4.01: Prerequisites to Simultaneous Linear Equations
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Learning Objectives
After successful completion of this lesson, you should be able to:
1) Define what a matrix is.
2) Identify special types of matrices.
What does a matrix look like?
Matrices are everywhere. If you have used a spreadsheet such as Excel or wrote numbers in a table, you have used a matrix. Matrices make the presentation of numbers clearer and make calculations easier to program. Look at the matrix below about the sale of tires in a Blowoutr’us store – given by quarter and make of tires.
\begin{matrix} Tirestone\\ Michigan\\ Copper\\ \end{matrix} \stackrel{\mbox{Q1. Q2. Q3. Q4.}}{\begin{bmatrix} 25 & 20 & 3 & 2 \\ 5 & 10 &15 &25 \\ 6 & 16 &7 & 27 \\ \end{bmatrix}} \nonumber
If one wants to know how many Copper tires were sold in Quarter 4, we go along the row Copper and column Q4 and find that it is 27.
So, what is a matrix?
A matrix is a rectangular array of elements. The elements can be symbolic expressions or/and numbers. Matrix \lbrack A\rbrack is denoted by
\displaystyle \lbrack A\rbrack = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\a_{21} & a_{22} &\cdots & a_{2n} \\ \vdots & & & \vdots \\a_{m1} & a_{m2} & \cdots & a_{\text{mn}} \\\end{bmatrix} \nonumber
Row i of \lbrack A\rbrack has n elements and is
\left\lbrack a_{i1} \ \ \ a_{i2} \ \ \ \ldots \ \ \ a_{\text{in}} \right\rbrack \nonumber
and column j of \lbrack A\rbrack has m elements and is
\begin{bmatrix} a_{1j} \\ a_{2j} \\ \vdots \\ a_{\text{mj}} \\\end{bmatrix} \nonumber
Each matrix has rows and columns, and this defines the size of the matrix. If a matrix \lbrack A\rbrack has m rows and n columns, the size of the matrix is denoted by m \times n. The matrix \lbrack A\rbrack may also be denoted by \lbrack A\rbrack_{m \times n} to show that \lbrack A\rbrack is a matrix with m rows and n columns.
Each entry in the matrix is called the entry or element of the matrix and is denoted by a_{\text{ij}} where i is the row number, and j is the column number of the element.
The matrix for the tire sales example could be denoted by the matrix A as
\lbrack A\rbrack = \begin{bmatrix} 25 & 20 & 3 & 2 \\ 5 & 10 & 15 & 25 \\ 6 & 16 & 7 & 27 \\ \end{bmatrix} \nonumber
There are 3 rows and 4 columns, so the size of the matrix is 3 \times 4. In the above \lbrack A\rbrack matrix, a_{34} = 27.
Audiovisual Lecture
Title: What is a Matrix?
Summary: Learn what a matrix is.
What are the special types of matrices?
Vector:
A vector is a matrix that has only one row or one column. There are two types of vectors – row vectors and column vectors.
Row vector
If a matrix \lbrack B\rbrack has one row, it is called a row vector \lbrack B\rbrack = \lbrack b_{1}\ b_{2}\ldots\ldots b_{n}\rbrack and n is the dimension of the row vector.
Give an example of a row vector.
Solution
\lbrack B \rbrack =\lbrack 25\ \ 20\ \ 3\ \ 2\ \ 0\rbrack \nonumber
is an example of a row vector of dimension 5.
Audiovisual Lecture
Title: What is a Row Vector
Summary: Learn what a row vector is.
Column vector
If a matrix \lbrack C\rbrack has one column, it is called a column vector
\lbrack C\rbrack = \begin{bmatrix} c_{1} \\ \vdots \\ \vdots \\ c_{m} \\ \end{bmatrix} \nonumber
and m is the dimension of the vector.
Give an example of a column vector.
Solution
\lbrack C\rbrack = \begin{bmatrix} 25 \\ 5 \\ 6 \\ \end{bmatrix} \nonumber
is an example of a column vector of dimension 3.
Audiovisual Lecture
Title: What is a Column Vector
Summary: Learn what a column vector is.
Submatrix
If some row(s) or/and column(s) of a matrix \lbrack A\rbrack are deleted (no rows or columns may be deleted), the remaining matrix is called a submatrix of \lbrack A\rbrack.
Audiovisual Lecture
Square matrix
If the number of rows m of a matrix is equal to the number of columns n of a matrix \lbrack A\rbrack, that is, m = n, then \lbrack A\rbrack is called a square matrix. The entries a_{11},a_{22},...,a_{nn} are called the diagonal elements of a square matrix. Sometimes the diagonal of the matrix is also called the principal or main of the matrix.
Give an example of a square matrix.
Solution
\lbrack A\rbrack = \begin{bmatrix} 25 & 20 & 3 \\ 5 & 10 & 15 \\ 6 & 15 & 7 \\ \end{bmatrix} \nonumber
is a square matrix as it has the same number of rows and columns, that is, 3. The diagonal elements of \lbrack A\rbrack are a_{11} = 25,\ \ a_{22} = 10,\ \ a_{33} = 7. \nonumber
Audiovisual Lecture
Title: What is a Square Matrix
Summary: Learn what a square matrix is.
Upper triangular matrix
A n \times n matrix for which a_{ij} = 0,\ \ i > j for all i,j is called an upper triangular matrix. That is, all the elements below the diagonal entries are zero.
Give an example of an upper triangular matrix.
Solution
\lbrack A\rbrack = \begin{bmatrix} 10 & - 7 & 0 \\ 0 & - 0.001 & 6 \\ 0 & 0 & 15005 \\ \end{bmatrix} \nonumber
is an upper triangular matrix.
Audiovisual Lecture
Lower triangular matrix
A n \times n matrix for which a_{ij} = 0,\ \ j > i for all i,j is called a lower triangular matrix. That is, all the elements above the diagonal entries are zero.
Give an example of a lower triangular matrix.
Solution
\lbrack A\rbrack = \begin{bmatrix} 1 & 0 & 0 \\ 0.3 & 1 & 0 \\ 0.6 & 2.5 & 1 \\ \end{bmatrix} \nonumber
is a lower triangular matrix.
Audiovisual Lecture
Title: What is a Lower Triangular Matrix
Summary: Learn what a lower triangular matrix is.
Diagonal matrix
A square matrix with all non-diagonal elements equal to zero is called a diagonal matrix, that is, only the diagonal entries of the square matrix can be non-zero (a_{ij} = 0,\ \ i \neq j).
Give examples of a diagonal matrix.
Solution
\lbrack A\rbrack = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 2.1 & 0 \\ 0 & 0 & 5 \\ \end{bmatrix} \nonumber
is a diagonal matrix.
Any or all the diagonal entries of a diagonal matrix can be zero. For example
\lbrack A\rbrack = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 2.1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \nonumber
is also a diagonal matrix.
Audiovisual Lecture
Identity matrix
A diagonal matrix with all diagonal elements equal to 1 is called an identity matrix, (a_{ij} = 0,\ \ i \neq j) for all i,j and a_{ii} = 1 for all i).
Give an example of an identity matrix.
Solution
\lbrack A\rbrack = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \nonumber
is an identity matrix.
Audiovisual Lecture
Title: What is an Identity Matrix
Summary: Learn what an identity matrix is.
Lesson 2: Binary Matrix Operations
Learning Objectives
After successful completion of this lesson, you should be able to:
1) Add one matrix to another
2) Subtract one matrix from another
3) Multiply one matrix by another
How do you add two matrices?
Two matrices \left\lbrack A \right\rbrack and \left\lbrack B \right\rbrack can be added if they are of the same size. The addition is then shown as
\left\lbrack C \right\rbrack = \left\lbrack A \right\rbrack + \left\lbrack B \right\rbrack \nonumber
where
c_{ij} = a_{ij} + b_{ij} \nonumber
Add the following two matrices.
\left\lbrack A \right\rbrack = \begin{bmatrix}5 & 2 & 3 \\1 & 2 & 7 \\\end{bmatrix} \nonumber \left\lbrack B \right\rbrack = \begin{bmatrix}6 & 7 & - 2 \\3 & 5 & 19 \\\end{bmatrix} \nonumber
Solution
\begin{split} \left\lbrack C \right\rbrack &= \left\lbrack A \right\rbrack + \left\lbrack B \right\rbrack\\ &= \begin{bmatrix} 5 & 2 & 3 \\ 1 & 2 & 7 \\ \end{bmatrix} + \begin{bmatrix} 6 & 7 & - 2 \\ 3 & 5 & 19 \\ \end{bmatrix}\\ &= \begin{bmatrix} 5 + 6 & 2 + 7 & 3 - 2 \\ 1 + 3 & 2 + 5 & 7 + 19 \\ \end{bmatrix}\\ &= \begin{bmatrix} 11 & 9 & 1 \\ 4 & 7 & 26 \\ \end{bmatrix} \end{split} \nonumber
Audiovisual Lecture
Titles: Adding two Matrices
Summary: Learn the theory of adding two matrices.
How do you subtract two matrices?
Two matrices \left\lbrack A \right\rbrack and \left\lbrack B \right\rbrack can be subtracted if they are the same size. The subtraction is then shown as
\left\lbrack D \right\rbrack = \left\lbrack A \right\rbrack - \left\lbrack B \right\rbrack \nonumber
where
d_{ij} = a_{ij} - b_{ij} \nonumber
Subtract matrix \left\lbrack B \right\rbrack from matrix \left\lbrack A \right\rbrack.
\left\lbrack A \right\rbrack = \begin{bmatrix} 5 & 2 & 3 \\ 1 & 2 & 7 \\ \end{bmatrix} \nonumber
\left\lbrack B \right\rbrack = \begin{bmatrix} 6 & 7 & - 2 \\ 3 & 5 & 19 \\ \end{bmatrix} \nonumber
Solution
\begin{split} \left\lbrack D \right\rbrack &= \left\lbrack A \right\rbrack - \left\lbrack B \right\rbrack\\ &= \begin{bmatrix} 5 & 2 & 3 \\ 1 & 2 & 7 \\ \end{bmatrix} - \begin{bmatrix} 6 & 7 & - 2 \\ 3 & 5 & 19 \\ \end{bmatrix}\\ &= \begin{bmatrix} \left( 5 - 6 \right) & \left( 2 - 7 \right) & \left( 3 - \left( - 2 \right) \right) \\ \left( 1 - 3 \right) & \left( 2 - 5 \right) & \left( 7 - 19 \right) \\ \end{bmatrix}\\ &= \begin{bmatrix} - 1 & - 5 & 5 \\ - 2 & - 3 & - 12 \\ \end{bmatrix}\end{split} \nonumber
Audiovisual Lecture
Titles: Adding two Matrices: Example
Summary: Learn the application of adding two matrices.
How do I multiply two matrices?
Two matrices \left\lbrack A \right\rbrack and \left\lbrack B \right\rbrack can be multiplied only if the number of columns of \left\lbrack A \right\rbrack is equal to the number of rows of \left\lbrack B \right\rbrack to give
\left\lbrack C \right\rbrack_{m \times n} = \left\lbrack A \right\rbrack_{m \times p}\left\lbrack B \right\rbrack_{p \times n} \nonumber
If \left\lbrack A \right\rbrack is a m \times p matrix and \left\lbrack B \right\rbrack is a p \times n matrix, the resulting matrix \left\lbrack C \right\rbrack is a m \times n matrix.
So how does one calculate the elements of \left\lbrack C \right\rbrack matrix?
\begin{split} c_{ij} &= \sum_{k = 1}^{p}{a_{ik}b_{kj}}\\ &= a_{i1}b_{1j} + a_{i2}b_{2j} + \ldots + a_{ip}b_{pj} \end{split} \nonumber
for each i = 1,\ 2,\ \ldots\ \ ,\ m and j = 1,\ 2,\ \ldots\ \ ,\ n.
To put it in simpler terms, the i^{th} row and j^{th} column of the \left\lbrack C \right\rbrack matrix in \left\lbrack C \right\rbrack = \left\lbrack A \right\rbrack\left\lbrack B \right\rbrack is calculated by multiplying the i^{th} row of \left\lbrack A \right\rbrack by the j^{th} column of \left\lbrack B \right\rbrack. That is,
\begin{split} c_{ij} &= \left\lbrack a_{i1} \ \ a_{i2}\ \ \ldots\ \ \ a_{ip} \right\rbrack \begin{bmatrix} \begin{matrix} b_{1j} \\ b_{2j} \\ \end{matrix} \\ \begin{matrix} \vdots \\ b_{pj} \\ \end{matrix} \\ \end{bmatrix}\\ &= a_{i1}b_{1j} + a_{i2}b_{2j} + \ldots + a_{ip}b_{pj}\\ &= \sum_{k = 1}^{p}{a_{ik}b_{kj}}\end{split} \nonumber
Given
\left\lbrack A \right\rbrack = \begin{bmatrix} 5 & 2 & 3 \\ 1 & 2 & 7 \\ \end{bmatrix} \nonumber
\left\lbrack B \right\rbrack = \begin{bmatrix} 3 & - 2 \\ 5 & - 8 \\ 9 & - 10 \\ \end{bmatrix} \nonumber
Find
\left\lbrack C \right\rbrack = \left\lbrack A \right\rbrack\left\lbrack B \right\rbrack \nonumber
Solution
c_{12} can be found by multiplying the first row of \left\lbrack A \right\rbrack by the second column of \left\lbrack B \right\rbrack,
\begin{split} c_{12} &= \begin{bmatrix} 5 & 2 & 3 \\ \end{bmatrix}\begin{bmatrix} - 2 \\ - 8 \\ - 10 \\ \end{bmatrix}\\ &= \left( 5 \right)\left( - 2 \right) + \left( 2 \right)\left( - 8 \right) + \left( 3 \right)\left( - 10 \right)\\ &= - 56\end{split} \nonumber
Similarly, one can find the other elements of \left\lbrack C \right\rbrack to give
\left\lbrack C \right\rbrack = \begin{bmatrix} 52 & - 56 \\ 76 & - 88 \\ \end{bmatrix} \nonumber
Audiovisual Lecture
Title: Multiplying Two Matrices
Summary: Learn the theory of multiplying two matrices.
Lesson 3: Setting Up Problems in Matrix Form
Learning Objectives
After successful completion of this lesson, you should be able to:
1) Develop simultaneous linear equations model from a physical problem
2) Set up simultaneous linear equations in matrix form
Matrix algebra is used for solving systems of equations. Can you illustrate this concept?
Matrix algebra is used to solve a system of simultaneous linear equations. In fact, for many mathematical procedures, such as the solution to a set of nonlinear equations, interpolation, integration, and differential equations, the solutions reduce to a set of simultaneous linear equations. Let us illustrate with an example for interpolation.
The upward velocity of a rocket is given at three different times on the following table.
Time, t | Velocity, v |
---|---|
\text{s} | (\text{m/s}) |
5 | 106.8 |
8 | 177.2 |
12 | 279.2 |
The velocity data is approximated by a polynomial as
v\left( t \right) = at^{2} + {bt} + c,\ 5 \leq t \leq 12\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{3.E1.1}) \nonumber
Set up the equations in matrix form to find the coefficients a,b,c of the velocity profile.
Solution
The polynomial is going through three data points \left( t_{1},v_{1} \right),\left( t_{2},v_{2} \right), \text{ and} \left( t_{3},v_{3} \right) where from Table \PageIndex{3.1}
t_{1} = 5,\ v_{1} = 106.8 \nonumber
t_{2} = 8,\ v_{2} = 177.2 \nonumber
t_{3} = 12,\ v_{3} = 279.2 \nonumber
Requiring that v\left( t \right) = at^{2} + {bt} + c passes through the three data points gives
\begin{split} v\left( t_{1} \right) &= v_{1} = at_{1}^{2} + bt_{1} + c\\ v\left( t_{2} \right) &= v_{2} = at_{2}^{2} + bt_{2} + c\\ v\left( t_{3} \right) &= v_{3} = at_{3}^{2} + bt_{3} + c \end{split} \nonumber
Substituting the data \left( t_{1},v_{1} \right),\left( t_{2},v_{2} \right),\ and\ \left( t_{3},v_{3} \right) gives
\begin{split} a\left( 5^{2} \right) + b\left( 5 \right) + c &= 106.8\\ a\left( 8^{2} \right) + b\left( 8 \right) + c &= 177.2\\ a\left( 12^{2} \right) + b\left( 12 \right) + c &= 279.2\end{split} \nonumber
or
\begin{split} 25a + 5b + c &= 106.8\\ 64a + 8b + c &= 177.2\\ 144a + 12b + c &= 279.2 \end{split} \nonumber
This set of equations can be rewritten in the matrix form as
\begin{bmatrix} 25a + & 5b + & c \\ 64a + & 8b + & c \\ 144a + & 12b + & c \\ \end{bmatrix} = \begin{bmatrix} 106.8 \\ 177.2 \\ 279.2 \\ \end{bmatrix} \nonumber
The above equation can be written as a linear combination as follows
a\begin{bmatrix} 25 \\ 64 \\ 144 \\ \end{bmatrix} + b\begin{bmatrix} 5 \\ 8 \\ 12 \\ \end{bmatrix} + c\begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix} = \begin{bmatrix} 106.8 \\ 177.2 \\ 279.2 \\ \end{bmatrix} \nonumber
and further using matrix multiplication gives
\begin{bmatrix} 25 & 5 & 1 \\ 64 & 8 & 1 \\ 144 & 12 & 1 \\ \end{bmatrix}\begin{bmatrix} a \\ b \\ c \\ \end{bmatrix} = \begin{bmatrix} 106.8 \\ 177.2 \\ 279.2 \\ \end{bmatrix} \nonumber
The above is an illustration of why matrix algebra is needed. The complete solution to the set of equations is given later in this chapter.
A general set of m linear equations and n unknowns,
\begin{split} &a_{11}x_{1} + a_{12}x_{2} + {......} + a_{1n}x_{n} = c_{1}\\ &a_{21}x_{1} + a_{22}x_{2} + {......} + a_{2n}x_{n} = c_{2}\\ &{.......................................}\\ & {.......................................}\\ &a_{m1}x_{1} + a_{m2}x_{2} + ...... + a_{mn}x_{n} = c_{m} \end{split} \nonumber
can be rewritten in the matrix form as
\begin{bmatrix} a_{11} & a_{12} & . & . & a_{1n} \\ a_{21} & a_{22} & . & . & a_{2n} \\ \vdots & & & & \vdots \\ \vdots & & & & \vdots \\ a_{m1} & a_{m2} & . & . & a_{mn} \\ \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \\ \vdots \\ \vdots \\ x_{n} \\ \end{bmatrix} = \begin{bmatrix} c_{1} \\ c_{2} \\ \vdots \\ \vdots \\ c_{m} \\ \end{bmatrix} \nonumber
Denoting the matrices by \left\lbrack A \right\rbrack, \left\lbrack X \right\rbrack, and \left\lbrack C \right\rbrack, the system of equation is \left\lbrack A \right\rbrack\ \left\lbrack X \right\rbrack = \left\lbrack C \right\rbrack, where \left\lbrack A \right\rbrack is called the coefficient matrix, \left\lbrack C \right\rbrack is called the right-hand side vector and \left\lbrack X \right\rbrack is called the solution vector.
Sometimes the \left\lbrack A \right\rbrack\ \left\lbrack X \right\rbrack = \left\lbrack C \right\rbrack system of equations is written in the augmented form, that is,
[A\ \vdots\ C]= \begin{bmatrix} a_{11} &a_{12} &\cdots &a_{1n}\ \ \vdots&c_1 \\ a_{21} &a_{22} &\cdots &a_{2n}\ \ \vdots&c_2 \\ \vdots&\vdots&\ddots&\ \ \ \ \ \ \ \vdots& \vdots\\ a_{m1} &a_{m2} &\cdots &a_{mn}\ \vdots&c_n \end{bmatrix} \nonumber
As an example, for the set of equations \begin{bmatrix} 25 & 5 & 1 \\ 64 & 8 & 1 \\ 144 & 12 & 1 \\ \end{bmatrix}\begin{bmatrix} a \\ b \\ c \\ \end{bmatrix} = \begin{bmatrix} 106.8 \\ 177.2 \\ 279.2 \\ \end{bmatrix} \nonumber the augmented matrix form is \begin{bmatrix} 25 & 5 & 1 & | & 106.8 \\ 64 & 8 & 1 & | & 177.2 \\ 144 & 12 & 1 & | & 279.2 \\ \end{bmatrix} \nonumber
Audiovisual Lecture
Title: Writing Problems in Matrix Form
Summary: Learn a real-life problem of setting up simultaneous linear equations in matrix form.
Lesson 4: Inverse of a Square Matrix
Learning Objectives
After successful completion of this lesson, you should be able to:
1) define the inverse of a matrix
2) know important statements about the inverse of a matrix
3) solve a set of equations where the inverse of the coefficient matrix is given
Can you divide two matrices?
If \lbrack A\rbrack\ \lbrack B\rbrack = \lbrack C\rbrack is defined, it might seem intuitive that \displaystyle \lbrack A\rbrack = \frac{\left\lbrack C \right\rbrack}{\left\lbrack B \right\rbrack}, but matrix division is not defined like that. However, an inverse of a matrix can be defined for certain types of square matrices. The inverse of a square matrix \lbrack A\rbrack, if existing, is denoted by \lbrack A\rbrack^{- 1} such that
\lbrack A\rbrack\ \lbrack A\rbrack^{- 1} = \lbrack I\rbrack = \lbrack A\rbrack^{- 1}\lbrack A\rbrack \nonumber
where \lbrack I\rbrack is the identity matrix.
In other words, let A be a square matrix. If \lbrack B\rbrack is another square matrix of the same size such that \lbrack B\rbrack\ \lbrack A\rbrack = \lbrack I\rbrack, then \lbrack B\rbrack is the inverse of \lbrack A\rbrack. \lbrack A\rbrack is then called to be invertible or nonsingular. If \lbrack A\rbrack^{- 1} does not exist, \lbrack A\rbrack is called noninvertible or singular.
If \lbrack A\rbrack and \lbrack B\rbrack are two n \times n matrices such that \lbrack B\rbrack\ \lbrack A\rbrack = \lbrack I\rbrack, then these statements are also true:
- [B] is the inverse of [A]
- [A] is the inverse of [B]
- [A] and [B] are both invertible
- [A] [B]= [I].
- [A] and [B] are both nonsingular
- all columns of [A] and [B] are linearly independent
- all rows of [A] and [B] are linearly independent.
Determine if
\lbrack B\rbrack = \begin{bmatrix} 3 & 2 \\ 5 & 3 \\ \end{bmatrix} \nonumber
is the inverse of
\lbrack A\rbrack = \begin{bmatrix} - 3 & 2 \\ 5 & - 3 \\ \end{bmatrix} \nonumber
Solution
\begin{split} \lbrack B\rbrack\lbrack A\rbrack &= \begin{bmatrix} 3 & 2 \\ 5 & 3 \\ \end{bmatrix}\begin{bmatrix} - 3 & 2 \\ 5 & - 3 \\ \end{bmatrix}\\ &= \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= \lbrack I\rbrack\end{split} \nonumber
Since
\left\lbrack B \right\rbrack\left\lbrack A \right\rbrack = \left\lbrack I \right\rbrack, \nonumber
\lbrack B\rbrack is the inverse of \lbrack A\rbrack, and \lbrack A\rbrack is the inverse of \lbrack B\rbrack.
But, we can also show that
\begin{split} \lbrack A\rbrack\lbrack B\rbrack &= \begin{bmatrix} - 3 & 2 \\ 5 & - 3 \\ \end{bmatrix}\begin{bmatrix} 3 & 2 \\ 5 & 3 \\ \end{bmatrix}\\ &= \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}\\ &= \lbrack I\rbrack \end{split} \nonumber
to prove that \lbrack A\rbrack is the inverse of \lbrack B\rbrack.
Audiovisual Lecture
Can I use the concept of the inverse of a matrix to find the solution of a set of equations [A][X] = [C]?
Yes, if the number of equations is the same as the number of unknowns, the coefficient matrix \lbrack A\rbrack is a square matrix.
Given
\lbrack A\rbrack\ \lbrack X\rbrack = \lbrack C\rbrack \nonumber
Then, if \lbrack A\rbrack^{- 1} exists, multiplying both sides by \lbrack A\rbrack^{- 1}.
\lbrack A\rbrack^{- 1}\lbrack A\rbrack\lbrack X\rbrack = \lbrack A\rbrack^{- 1}\lbrack C\rbrack \nonumber
\lbrack I\rbrack\ \lbrack X\rbrack = \lbrack A\rbrack^{- 1}\lbrack C\rbrack \nonumber
\lbrack X\rbrack = \lbrack A\rbrack^{- 1}\lbrack C\rbrack \nonumber
This implies that if we are able to find \lbrack A\rbrack^{- 1}, the solution vector of \lbrack A\rbrack\ \lbrack X\rbrack = \lbrack C\rbrack is simply a multiplication of \lbrack A\rbrack^{- 1} and the right-hand side vector, \lbrack C\rbrack.
Audiovisual Lecture
Title: Using Concepts of Inverse to Solve a Set of Equations
Summary: Learn via an example how to find a solution to a set of equations if the inverse of the matrix is given.
How do I find the inverse of a matrix?
If \lbrack A\rbrack is a n \times n matrix, then \lbrack A\rbrack^{- 1} is a n \times n matrix, and according to the definition of inverse of a matrix
\lbrack A\rbrack\ \lbrack A\rbrack^{- 1} = \lbrack I\rbrack \nonumber
Denoting
\lbrack A\rbrack = \begin{bmatrix} a_{11} & a_{12} & \cdot & \cdot & a_{1n} \\ a_{21} & a_{22} & \cdot & \cdot & a_{2n} \\ \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot \\ a_{n1} & a_{n2} & \cdot & \cdot & a_{nn} \\ \end{bmatrix} \nonumber
\lbrack A\rbrack^{- 1} = \begin{bmatrix} a_{11}^{\prime} & a_{12}^{\prime} & \cdot & \cdot & a_{1n}^{\prime} \\ a_{21}^{\prime} & a_{22}^{\prime} & \cdot & \cdot & a_{2n}^{\prime} \\ \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot \\ a_{n1}^{\prime} & a_{n2}^{\prime} & \cdot & \cdot & a_{nn}^{\prime} \\ \end{bmatrix} \nonumber
\lbrack I\rbrack = \begin{bmatrix} 1 & 0 & \cdot & \cdot & \cdot & 0 \\ 0 & 1 & & & & 0 \\ 0 & & \cdot & & & \cdot \\ \cdot & & & 1 & & \cdot \\ \cdot & & & & \cdot & \cdot \\ 0 & \cdot & \cdot & \cdot & \cdot & 1 \\ \end{bmatrix} \nonumber
Using the definition of matrix multiplication, the first column of the \lbrack A\rbrack^{- 1} matrix can then be found by solving
\begin{bmatrix} a_{11} & a_{12} & \cdot & \cdot & a_{1n} \\ a_{21} & a_{22} & \cdot & \cdot & a_{2n} \\ \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot \\ a_{n1} & a_{n2} & \cdot & \cdot & a_{nn} \\ \end{bmatrix}\begin{bmatrix} a_{11}^{\prime} \\ a_{21}^{\prime} \\ \cdot \\ \cdot \\ a_{n1}^{\prime} \\ \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ \cdot \\ \cdot \\ 0 \\ \end{bmatrix} \nonumber
Similarly, one can find the other columns of the \lbrack A\rbrack^{- 1} matrix by changing the right-hand side accordingly.
Audiovisual Lecture
The upward velocity of the rocket is given by
Time, t \ (\text{s}) | Velocity, v \ (\text{m}/\text{s}) |
---|---|
5 | 106.8 |
8 | 177.2 |
12 | 279.2 |
In an earlier example, we wanted to approximate the velocity profile by
v\left( t \right) = at^{2} + {bt} + c,\ 5 \leq t \leq 12 \nonumber
We found that the coefficients a,\ b,\text{ and }\ c in v\left( t \right) are given by solving
\begin{bmatrix} 25 & 5 & 1 \\ 64 & 8 & 1 \\ 144 & 12 & 1 \\ \end{bmatrix}\begin{bmatrix} a \\ b \\ c \\ \end{bmatrix} = \begin{bmatrix} 106.8 \\ 177.2 \\ 279.2 \\ \end{bmatrix} \nonumber
First, find the inverse of
\left\lbrack A \right\rbrack = \begin{bmatrix} 25 & 5 & 1 \\ 64 & 8 & 1 \\ 144 & 12 & 1 \\ \end{bmatrix} \nonumber
and then use the definition of inverse to find the coefficients a,\ b,\ \text{and}\ c, and the velocity profile.
Solution
If
\left\lbrack A \right\rbrack^{- 1} = \begin{bmatrix} a_{11}^{\prime} & a_{12}^{\prime} & a_{13}^{\prime} \\ a_{21}^{\prime} & a_{22}^{\prime} & a_{23}^{\prime} \\ a_{31}^{\prime} & a_{32}^{\prime} & a_{33}^{\prime} \\ \end{bmatrix} \nonumber
is the inverse of \lbrack A\rbrack, then
\begin{bmatrix} 25 & 5 & 1 \\ 64 & 8 & 1 \\ 144 & 12 & 1 \\ \end{bmatrix}\begin{bmatrix} a_{11}^{\prime} & a_{12}^{\prime} & a_{13}^{\prime} \\ a_{21}^{\prime} & a_{22}^{\prime} & a_{23}^{\prime} \\ a_{31}^{\prime} & a_{32}^{\prime} & a_{33}^{\prime} \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \nonumber
gives three sets of equations
\begin{bmatrix} 25 & 5 & 1 \\ 64 & 8 & 1 \\ 144 & 12 & 1 \\ \end{bmatrix}\begin{bmatrix} a_{11}^{\prime} \\ a_{21}^{\prime} \\ a_{31}^{\prime} \\ \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix} \nonumber
\begin{bmatrix} 25 & 5 & 1 \\ 64 & 8 & 1 \\ 144 & 12 & 1 \\ \end{bmatrix}\begin{bmatrix} a_{12}^{\prime} \\ a_{22}^{\prime} \\ a_{32}^{\prime} \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0 \\ \end{bmatrix} \nonumber
\begin{bmatrix} 25 & 5 & 1 \\ 64 & 8 & 1 \\ 144 & 12 & 1 \\ \end{bmatrix}\begin{bmatrix} a_{13}^{\prime} \\ a_{23}^{\prime} \\ a_{33}^{\prime} \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \\ \end{bmatrix} \nonumber
Solving the above three sets of equations separately gives
\begin{bmatrix} a_{11}^{\prime} \\ a_{21}^{\prime} \\ a_{31}^{\prime} \\ \end{bmatrix} = \begin{bmatrix} 0.04762 \\ - 0.9524 \\ 4.571 \\ \end{bmatrix} \nonumber
\begin{bmatrix} a_{12}^{\prime} \\ a_{22}^{\prime} \\ a_{32}^{\prime} \\ \end{bmatrix} = \begin{bmatrix} - 0.08333 \\ 1.417 \\ - 5.000 \\ \end{bmatrix} \nonumber
\begin{bmatrix} a_{13}^{\prime} \\ a_{23}^{\prime} \\ a_{33}^{\prime} \\ \end{bmatrix} = \begin{bmatrix} 0.03571 \\ - 0.4643 \\ 1.429 \\ \end{bmatrix} \nonumber
Hence
\lbrack A\rbrack^{- 1} = \begin{bmatrix} 0.04762 & - 0.08333 & 0.03571 \\ - 0.9524 & 1.417 & - 0.4643 \\ 4.571 & - 5.000 & 1.429 \\ \end{bmatrix} \nonumber
Now
\left\lbrack A \right\rbrack\left\lbrack X \right\rbrack = \left\lbrack C \right\rbrack \nonumber
where
\left\lbrack X \right\rbrack = \begin{bmatrix} a \\ b \\ c \\ \end{bmatrix} \nonumber
\left\lbrack C \right\rbrack = \begin{bmatrix} 106.8 \\ 177.2 \\ 279.2 \\ \end{bmatrix} \nonumber
Using the definition of \left\lbrack A \right\rbrack^{- 1},
\left\lbrack A \right\rbrack^{- 1}\left\lbrack A \right\rbrack\left\lbrack X \right\rbrack = \left\lbrack A \right\rbrack^{- 1}\left\lbrack C \right\rbrack \nonumber
\left\lbrack X \right\rbrack = \left\lbrack A \right\rbrack^{- 1}\left\lbrack C \right\rbrack \nonumber
\begin{bmatrix} a \\ b \\ c \\ \end{bmatrix} =\begin{bmatrix} 0.04762 & - 0.08333 & 0.03571 \\ - 0.9524 & 1.417 & - 0.4643 \\ 4.571 & - 5.000 & 1.429 \\ \end{bmatrix}\begin{bmatrix} 106.8 \\ 177.2 \\ 279.2 \\ \end{bmatrix} \nonumber
Conducting matrix multiplication of the right hand side gives
\begin{bmatrix} a \\ b \\ c \\ \end{bmatrix} = \begin{bmatrix} 0.2905 \\ 19.69 \\ 1.086 \\ \end{bmatrix} \nonumber
So
v\left( t \right) = 0.2905t^{2} + 19.69t + 1.086,\ 5 \leq t \leq 12 \nonumber
Multiple Choice Test
(1). Given
[A] =\begin{bmatrix} 6 & 2 & 3 & 9 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 4 & 5 \\ 0 & 0 & 0 & 6 \\ \end{bmatrix} \nonumber
then [A] is a ______________ matrix.
(A) diagonal
(B) identity
(C) lower triangular
(D) upper triangular
(2). A square matrix [A] is lower triangular if
(A) a_{ij} = 0,j > i
(B) a_{ij} = 0,i > j
(C) a_{ij} \neq 0,i > j
(D) a_{ij} \neq 0,j > i
(3). Given
\lbrack A\rbrack = \begin{bmatrix} 12.3 & - 12.3 & 20.3 \\ 11.3 & - 10.3 & - 11.3 \\ 10.3 & - 11.3 & - 12.3 \\ \end{bmatrix},\ \lbrack B\rbrack = \begin{bmatrix} 2 & 4 \\ - 5 & 6 \\ 11 & - 20 \\ \end{bmatrix} \nonumber
if [C] = [A] [B], then c_{31}= _____________________
(A) -58.2
(B) -37.6
(C) 219.4
(D) 259.4
(4). The following system of equations has ____________ solution(s).
x + y = 2 \nonumber
6x+6y =12 \nonumber
(A) infinite
(B) no
(C) two
(D) unique
(5). Consider there are only two computer companies in a country. The companies are named Dude and Imac. Each year, company Dude keeps 1/5th of its customers, while the rest switch to Imac. Each year, Imac keeps 1/3rd of its customers, while the rest switch to Dude. If in 2003, Dude had 1/6th of the market and Imac had 5/6th of the market, what will be the share of Dude computers when the market becomes stable?
(A) 37/90
(B) 5/11
(C) 6/11
(D) 53/90
(6). Three kids, Jim, Corey, and David, receive an inheritance of \$2,253,453. The money is put in three trusts but is not divided equally, to begin with. Corey’s trust is three times that of David’s because Corey made an A in Dr. Kaw’s class. Each trust is put in an interest-generating investment. The three trusts of Jim, Corey, and David pay an interest of 6\%, 8\%, 11\%, respectively. The total interest of all the three trusts combined at the end of the first year is \$190,740.57. The equations to find the trust money of Jim (J), Corey (C), and David (D) in a matrix form is
(A) \begin{bmatrix} 1 & 1 & 1 \\ 0 & 3 & - 1 \\ 0.06 & 0.08 & 0.11 \\ \end{bmatrix}\begin{bmatrix} J \\ C \\ D \\ \end{bmatrix} = \begin{bmatrix} 2,253,453 \\ 0 \\ 190,740.57 \\ \end{bmatrix}
(B) \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & - 3 \\ 0.06 & 0.08 & 0.11 \\ \end{bmatrix}\begin{bmatrix} J \\ C \\ D \\\end{bmatrix} = \begin{bmatrix} 2,253,453 \\ 0 \\ 190,740.57 \\ \end{bmatrix}
(C) \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & - 3 \\ 6 & 8 & 11 \\ \end{bmatrix}\begin{bmatrix} J \\ C \\ D \\ \end{bmatrix} = \begin{bmatrix} 2,253,453 \\ 0 \\ 190,740.57 \\ \end{bmatrix}
(D) \begin{bmatrix} 1 & 1 & 1 \\ 0 & 3 & - 1 \\ 6 & 8 & 11 \\ \end{bmatrix}\begin{bmatrix} J \\ C \\ D \\ \end{bmatrix} = \begin{bmatrix} 2,253,453 \\ 0 \\ 19,074,057 \\ \end{bmatrix}
For complete solution, go to
http://nm.mathforcollege.com/mcquizzes/04sle/quiz_04sle_background_solution.pdf