish to solve for the triple (x1,x2,x3)(x_1,x_2,x_3) of real numbers that satisfy this system
Practice: Solving a System of Equations
Practice: Solving a System of Equations
Solve the system of equations 2x+2y−2z=2x+2z−7=03y=−7−2z\begin{array}{rcl} 2x+2y-2z&=&2\\ x+2z-7&=&0\\ 3y&=&-7-2z \end{array}
Linear algebra
For which values of aaand bbdoes the system x+3y=2b3x−3ay=a\begin{array}{l}x+3y = 2b\\3x-3ay = a\end{array}have infinitely many solutions?
For which values of a and b does the system x+3y=2b3x−3ay=a\begin{array}{l}x+3y = 2b\\3x-3ay = a\end{array}have infinitely many solutions?
For which values of a and b does the system 2x−4y=a3x−ay=3b\begin{array}{l}2x-4y = a\\3x-ay = 3b\end{array}have no solutions?
Write the general solution to the system of linear equations
x−y−3z=−42x−3y+z=0\begin{array}{l} x - y - 3z = -4\\ 2x - 3y + z = 0 \end{array}
System of Linear Equations
Write the general solution to the system of linear equations
x−y−3z=−42x−3y+z=0\begin{array}{l} x - y - 3z = -4\\ 2x - 3y + z = 0 \end{array}
Solve the following linear system of equations.
2x+3y−z=7x−2y+5z=27y−11z=6\begin{array}{c} 2x + 3y - z = 7\\ x - 2y + 5z = 2\\ 7y - 11z = 6 \end{array}
Solving Linear Systems
Solve the following homogeneous system of linear equations. What type of solution is it?
2x1−x2+2x4=02x2+4x3−x4=0\begin{aligned} 2x_1-x_2+2x_4 &= 0\\ 2x_2 + 4x_3 − x_4 &= 0 \end{aligned}
Solving Linear Systems
Solve the following system of linear equations. What type of solution is it?
3x1+6x2=12x1+4x2=2x2+x3=3\begin{aligned} 3x_1 + 6x_2&= 1\\ 2x_1 + 4x_2 &=2\\ x_2 + x_3 &= 3 \end{aligned}
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}