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4.01: Prerequisites to Simultaneous Linear Equations

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Lesson 1: Definition of Matrices and Special Matrices

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.

Example \PageIndex{1.1}

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.

Example \PageIndex{1.2}

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.

Example \PageIndex{1.3}

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.

Example \PageIndex{1.4}

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.

Example \PageIndex{1.5}

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).

Example \PageIndex{1.6}

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).

Example \PageIndex{1.7}

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

Example \PageIndex{2.1}

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

Example \PageIndex{2.2}

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

Example \PageIndex{2.3}

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.

Example \PageIndex{3.1}

The upward velocity of a rocket is given at three different times on the following table.

Table \PageIndex{3.1}. Velocity vs. time data for a rocket
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:

  1. [B] is the inverse of [A]
  2. [A] is the inverse of [B]
  3. [A] and [B] are both invertible
  4. [A] [B]= [I].
  5. [A] and [B] are both nonsingular
  6. all columns of [A] and [B] are linearly independent
  7. all rows of [A] and [B] are linearly independent.
Example \PageIndex{4.1}

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

Example \PageIndex{4.2}

The upward velocity of the rocket is given by

Table \PageIndex{4.1}. Velocity vs. time data for a rocket
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


This page titled 4.01: Prerequisites to Simultaneous Linear Equations is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Autar Kaw via source content that was edited to the style and standards of the LibreTexts platform.

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