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Section 1.1 Matrices and Systems of Linear Equations

Subsection 1.1.1 Systems of linear equations

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Definition 1.1.1. Linear equations.

A linear equation in the \(n\) variables \(x_1, x_2, \cdots, x_n \) is an equation that can be written in the form

\begin{equation*} a_1\,x_1 + a_2\,x_2 + \cdots + a_n\,x_n = b \end{equation*}

where the coefficients \(a_1, a_2,\cdots, a_n \) and the independent variable \(b\) are constants.

The equation of the line \(y = 3x + 2 \) can be written as

\begin{equation*} -3x_1 + x_2 = 2 \end{equation*}

The equation of a plane is also considered a linear equation. For example

\begin{equation*} 3x_1 - x_2 + x_3 = 2 \end{equation*}

Determine if the given equation is linear (input y or n).

  1. Is \(x = 3y - 2\) Linear?

  2. Is \(y + \sin(x) = 3\) Linear?

  3. Is \(a + b - 3c = 4\) Linear?

  4. Is \(x_1 - x_3+ x_5 = 0\) Linear?

  5. Is \(x_1^2 + x_2 = 3\) Linear?

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Answer 1

\(\text{Yes}\)

Answer 2

\(\text{No}\)

Answer 3

\(\text{Yes}\)

Answer 4

\(\text{Yes}\)

Answer 5

\(\text{No}\)

Solution

1, 3 and 4 are linear. 2 and 5 are non-linear.

The name of the variable does not matter (see 1, 3 and 4) as long as the variables are in linear form the equation is consider linear. The second equation is not linear because the trigonometric function and the last equation is not linear because the first variable is squared.

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Definition 1.1.5. Standard form.

In the standard form of a linear equation all terms containing unknowns are placed in the left-hand-side of the equation and all constant terms in the right-hand-side.

  • \begin{equation*} 3x_1 - x_2 + x_3 = 2 \end{equation*}
  • \begin{equation*} a_1 x_1 + a_2 x_2 + a_3 x_3 + \cdots + a_n x_n = b \end{equation*}

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Definition 1.1.7. Solutions of linear equations.

A point that satisfies a linear equation is considered a solution to that equation.

Consider the equation \(3x_1 + 4x_2 = 21. \)

  1. Is the point \(\left(3, -2 \right)\) a solution to this equation?

  2. Is the point \(\left(0, 2 \right)\) a solution to this equation?

  3. Is the point \(\left(3, 3 \right)\) a solution to this equation?

  4. Is the point \(\left(7, 0 \right)\) a solution to this equation?

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Answer 1

\(\text{No}\)

Answer 2

\(\text{No}\)

Answer 3

\(\text{Yes}\)

Answer 4

\(\text{Yes}\)

Solution

Substitute the first coordinate for the first variable \(\left(x_1\right)\) and the second coordinate for the second variable \(\left(x_2\right)\) and if the result is \(21\) then the point is a solution to the linear equation.

Consider the equation \(x_1 - 4x_2 + 3x_3 - 2x_4 = -1. \)

  1. Is the point \(\left(1, 0, 0, 1 \right)\) a solution to this equation?

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Answer

\(\text{Yes}\)

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Definition 1.1.10. Systems of linear equations.

A system of linear equations consist of two or more linear equations such that all equations in the system are considered simultaneously.

The system of linear equations below consists of \(m\) equations of \(n \) variables,

\begin{equation*} \left\{\begin{array}{ccccccccccc} a_{11}\, x_1 \amp+\amp a_{12}\, x_2 \amp+\amp a_{13}\, x_3 \amp+\amp \cdots \amp+\amp a_{1n}\, x_n \amp= \amp b_1\\ a_{21}\, x_1 \amp+\amp a_{22}\, x_2 \amp+\amp a_{23}\, x_3 \amp+\amp \cdots \amp+\amp a_{2n}\, x_n \amp= \amp b_2\\ a_{31}\, x_1 \amp+\amp a_{32}\, x_2 \amp+\amp a_{33}\, x_3 \amp+\amp \cdots \amp+\amp a_{3n}\, x_n \amp= \amp b_3\\ \vdots\amp+\amp \vdots \amp+\amp \vdots \amp+\amp \vdots \amp+\amp \vdots \amp=\amp \vdots\\ a_{m1}\, x_1 \amp+\amp a_{m2}\, x_2 \amp+\amp a_{m3}\, x_3 \amp+\amp \cdots \amp+\amp a_{mn}\, x_n \amp= \amp b_m\\ \end{array} \right. \end{equation*}

Here the \(x_i\)'s are the \(\textit{unknowns}\text{,}\) the \(a_{i j}\)'s are the \(\textit{coefficients}\text{,}\) and the \(b_i\)'s are the \(\textit{independent variables}\) or right-hand-side.

Definition 1.1.11. Homogeneous systems of linear equations.

A system of equations is called homogeneous if each equation in the system is equal to 0. A homogeneous system has the form

\begin{equation*} \left\{\begin{array}{ccccccccccc} a_{11}\, x_1 \amp+\amp a_{12}\, x_2 \amp+\amp a_{13}\, x_3 \amp+\amp \cdots \amp+\amp a_{1n}\, x_n \amp= \amp 0\\ a_{21}\, x_1 \amp+\amp a_{22}\, x_2 \amp+\amp a_{23}\, x_3 \amp+\amp \cdots \amp+\amp a_{2n}\, x_n \amp= \amp 0\\ a_{31}\, x_1 \amp+\amp a_{32}\, x_2 \amp+\amp a_{33}\, x_3 \amp+\amp \cdots \amp+\amp a_{3n}\, x_n \amp= \amp 0\\ \vdots\amp+\amp \vdots \amp+\amp \vdots \amp+\amp \vdots \amp+\amp \vdots \amp=\amp \vdots\\ a_{m1}\, x_1 \amp+\amp a_{m2}\, x_2 \amp+\amp a_{m3}\, x_3 \amp+\amp \cdots \amp+\amp a_{mn}\, x_n \amp= \amp 0\\ \end{array} \right. \end{equation*}
Definition 1.1.12. Solutions to systems of linear equations.

A solution to a system of linear equations are the points that satisfy all equations in the linear system at the same time.

Consider the system of equations

\begin{equation*} \left\{\begin{array}{lcr} \hspace{0.05cm}x_1 + x_2 \amp=\amp 3\\ 2x_1 - x_2 +\amp=\amp 3 \end{array} \right. \end{equation*}

A solution to the first equation would be \(x_1 = 1, x_2 = 2 \) since

\begin{equation*} (1)+ (2) = 3 \text{.} \end{equation*}

However, this is not a solution to the second equation, since

\begin{equation*} 2(1) - (2) - (2) \ne 3 \text{.} \end{equation*}

Similarly, \(x_1 = 1, x_2 =-1 \) is a solution to the second equation but not to the first. Check!

Then, according to our definition neither set of values is a solution to the system of equations.

A solution to the system of equations is \(x_1 = 2, x_2 = 1 \text{,}\) since

\begin{equation*} \begin{array}{lcr} \hspace{0.05cm}(2) + (1) \amp=\amp 3\\ \amp\text{and}\amp\\ 2(2) - (1) \amp=\amp 3 \end{array} \end{equation*}

In this case the system has a unique solution.

Consider the system of linear equations depicted in the figure below.

Does the system have unique solution, infinitely many solutions, or no solution?

  • Unique solution

  • Infinitely many solutions

  • No solution

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Answer

\(\text{No solution}\)

Solution

For the system to have a solution ALL three lines should intersect, so that the solution satisfies all three equations simultaneously.

Consider the system of linear equations depicted in the figure below.

Does the system have unique solution, infinitely many solutions, or no solution?

  • Unique solution

  • Infinitely many solutions

  • No solution

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Answer

\(\text{Infinitely ... solutions}\)

Solution

Lines overlap each other, meaning there are infinitely many points that satisfy both equations simultaneously.

In previous problems we saw that, in two dimensions, solving a system of linear equations is equivalent to finding the intersection of all the lines within the system. We also know that linear equations in three dimensions correspond to planes. Using what you learned in previous problems and examples, can you determine whether the system depicted below has a unique solution, no solution or infinitely many solutions?

  • Unique solution

  • Infinitely many solutions

  • No solution

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Answer

\(\text{Infinitely ... solutions}\)

Solution

The two planes intersect at a line, meaning every point that satisfies that line is a solution to the system of linear equations.

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Definition 1.1.17. Consistent systems of linear equations.

A system of linear equations is called consistent if there exists at least one solution. It is called inconsistent if there is no solution.

Insight 1.1.18. Summary of classification of systems of linear equations.

Subsection 1.1.2 Matrices

In linear algebra we are interested in solving linear systems of equations, however, the geometric approach used in the last section becomes cumbersome for systems with more than 3 variables, which is often the case in applications. Matrices offer an alternative for manipulating large systems of linear equations in a concise way.

Definition 1.1.19. Matrix.

A matrix is an array of numbers organized in rows and columns.

\begin{equation*} \left[ \begin{array}{cccc} \hspace{0.75cm} 3 \hspace{0.75cm}\amp \hspace{0.75cm} 2 \hspace{0.75cm}\amp \hspace{0.75cm} -4 \hspace{0.75cm}\amp \hspace{0.75cm} 1\hspace{0.75cm}\\ 1 \amp 1 \amp 2 \amp 0\\ 1 \amp -3 \amp -1 \amp 2\\ \end{array}\right] \begin{array}{cc} \rightarrow \amp \text{row 1}\\ \rightarrow \amp \text{row 2}\\ \rightarrow \amp \text{row 3}\\ \end{array} \end{equation*}
\begin{equation*} \hspace{-3cm}\begin{array}{cccc} \uparrow \amp \uparrow \amp \uparrow \amp \uparrow\\ \text{column 1} \amp \text{column 2} \amp \text{column 3} \amp \text{column 4}\\ \end{array} \end{equation*}

For now we are only interested in matrices as representations of systems of linear equations. We will come back to matrices properties and operations in [cross-reference to target(s) "chapter2" missing or not unique].

Subsection 1.1.3 Matrix representation of systems of linear equations

A system of linear equations can be represented in matrix form using its coefficient matrix (Subsubsection 1.1.3.1), a variable vector (Subsubsection 1.1.3.2), and a constant vector (Subsubsection 1.1.3.3).

Insight 1.1.20. Video summary.

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Subsubsection 1.1.3.1 Coefficient matrix

The coefficient matrix of a system of linear equations contains all the coefficients in the system of equations. In this matrix representation the rows of the matrix correspond to each equation in the system and the columns to each variable. For a system with \(m \) equations and \(n \) variables the size of the coefficient matrix is \(m \times n \text{.}\)

The system of linear equations

\begin{equation*} \left\{ \begin{array}{rrrrrrrrr} 3x_1 \amp + \amp 2x_2 \amp - \amp 4x_3 \amp + \amp x_4 \amp = \amp 8\\ x_1 \amp + \amp x_2 \amp + \amp 2x_3 \amp \amp \amp = \amp 1\\ x_1 \amp - \amp 3x_2 \amp - \amp x_3 \amp + \amp 2x_4 \amp = \amp 2 \end{array}\right. \end{equation*}

has the coefficient matrix:

\begin{equation*} \left[ \begin{array}{rrrr} 3 \amp 2 \amp -4 \amp 1\\ 1 \amp 1 \amp 2 \amp 0\\ 1 \amp -3 \amp -1 \amp 2\\ \end{array}\right]. \end{equation*}

Find the coefficient matrix for the given system of equations,

\begin{equation*} \begin{array}{rrrrrrr} {7} x_1 \amp + \amp {6} x_2 \amp - \amp {6} x_3 \amp = \amp {-1}\\ {2} x_1 \amp + \amp {8} x_2 \amp + \amp {3} x_3 \amp = \amp {-1}\\ {3} x_1 \amp - \amp {3} x_2 \amp - \amp {1} x_3 \amp = \amp {-2} \end{array} \end{equation*}
\(\,\) \(\lceil\)


\(\rceil\)
\(\boldsymbol{A} =\) \(\vert\)


\(\vert\)
\(\,\) \(\lfloor\)


\(\rfloor\)
\(\,\) \(\,\) \(\,\)

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Answer 1

\(7\)

Answer 2

\(6\)

Answer 3

\(-6\)

Answer 4

\(2\)

Answer 5

\(8\)

Answer 6

\(3\)

Answer 7

\(3\)

Answer 8

\(-3\)

Answer 9

\(-1\)

Solution
\begin{equation*} \left[\begin{array}{ccc} {7} \amp {6} \amp {-6}\\ {2} \amp {8} \amp {3}\\ {3} \amp {-3} \amp {-1}\\ \end{array}\right] \end{equation*}

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Subsubsection 1.1.3.2 Variable vector

For a system with \(m \) equations and \(n \) variables the variable vector is a \(1 \times n \) matrix containing the unknowns of the system.

The system of linear equations

\begin{equation*} \left\{ \begin{array}{rrrrrrrrr} 3x_1 \amp + \amp 2x_2 \amp - \amp 4x_3 \amp + \amp x_4 \amp = \amp 8\\ x_1 \amp + \amp x_2 \amp + \amp 2x_3 \amp \amp \amp = \amp 1\\ x_1 \amp - \amp 3x_2 \amp - \amp x_3 \amp + \amp 2x_4 \amp = \amp 2 \end{array}\right. \end{equation*}

has the variable vector:

\begin{equation*} \left[ \begin{array}{c} x_1 \\ x_2 \\ x_3\\ x_4 \end{array}\right]. \end{equation*}

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Subsubsection 1.1.3.3 Constant vector

When a system of linear equations is written in standard form (Definition 1.1.5) the constant vector contains the right-hand-side of the equations. For a system with \(m \) equations and \(n \) variables the constant vector is a \(1 \times m \) matrix.

The system of linear equations

\begin{equation*} \left\{ \begin{array}{rrrrrrrrr} 3x_1 \amp + \amp 2x_2 \amp - \amp 4x_3 \amp + \amp x_4 \amp = \amp 8\\ x_1 \amp + \amp x_2 \amp + \amp 2x_3 \amp \amp \amp = \amp 1\\ x_1 \amp - \amp 3x_2 \amp - \amp x_3 \amp + \amp 2x_4 \amp = \amp 2 \end{array}\right. \end{equation*}

has the constant vector:

\begin{equation*} \left[ \begin{array}{c} 8 \\ 1 \\ 2 \end{array}\right]. \end{equation*}

Consider the following system of equations,

\begin{equation*} \begin{array}{rrrrrrr} {8} u \amp - \amp {3} v \amp - \amp {9} w \amp = \amp {-8}\\ {7} u \amp + \amp {8} v \amp \amp \amp = \amp {-3}\\ {3} u \amp - \amp {8} v \amp + \amp {6} w \amp = \amp {9} \end{array} \end{equation*}

The coefficient matrix for the system is

\(\,\) \(\lceil\)


\(\rceil\)
\(A =\) \(\vert\)


\(\vert\)
\(\,\) \(\lfloor\)


\(\rfloor\)
\(\,\) \(\,\) \(\,\)

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The vector of variables is

\(\,\) \(\lceil\)
\(\rceil\)
\(x =\) \(\vert\)
\(\vert\)
\(\,\) \(\lfloor\)
\(\rfloor\)
\(\,\) \(\,\) \(\,\) \(\,\)

The constant vector for the system is

\(\,\) \(\lceil\)
\(\rceil\)
\(b =\) \(\vert\)
\(\vert\)
\(\,\) \(\lfloor\)
\(\rfloor\)
\(\,\) \(\,\) \(\,\) \(\,\)

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Answer 1

\(8\)

Answer 2

\(-3\)

Answer 3

\(-9\)

Answer 4

\(7\)

Answer 5

\(8\)

Answer 6

\(0\)

Answer 7

\(3\)

Answer 8

\(-8\)

Answer 9

\(6\)

Answer 10

\(\text{u}\)

Answer 11

\(\text{v}\)

Answer 12

\(\text{w}\)

Answer 13

\(-8\)

Answer 14

\(-3\)

Answer 15

\(9\)

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Subsubsection 1.1.3.4 Augmented matrix

A system of equations can be abbreviated using a augmented matrix. To obtain the augmented matrix for a system we combine the coefficient matrix and the constant vector.

The system of linear equations

\begin{equation*} \left\{ \begin{array}{rrrrrrrrr} 3x_1 \amp + \amp 2x_2 \amp - \amp 4x_3 \amp + \amp x_4 \amp = \amp 8\\ x_1 \amp + \amp x_2 \amp + \amp 2x_3 \amp \amp \amp = \amp 1\\ x_1 \amp - \amp 3x_2 \amp - \amp x_3 \amp + \amp 2x_4 \amp = \amp 2 \end{array}\right. \end{equation*}

has the augmented matrix:

\begin{equation*} \left[ \begin{array}{rrrrr} 3 \amp 2 \amp -4 \amp 1 \amp 8\\ 1 \amp 1 \amp 2 \amp 0 \amp 1\\ 1 \amp -3 \amp -1 \amp 2 \amp 2\\ \end{array}\right]. \end{equation*}

Note: sometimes a bar is added to the augmented matrix to differentiate the coefficient matrix from the independent vector.

\begin{equation*} \left[ \begin{array}{rrrr|r} 3 \amp 2 \amp -4 \amp 1 \amp 8\\ 1 \amp 1 \amp 2 \amp 0 \amp 1\\ 1 \amp -3 \amp -1 \amp 2 \amp 2\\ \end{array}\right]. \end{equation*}

Find the augmented matrix for the system of linear equations,

\begin{equation*} \begin{array}{rrrrrrr} {9} x_1 \amp - \amp {9} x_2 \amp + \amp {5} x_3 \amp = \amp {-5}\\ {4} x_1 \amp + \amp {7} x_2 \amp - \amp {3} x_3 \amp = \amp {-4}\\ {-6} x_1 \amp + \amp {1} x_2 \amp + \amp {6} x_3 \amp = \amp {-8} \end{array} \end{equation*}
\(\,\) \(\lceil\)



\(\rceil\)
\(\boldsymbol{A} =\) \(\vert\)



\(\vert\)
\(\,\) \(\lfloor\)



\(\rfloor\)
\(\,\) \(\,\) \(\,\)

\(\,\)
Answer 1

\(9\)

Answer 2

\(-9\)

Answer 3

\(5\)

Answer 4

\(-5\)

Answer 5

\(4\)

Answer 6

\(7\)

Answer 7

\(-3\)

Answer 8

\(-4\)

Answer 9

\(-6\)

Answer 10

\(1\)

Answer 11

\(6\)

Answer 12

\(-8\)

Solution
\begin{equation*} A = \left[\begin{array}{cccc} {9} \amp {-9}\amp {5} \amp {-5}\\ {4} \amp {7} \amp {-3} \amp {-4}\\ {-6} \amp {1} \amp {6} \amp {-8} \end{array} \right] \end{equation*}

Convert the augmented matrix

\begin{equation*} A = \begin{array}{ccc} \huge{\lceil} \normalsize{\, {-7} }\amp {6} \amp {8} \,\, \huge{\rceil}\\ \huge{\vert} \normalsize{\, {-2}} \amp {7} \amp {7} \,\, \huge{\vert}\\ \huge{\lfloor} \normalsize{\, {-4}} \amp {5} \amp {8} \,\, \huge{\rfloor} \end{array} \end{equation*}

to the equivalent linear system. Use x1 and x2 to enter the variables \(x_1\) and \(x_2 \text{.}\)

=

=

=

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Answer 1

\(6x_2-7x_1\)

Answer 2

\(8\)

Answer 3

\(7x_2-2x_1\)

Answer 4

\(7\)

Answer 5

\(5x_2-4x_1\)

Answer 6

\(8\)

Solution
\begin{equation*} \begin{array}{ccc} {-7} x_1 + {6} x_2 \amp = \amp {8}\\ {-2} x_1 + {7} x_2 \amp = \amp {7}\\ {-4} x_1 + {5} x_2 \amp = \amp {8}\\ \end{array} \end{equation*}

Determine the value of \(h \) such that the matrix is the augmented matrix of a consistent linear system.

See Definition 1.1.17

\begin{equation*} \left[ \begin{array} {rr} {4} \amp {-6} \cr {-8} \amp {12} \end{array} \begin{array}{r} h \cr {3} \end{array} \right] \end{equation*}

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\(h =\)

\(\,\)

Answer

\(\frac{-3}{2}\)

Solution

The system of equations corresponding to the augmented matrix is

\begin{equation*} \left\{ \begin{array}{rcr} {4} x_1 - {6} x_2 \amp = \amp h \cr {-8} x_1 + {12} x_2 \amp = \amp {3} \end{array}\right. \end{equation*}

From the second equation

\begin{equation*} x_1 = {-{\textstyle\frac{3}{8}}} \,+ {{\textstyle\frac{3}{2}}}\,x_2 \end{equation*}

Substituing into the first equation gives

\begin{equation*} {4} \left( {-{\textstyle\frac{3}{8}}} + {{\textstyle\frac{3}{2}}}\,x_2 \right) - {6} x_2 = h \end{equation*}

Simplifying

\begin{equation*} \begin{array}{rcr} {-{\textstyle\frac{3}{2}}} + {6} x_2 - {6} x_2 \amp = \amp h \\ {-{\textstyle\frac{3}{2}}} \amp = \amp h \end{array} \end{equation*}

Subsection 1.1.4 Homework

Link to webwork: HW 1.1